-Kinematics is the branch of mechanics concerned with the motion of objects without reference to the forces that cause the motion.

-Dynamics is the branch of mechanics concerned with the motion of bodies under the action of forces.

-Dynamics is the branch of mechanics concerned with the motion of bodies under the action of forces.

This is chapter 1 and we will be studying kinematics and dynamics. Lets first start with the question what is kinematics and what is dynamics?

During the MCAT you really should be concern with metric system. Where it will be dealing with meters, kilometers and seconds.

There are other systems in there as well like: FPS→Food-Pound-second or centimeters, CGS→grams and seconds.

There are other systems in there as well like: FPS→Food-Pound-second or centimeters, CGS→grams and seconds.

Lets start with the units that we will be using in physics:

meters (m)

In metric system length will be given in which unit?

Kilograms (Kg)

In metric system Mass (not weight) will be given in which unit?

Seconds (s)

In metric system Time will be given in which unit?

Ampere (A)→ (coulomb/second)

In metric system Current will be given in which unit?

Mole (mol)

In metric system Amount of substence will be given in which unit?

Kelvin (K)

In metric system Temeperature will be given in which unit?

Candela (cd)

In metric system Luminous Intensity will be given in which unit?

-Base Units are units that are: seconds, meters…

-Derived Units are units that are derived from the base units divided, subtracted, multiplied together.

Example is 1 Newton that is derived from: kilograms, meters and seconds.

-Derived Units are units that are derived from the base units divided, subtracted, multiplied together.

Example is 1 Newton that is derived from: kilograms, meters and seconds.

So you will be dealing with two types of units:

-Slug which is a unit of mass associated with Imperial system (food-pound-second)

To help you on the test day memorize some of the units:

Mass:

Mass:

Newtons

Force units in metric system:

Joules

Work and energy units in metric system:

Watt (kg×m)/s³

Power units in metric system:

-length might be given in Angstroms 1A=10⁻¹⁰m

nanometers 1am=10⁻⁹m

nanometers 1am=10⁻⁹m

AT the molecular and atomic level there will be other units that will be used:

Newtown is the unit of force. And force will obey the same relationship with mass and acceleration regardless of the unit.

Force is always produce of mass and acceleration.

So one pound (lb) must be equal one (slug x ft)/s²

Force is always produce of mass and acceleration.

So one pound (lb) must be equal one (slug x ft)/s²

If the newton is the product of kilograms and meters/second₂ what units comprise the pound?

angstrom, centimeters, inch, foot, mile

Write the fallowing units in the order of the smallest to the largest:

centimeter, angstrom, inch, mile, foot

centimeter, angstrom, inch, mile, foot

Vectors have magnitude (size) and direction. Examples are: displacement, velocity, acceleration and force.

Scalars only have magnitude. Examples are: distance, speed, energy, pressure, and mass.

Scalars only have magnitude. Examples are: distance, speed, energy, pressure, and mass.

Lets move on to vectors and scalars:

-The direction of the arrow represents the direction of the vector.

-The length of the vector represents the magnitude of the vector

-The length of the vector represents the magnitude of the vector

Vectors have magnitude and direction. How do we represent these?

So there are two ways to write vectors:

-letter with arrow on top. Just like the one is the picture

-bold letter (this is what Kaplan uses)

Note: scalar quantities are represented with small letters

-letter with arrow on top. Just like the one is the picture

-bold letter (this is what Kaplan uses)

Note: scalar quantities are represented with small letters

Common notations of vectors? arrow on top or blood letter.

You have brick on the ground and you pick up the brick and move it. You then take a ruler out and measure the distance and find out that you moved the brick 5meters.

Example #1 Brick: this will show you difference between vector and scaler

Its a scalar since the inly information we have is the length which is 5meters.

We do not know which direction you moved the brick.

This is distance.

We do not know which direction you moved the brick.

This is distance.

Question #1 for the example #1: is the measurement of 5 meters a vector or scalar?

This is vector since we have magnitude-5 meters and direction-to the right.

This is going to be called DISPLACEMENT since we are talking about vector.

This is going to be called DISPLACEMENT since we are talking about vector.

What if I tell you know that the brick was moved 5 meters to the right?

Lets start with speed:

Lets say that the brick was moved 5 meters in 2 seconds.

What will be the speed?

5meters/2seconds=2.5meters/second

Is this vector or scaler?

This is scalar quality since no direction has been given. And the scalar of how fast something is going is called speed.

Lets say that the brick was moved 5 meters in 2 seconds.

What will be the speed?

5meters/2seconds=2.5meters/second

Is this vector or scaler?

This is scalar quality since no direction has been given. And the scalar of how fast something is going is called speed.

Lets now look at speed and velocity. Which one is vector and which one is scaler?

We will get 2.5m/second to the right.

When we talk how fast objects are moving in vector quantities we call it VELOCITY.

When we talk how fast objects are moving in vector quantities we call it VELOCITY.

But what is we say that the brick went 5 meters to the right in 2 seconds?

-When you specify the magnitude and direction you have vector quantitate called VELOCITY

-When you only specify magnitude and not direction you have SPEED

-When you only specify magnitude and not direction you have SPEED

Main think to remember in here: Speed vs Velocity

When you will be dealing with more then one domination you will be dealing with 2D vectors.

Lets make sure we can understand 2D vectors.

Lets make sure we can understand 2D vectors.

So far we have been talking about object moving in two directions: up down or left right. But what if we want to talk about 2D or 3D?

One thing to remember in here is that you can move vectors around as long as you keep their magnitude and direction.

Its does not have to be tail to tail in here. This will help you add the vectors.

Its does not have to be tail to tail in here. This will help you add the vectors.

In 2D we can add vectors: you have vector a and vector b. What happens when you add them.

You draw the vector a but you but the vector b tail to the head of vector a.

Then you connect the tail of a and head of b and you call that vector c that will be the sum of a+b=c

Then you connect the tail of a and head of b and you call that vector c that will be the sum of a+b=c

So lets add the vectors a+b

So if you are displaced in the direction of a and its magnitude and then direction of b and its magnitude then the total displacement is c.

Lets connect this example with displacement:

You have vector X that is the sum of the pink vector and green vector.

Notice that X vector and green vector are both head to head (its very important to keep track of the directions in which arrows point).

Why would you want to brake down the vector?

It brakes it down to its vertical and horizontal components.

So vector X can be expressed as the sum of the vertical Xv and horizontal components Xh: X=Xv+Xh (vector X is the sum of its horizontal and vertical components)

Notice that X vector and green vector are both head to head (its very important to keep track of the directions in which arrows point).

Why would you want to brake down the vector?

It brakes it down to its vertical and horizontal components.

So vector X can be expressed as the sum of the vertical Xv and horizontal components Xh: X=Xv+Xh (vector X is the sum of its horizontal and vertical components)

You can use the same idea to brake down any vector in 2D to its components:

Vectors vertical and horizontal components:

How to draw a vertical and horizontal components:

Because this can brake down 2D problem into 1D problem. One vector will be in vertical direction and the other in horizontal direction.

Why would you want to brake down the vector into its vertical and horizontal components?

-Vector |A| length is 5 (notice the brackets around the vector which shows vector’s magnitude)

-We will show the vector direction by the angle agains the X axis. The angle is 36.8699°

We want to find out the vertical and horizontal components of vector A.

-We will show the vector direction by the angle agains the X axis. The angle is 36.8699°

We want to find out the vertical and horizontal components of vector A.

Lets look at the example #2 Braking Down Components of Vector

Vertical component is in purple Ay

Horizontal component is in green Ax

Lets fund the Ax and Ay (the horizontal and vertical components)

Horizontal component is in green Ax

Lets fund the Ax and Ay (the horizontal and vertical components)

Lets now draw a vertical and horizontal components of vector |A|

So as you can see this is a right triangle and we will use Pythagorean Theorem: a²+b²=c²

Finding horizontal and vertical components of vector |A|:

SOH CAN TOA

SOH stands for Sine equals Opposite over Hypotenuse. CAH stands for Cosine equals Adjacent over Hypotenuse. TOA stands for Tangent equals Opposite over Adjacent.

Notice that this directions (opposite and adjacent) are in reference to the angle.

SOH stands for Sine equals Opposite over Hypotenuse. CAH stands for Cosine equals Adjacent over Hypotenuse. TOA stands for Tangent equals Opposite over Adjacent.

Notice that this directions (opposite and adjacent) are in reference to the angle.

Lets brush on the basic trigonometry and Pythagorean Theorem:

We have the angle=36.8699°

We have length of hypotenuse=5

Lets find x and y components of the vector.

-Finding the vertical component:

sin 36.8699°=|Ay|/5

Now solve for |Ay| (multiply both sided by 5)

5 x sin 36.8699°=|Ay|

|Ay|=5

-Finding the horizontal components of vector:

cos 36.8699°=|Ax|/5

5 x cos 36.8699°=|Ax|

|Ax|=3.9~4

We have length of hypotenuse=5

Lets find x and y components of the vector.

-Finding the vertical component:

sin 36.8699°=|Ay|/5

Now solve for |Ay| (multiply both sided by 5)

5 x sin 36.8699°=|Ay|

|Ay|=5

-Finding the horizontal components of vector:

cos 36.8699°=|Ax|/5

5 x cos 36.8699°=|Ax|

|Ax|=3.9~4

Lets now get back into the problem of the vector:

Lets say a car has velocity of 5m/s remember that velocity is vector component and it has magnitude and direction. So you can brake down the vector into x and y components and say that the car is going upwards direction 3m/s and its going horizontal direction 4m/s.

How is this problem applying to physics?

What is unit vector?

A unit vector is just a vector that goes in a particular direction that has a magnitude of one. You also have the y and x components but the magnitude is one.

How do you notate unit vector? You place ∧ on top of the letter.

So unit vector had magnitude of one an you draw it on the x and y axis.

A unit vector is just a vector that goes in a particular direction that has a magnitude of one. You also have the y and x components but the magnitude is one.

How do you notate unit vector? You place ∧ on top of the letter.

So unit vector had magnitude of one an you draw it on the x and y axis.

Lets move on to the UNIT VECTOR NOTATION:

If we are dealing with 2D we can define any vector in terms of the sum of the two vectors.

Why do we use UNIT VECTOR?

So normally when you have vector you can find its x and y components by adding the vectors (head to tail). Then you can use SOH CAN TOA to find vertical and horizontal components.

But you must remember that: |V|=Vx+Vy (the value of the V vector is the horizontal and vertical components).

Now we can take |V|=Vx+Vy and define Vx and Vy:

Lets define Vx this is 10cos30° and Vy=10sin30° (we got this from SOH CAN TOA)

Since the unit vector goes in the same direction that vertical and horizontal components but has magnitude of 1 we can multiply the vertical and horizontal components by unit vector.

Vx=10cos30°× i (i is unit vector)→5√3i

Vy=10sin30° x j (j is unit vector)→5j

|V|=Vx+Vy=5√3i×5j

But you must remember that: |V|=Vx+Vy (the value of the V vector is the horizontal and vertical components).

Now we can take |V|=Vx+Vy and define Vx and Vy:

Lets define Vx this is 10cos30° and Vy=10sin30° (we got this from SOH CAN TOA)

Since the unit vector goes in the same direction that vertical and horizontal components but has magnitude of 1 we can multiply the vertical and horizontal components by unit vector.

Vx=10cos30°× i (i is unit vector)→5√3i

Vy=10sin30° x j (j is unit vector)→5j

|V|=Vx+Vy=5√3i×5j

Example #2 Vector |V|=10 and the angle=30°

We can define any vector in a simple formula and not need to draw a vector each time:

r=a×i+b×j where i and j are unit vectors

r=a×i+b×j where i and j are unit vectors

So what does the unit vector gives us?

Lets say that vector a=2i + 3j and you have another vector b=10i + 2j

What is the sum of the 2 vectors?

a+b= (2+10)i + (3+2)j=12i + 5j

What is the sum of the 2 vectors?

a+b= (2+10)i + (3+2)j=12i + 5j

Finding the sum of the two vectors using unit vector:

A+B=-1i + 4j

Example #3: you have 2 vectors:

A=-3i + 2j

B=2i + 2j

Lets add them up:

A=-3i + 2j

B=2i + 2j

Lets add them up:

A=-3i + 2j

Remember that -3i is on the X axis

2j is on Y axis.

So you draw the two vectors and put them head to tail and you get vector A.

Remember that -3i is on the X axis

2j is on Y axis.

So you draw the two vectors and put them head to tail and you get vector A.

Now lets visualize vector A:

B=2i+2j

Remember that 2i is on the x axis and 2j is on the y axis.

Remember that 2i is on the x axis and 2j is on the y axis.

Now lets visualize vector B:

remember head to tails!

Now trace -1i + 6j which represents the A+B vector.

So as you can see you don’t have to draw them which takes forever. Just use unit vector on MCAT 🙂

Now trace -1i + 6j which represents the A+B vector.

So as you can see you don’t have to draw them which takes forever. Just use unit vector on MCAT 🙂

Now lest add vector A and B together:

Notice that displacement is vector quantity so it takes into account direction of movement.

V=S/∆t

V=S/∆t

Formula to remember: VELOCITY

Velocity is vector quantity and it takes into account magnitude and direction.

V=S/∆t (where S is dispalacment)

Speed does not take direction into account-only magnitude.

Speed or rate r=d/∆t (where d is distance)

V=S/∆t (where S is dispalacment)

Speed does not take direction into account-only magnitude.

Speed or rate r=d/∆t (where d is distance)

What is the difference between Velocity and speed?

In physics classes they usually don’t use east or west but they do use + or negative direction to indicate direction of velocity.

Speed or rate r=d/∆t (where d is distance)

Formula to remember: Speed

They are giving us magnitude of 5km and direction north. So this is vector quantity.

He was displaced 5km in 1h.

Another thing to notice in here is the world AVRAGE. So while he was driving his velocity was changing but we are calculating average.

V=5km North/1h=5kmNorth/h

So his average velocity was 5km/h North (remember to put direction for velocity) if you don’t than you are talking about speed.

His average speed is r=d/∆t=5km/h (no direction)

He was displaced 5km in 1h.

Another thing to notice in here is the world AVRAGE. So while he was driving his velocity was changing but we are calculating average.

V=5km North/1h=5kmNorth/h

So his average velocity was 5km/h North (remember to put direction for velocity) if you don’t than you are talking about speed.

His average speed is r=d/∆t=5km/h (no direction)

Lets now apply the knowledge of vectors to calculating Speed and Velocity:

If Mark was able to travel 5km north in 1 hour in his car what was his average velocity?

If Mark was able to travel 5km north in 1 hour in his car what was his average velocity?

There are 1000m per km.

There are 60 s per 1 minute and 60 minutes per 1h.

There are 60 s per 1 minute and 60 minutes per 1h.

How can we now convert 5km/h into meters/seconds? This is conversion unit problem.

This is vector quantity since they give us magnitude 3m/s and direction to the east. So this is velocity.

How long will it take him to travel 720m to the east?

You can calculate the time from scaler formula:

r=d/t

Notice at the end when you have 720m/(3m/s) to cancel out meters you multiply by receptacles of units.

How long will it take him to travel 720m to the east?

You can calculate the time from scaler formula:

r=d/t

Notice at the end when you have 720m/(3m/s) to cancel out meters you multiply by receptacles of units.

Another problem: Ben has been running at constant velocity of 3m/s to the east. How long will intake him to travel 720meters?

This is vector quantity.

V=S/t

S=V x t

S=5m/s x 1 minute= 5m/s x 1 minute x 60s/minute=5×60 meters to the south=300 meters to the south.

V=S/t

S=V x t

S=5m/s x 1 minute= 5m/s x 1 minute x 60s/minute=5×60 meters to the south=300 meters to the south.

Another problem: If Marcia travels for 1 minute at 5m/s to the south how much will she be displaced?

Imagine that you are running home, at first you run fast then you slow down and then you speed up agin. These numbers are Instantaneous Speed.

Instantaneous speed is the speed of the object at particular time.

Instantaneous speed is the speed of the object at particular time.

What is INSTANTANEOUS SPEED?

If you include the direction in the speed of the object at particular time its a instantiates velocity.

8m/s to the right is velocity.

8m/s to the right is velocity.

What about INSTANTANEOUS VELOCITY?

If your home was 1000meters away and it would took you 200s to get there your average velocity would be 5m/s. But this value does not reflect INSTANTANEOUS velocities (velocities at specific point of time).

Notice that INSTANTANEOUS velocity is different then INSTANTANEOUS speed:

-Your average velocity will be: 60m/15s=4m/s+

Lets say you jobbed 60 meters in time of 15 seconds. During this time you were speeding up and slowing down.

So instead you can use a graph of time vs distance and the slope at the specific time will tell you INSTANTANEOUS velocity.

Lets say you want to know INSTANTANEOUS velocity at particular point of time: this is very hard and you need calculus since to find a velocity of the object at a moment of time you need to use very small distance and very very small time which is super hard to measure.

Its change of velocity over time. This is vector quality.

Lets talk now about Acceleration. What is acceleration?

When you are buying a car they will give you change of acceleration over time. Porsche 911 will go 0-60miles per hour in 3 seconds.

So what is acceleration in here?

a=(60-0 miles)/3seconds=20miles per hour/second

Look at the units in here: miles per hour/second

So what is acceleration in here?

a=(60-0 miles)/3seconds=20miles per hour/second

Look at the units in here: miles per hour/second

Problem #1 Acceleration:

-miles per hour/second (we got these from the previous example)

-miles/second²

Lets investigate how you got the miles/seconds²:

So we are left with miles per hour/second. To get rid of hours you need to multiply by hours/seconds.

Every hour has 3600 seconds. Look at the math in the picture.

-miles/second²

Lets investigate how you got the miles/seconds²:

So we are left with miles per hour/second. To get rid of hours you need to multiply by hours/seconds.

Every hour has 3600 seconds. Look at the math in the picture.

Units of acceleration:

-the take off velocity is 280km/h +

-it has constant acceleration of 1.0m per second/second

-it has constant acceleration of 1.0m per second/second

Another problem: Example is Airbus. And we want to know how long will it take this aircraft to take off?

How long does take off last?

As you can see the units are not the same so lets convert the units so they are the same. Lets convert everything into meters/second.

As you can see the units are not the same so lets convert the units so they are the same. Lets convert everything into meters/second.

Lets solve the problem of the airbus:

-the take off velocity is 280km/h +

-it has constant acceleration of 1.0m per second/second

-the take off velocity is 280km/h +

-it has constant acceleration of 1.0m per second/second

Acceleration= ∆V/∆t both sided multiplied by ∆t

∆t × a=∆V

∆t=∆V/a

∆t=(78m/s)/1m/s²=78s

∆t × a=∆V

∆t=∆V/a

∆t=(78m/s)/1m/s²=78s

Now that we have the same units lets find how long does the takeoff last?

In here we are assuming that we have constant acceleration so we can calculate average velocity: Va=(78m/s+0m/s)/2=39m/s

Now we found the average velocity we can find displacement S by multiplying average velocity by time.

S=Va x t=39m/s x 78s=3042m

Now we found the average velocity we can find displacement S by multiplying average velocity by time.

S=Va x t=39m/s x 78s=3042m

Now lets continue with the same example but lets find out how long of the runway does the airbus needs to take off? You want to find out the displacement.

The purple slow is constant acceleration. Starting with velocity =0

The distance traveled is the area under the slope up to the point of 78 seconds because this is how long it takes the plane to take off.

The distance traveled is the area under the slope up to the point of 78 seconds because this is how long it takes the plane to take off.

Lets look at the graph of the plane. As you know the plane is constantly accelerating:

Its some velocity that is between our final and initial velocity and if you take the average velocity for the same amount of time 78 seconds you would get the same area under the curve-the same distance.

So what is average velocity?

Lets say you have object that is moving to the right at constant velocity of 5m/s.

Lets plot the velocity vs time graph.

You can see that the velocity does not changes so the slope of the green line is the same.

Now ask yourself how far will the object travel after 5 seconds? You can think about that is two ways:

-You know that velocity is displacement/∆time

V=S/∆t when you multiply by ∆t you get S=V×∆t

So lets calculate it:

S=V×∆t=5m/s×5s=25m→this is the displacement.

Now look at the area of the rectangle in the picture. That rectangle area will be also 25 since 5×5=25

Lets plot the velocity vs time graph.

You can see that the velocity does not changes so the slope of the green line is the same.

Now ask yourself how far will the object travel after 5 seconds? You can think about that is two ways:

-You know that velocity is displacement/∆time

V=S/∆t when you multiply by ∆t you get S=V×∆t

So lets calculate it:

S=V×∆t=5m/s×5s=25m→this is the displacement.

Now look at the area of the rectangle in the picture. That rectangle area will be also 25 since 5×5=25

Lets look deeper into the question why distance is area under the velocity time line?

Since displacement is velocity x change of time.

So the main rule in here to remember is that when you plot velocity (or speed) vs. time the area under the line in axis is equal to displacement or distance.

You have constant acceleration of 1m/s²

We have initial velocity of zero but after each second you are going 1 meter. So look at the slope.

Acceleration is=∆V/∆t so the slope of the line is acceleration. We assume that acceleration is constant so we have constant slope.

We have initial velocity of zero but after each second you are going 1 meter. So look at the slope.

Acceleration is=∆V/∆t so the slope of the line is acceleration. We assume that acceleration is constant so we have constant slope.

Now lets look at the example when velocity is changing?

The area under the slope will be the distance. So you need to calculate the area under the line.

The formula for the area under the line is:

A=½ x b x h (base times height)

Displacement=½ x 5s ×5m/s=12.5m

The formula for the area under the line is:

A=½ x b x h (base times height)

Displacement=½ x 5s ×5m/s=12.5m

So now lets say we have been accelerating 1m/s for 5seconds. How far have we traveled?

-the slope will be acceleration

-the area under the slope will be the distance traveled

-the area under the slope will be the distance traveled

The things you need to remember when you plot time vs velocity:

We are starting with the initial velocity of 5m/s and you have constant acceleration of 2m/s² and we do this for duration of 4 seconds. Notice Vi (initial) and V (final).

How fast do we travel after 4 seconds and how far do we travel over the 4 seconds.

Lets draw a graph.

Notice that the slope of this graph is the constant acceleration 2m/s²

How fast do we travel after 4 seconds and how far do we travel over the 4 seconds.

Lets draw a graph.

Notice that the slope of this graph is the constant acceleration 2m/s²

Lets now move on to the next topic of average velocity for constant acceleration:

We started wit velocity of 5m/s² after that we are accelerating 2m/s².

Vi + (∆t) Ac (constant acceleration)=Vf

5m/s² + 4sx2m/s²=Vf

13m/s=Vf this is the final velocity after the 4 seconds.

Vi + (∆t) Ac (constant acceleration)=Vf

5m/s² + 4sx2m/s²=Vf

13m/s=Vf this is the final velocity after the 4 seconds.

Lets start calculating how fast are we going? What is the final velocity after the 4 seconds?

Must remember the formula that calculates final velocity for constant acceleration?

Remember that the distance is the area under the curve. You can brake this area into rectangle and triangle and find out the distance.

Rectangle=4×5=20meters

Triangle= ½ a × b=½ x 4 (base) × height (final velocity-initial velocity)= ½ x 4 × (13-5)=

½ x 4 x 8=16m

Total distance=20 meters+ 16 meters=36 meters

Rectangle=4×5=20meters

Triangle= ½ a × b=½ x 4 (base) × height (final velocity-initial velocity)= ½ x 4 × (13-5)=

½ x 4 x 8=16m

Total distance=20 meters+ 16 meters=36 meters

Now lets calculate the total distance that we would traveled? Calculating displacement.

In here you can see its really average of the initial velocity and final velocity times time.

This can only apply if the acceleration is constant (so slop of the line is strait and not a curve).

D= 4s × ½ (5m/s+13m/s)=

4s × ½(18m/s)= 4s x 9m/s=36m

Just remember this formula.

This can only apply if the acceleration is constant (so slop of the line is strait and not a curve).

D= 4s × ½ (5m/s+13m/s)=

4s × ½(18m/s)= 4s x 9m/s=36m

Just remember this formula.

Now we can calculate the displacement just like we did in the previous example but now we can just use formula instead of calculating area under the curve:

There are 3 laws of motion that newton came up with.

Lets now move on to the Newton’s Laws:

If object is at rest it will stay at rest or it will be moving at constant velocity unless its acted on by force.

The first part tells us that: if we have object at rest it will stay there. It will not move unless unbalanced force acts on it. Why unbalanced? If you have two forces but they are in opposite direction and same magnitude the object still will not move unless the force on one side is bigger (unbalanced).

The second part of the fist law tells us: if something is moving at constant velocity will continue have that velocity indefinitely unless force will act on it.

The first part tells us that: if we have object at rest it will stay there. It will not move unless unbalanced force acts on it. Why unbalanced? If you have two forces but they are in opposite direction and same magnitude the object still will not move unless the force on one side is bigger (unbalanced).

The second part of the fist law tells us: if something is moving at constant velocity will continue have that velocity indefinitely unless force will act on it.

Newton’s First Law:

In here we are talking about Object in motions. Object in motions with constant velocity would go on forever but we know that there are unbalanced forced that will act on the object and the object will stop, or change direction due to force acting on it.

Friction is one force, Air resistance, gravity force…

Newton’s First Law describes a World where there is no forces and objects go on forever.

Friction is one force, Air resistance, gravity force…

Newton’s First Law describes a World where there is no forces and objects go on forever.

What do you need to understand about Newton’s First Law:

Balanced vs. Unbalanced force:

True.

This is Newton’s First Law.

This is Newton’s First Law.

True or False: If the Net force on a body will be zero the body velocity will not change.

In here we have speed not velocity.

Example: Imagine ice skater that is going strait at constant velocity but then the ice skater grabs a rope and start going in the circle instead of going strait.

We can see that the direction of movement changed but the speed might be the same.

Many times direction and speed (which when combine is velocity) changes but not always. So sometimes only direction changes or speed. So one or the other.

Example: Imagine ice skater that is going strait at constant velocity but then the ice skater grabs a rope and start going in the circle instead of going strait.

We can see that the direction of movement changed but the speed might be the same.

Many times direction and speed (which when combine is velocity) changes but not always. So sometimes only direction changes or speed. So one or the other.

True or False: An unbalanced forced on the body will always impact the object speed.

True.

True or False: The reason why initially moving objects tend to come to rest in our everyday life is because they are being acted on by unbalanced forces.

False.

If a block is moving in + direction and you push it (give it move energy) you can change the speed (accelerate it) and the object will be moving faster but you might not changed direction.

If a block is moving in + direction and you push it (give it move energy) you can change the speed (accelerate it) and the object will be moving faster but you might not changed direction.

True or False: The unbalanced forced on an object will always change the object direction.

The Newton’s second law deals with the question of how does the net force affect the objects constant velocity?

Remember the first law just stated that the object would be moving at the constant velocity infinitely if the net force would not act on it. So now we will look how the velocity will change when net forces act on the object.

Remember the first law just stated that the object would be moving at the constant velocity infinitely if the net force would not act on it. So now we will look how the velocity will change when net forces act on the object.

Lets now move on to Newton’s Second Law:

Lets say you have brick and you apply a net force on the left side of the object the object’s acceleration will change in the same direction.

What Newton’s Second Law tells us is that the acceleration is proportional to the Force that is applied.

What Newton’s Second Law tells us is that the acceleration is proportional to the Force that is applied.

Newton’s second law gives up the formula: F=m×a

Force=mass x acceleration.

Force and acceleration are vector quantities.

Force=mass x acceleration.

Force and acceleration are vector quantities.

Notice that mass is different then weight:

Mass is a measurement of the amount of matter something contains, while Weight is the measurement of the pull of gravity on an object.

Newton

What is the unit of force?

Force=10N and its acting on mass of 2kg and you want to know acceleration:

F=m × a

F=m × a

Example #1 Illustrating Newton’s Second Law:

Look at the calculation in the picture.

The acceleration is now 10m/s²

The acceleration is now 10m/s²

Now what if you double the force to 20N?

When you double the force you will double the acceleration!!! From 5m/s² → 10m/s₂

So acceleration is PROPORTIONAL to force-must remember.

So acceleration is PROPORTIONAL to force-must remember.

What you need to notice about this example?

Notice that to find acceleration from this formula F=m×a you must divide by mass.

a=F/m

The larger the mass you have of object the smaller the acceleration and the bigger the force you must applied. In the picture you have Force of 10N acting on the 2kg object. Now look how 10N force (the same magnitude) is acting on 4kg object (double the weight). Find acceleration?

a=F/m=(10 kg m/s²)/4kg=2.5m/s²

The acceleration is 1/2!!!

a=F/m

The larger the mass you have of object the smaller the acceleration and the bigger the force you must applied. In the picture you have Force of 10N acting on the 2kg object. Now look how 10N force (the same magnitude) is acting on 4kg object (double the weight). Find acceleration?

a=F/m=(10 kg m/s²)/4kg=2.5m/s²

The acceleration is 1/2!!!

How will mass affect acceleration?

False

If ball is rolling down the grass it will eventually slow down unless you give the ball a push and then it will keep going. Since there is friction, gravity and air resistance that is lowing down the ball. But if you apply a force by pushing the forward the ball will keep going.

If ball is rolling down the grass it will eventually slow down unless you give the ball a push and then it will keep going. Since there is friction, gravity and air resistance that is lowing down the ball. But if you apply a force by pushing the forward the ball will keep going.

True/False: An object in motion will slow down unless it is acted on by an unbalanced force in the direction of motion:

True

True/False: An object in motion will maintain its speed and direction forever unless acted on by unbalanced force.

True

True/False: An object at rest will stay at rest unless acted on by unbalanced force.

True.

True/False: An object acted by unbalanced force will always accelerate in direction of the unbalance force.

This is know as: To every action there is an equal and opposite reaction. Or the forces of two bodies on each other are always equal and are directed in opposite direction.

Lets now move on to Newton’s 3rd Law:

What Newton is saying is that you can’t have a force just acting on the body without that body having opposite force acting on the first body.

Understanding of the 3rd Law:

Imagine that you are pushing on the block and exert a net force that will move the block to the right.

So if you push hard enough and apply big force you might overcome the friction and the block will move. But as you are moving the block and exerting force on block that block exerts force on you as well that is equal and opposite. The evidence is that your hand will get compressed.

So if you push hard enough and apply big force you might overcome the friction and the block will move. But as you are moving the block and exerting force on block that block exerts force on you as well that is equal and opposite. The evidence is that your hand will get compressed.

Example #1 understanding of the Newton’s 3rd Law:

Imagine that you are walking on the beach. You are exerting a force on the sand since you can see a food prints and sand is getting compressed. But the sand is exerting the equal but opposite force.

The evidence is: remember Newton second law? You have gravitational force that is pulling you down, so you should be accelerating downwards through sand unless there is other force that balances it out.

So the gravitational force pulls you down but the force exerted from sand back on your foot stops you from accelerating. So there is zero net force on you and you stop.

The evidence is: remember Newton second law? You have gravitational force that is pulling you down, so you should be accelerating downwards through sand unless there is other force that balances it out.

So the gravitational force pulls you down but the force exerted from sand back on your foot stops you from accelerating. So there is zero net force on you and you stop.

Example #2 understanding of the Newton’s 3rd Law:

When a rocket is trying to take off into the space there is nothing to push off. Thats why you use gasoline that will create gas that will come out of the back of your rocket and push backwards while the engine push forward.

So you have two opposite forces in two different directions.

So you have two opposite forces in two different directions.

Example #3 understanding of the Newton’s 3rd Law:

Imagine that you are astronaut in space and you got detached from the ship. You are starting to drift away into space. What can the astronaut do to to change the direction of movement so he can drift back to spec station?

There is nothing to push off but he can take off something that is heavy and throw it in the opposite direction to push off it.

There is nothing to push off but he can take off something that is heavy and throw it in the opposite direction to push off it.

Example #4 understanding of the Newton’s 3rd Law:

What is Center of Mass? Its a point on to object or not need to be on the object. We can pretend that the entire mass exist at that point.

Example is a ruler that has mass of 10kg. And if the Force is applied at the center of mass of 10N this ruler will accelerate the same way as would a POINT MASS. Which is a small dot that would also have mass of 10kg.

Example is a ruler that has mass of 10kg. And if the Force is applied at the center of mass of 10N this ruler will accelerate the same way as would a POINT MASS. Which is a small dot that would also have mass of 10kg.

Lets now move on to the next concept in this chapter that is CENTER OF MASS.

Ruler?

F/M=a

10N/10kg=1m/s²

Point mass?

F/M=a

10N/10kg=1m/s²

So both would accelerate upwards at 1m/s².

F/M=a

10N/10kg=1m/s²

Point mass?

F/M=a

10N/10kg=1m/s²

So both would accelerate upwards at 1m/s².

What would be the acceleration of ruler vs. point mass?

If you have some really crazy shaped object and you want to find out what it will do and how would it move you can find center of mass and see how it would move without worrying about shape of the object.

Why do we need Center of mass?

-If the object has uniform distribution of mass which means its made out of the same material and its density is the same through the object. Then the center of mass will be the object geometric center.

Example why its important to find center of mass?

Imagine now that you have square and ½ of the square is made out of lead and ½ is made out of stariaphone.

In this case the center of mass will not be the object geometric center since the object is made out of the two different materials.

In this case the center of mass will not be the object geometric center since the object is made out of the two different materials.

When center of mass is not in the center:

Imagine you have a horse and that horse has center of mass. If we apply a force on the center of mass the horse will move in the direction of that force with some acceleration.

To find acceleration you use F=m x a

You divide F by mass and get acceleration.

To find acceleration you use F=m x a

You divide F by mass and get acceleration.

So far we made simplifying assumption that the force acts on the center of mass:

Imagine that you have a ruler that has center of mass but now you apply force away from center of mass what will happen to object? It will rotate around the center of mass.

But what happens when a force acts away from the center of mass?

This happens when you apply force in the area that is away from the center of mass.

When you trow a basbal in air it will spin around the center of mass.

So when you trow object in the air the point at which it rotates around its center of mass.

Center of mass of rotating object:

-When you have a force that acts on the center of mass you will have shifting motion

-When you have a force that acts away from venter of mass you will have rotating motion

-When you have a force that acts away from venter of mass you will have rotating motion

So to recap:

Imagine a guy that wants to jump over a bar. The guy center of mass is at the gut.

There are two ways to jump:

-You can jump over the bar with your feet underneath.

-But when you watch Olympics you can see that people jump with their back almost touching a bar. But in this case the center of mass in this case travels BELOW the bar (since you take legs, arms and head into account). And because of that you can clear a bar without having center of mass going above the bar so you need less force to do it. So with the same force you can clear a higher bar!

There are two ways to jump:

-You can jump over the bar with your feet underneath.

-But when you watch Olympics you can see that people jump with their back almost touching a bar. But in this case the center of mass in this case travels BELOW the bar (since you take legs, arms and head into account). And because of that you can clear a bar without having center of mass going above the bar so you need less force to do it. So with the same force you can clear a higher bar!

Everyday application of center of mass:

-So we have learned so far that if you have a center of mass and you apply a force to the center of mass you will cause the object in this case a horse to shift in the direction of force. You can find acceleration using a formula: F=m×a

-But what is you apply force away from the center of mass? Then the object will be rotated AROUND the center of mass.

-But what is you apply force away from the center of mass? Then the object will be rotated AROUND the center of mass.

Lets now talk about TORQUE:

The object will rotate around the attachment point. Just like wskazowki na zegarku.

What if you apply a force to a object and the object is attached to some structure? You are applying a force away from the attachment?

A force that is applied a distance away from the attachment point is called torque.

What is torque?

T=F×d

Torque=Force × Distance

Torque=Force × Distance

What is the letter for torque? and what is the formula?

Its strange looking T.

Its strange looking T.

-Which force we are talking about? Force that is perpendicular to the distance vector.

Lets look at the formula of torque in details?

T=F x D

Force is in Netons

Distance is in meters

T=N x m

Side note: this is not work and we will not use Jules in here but rather we will use

N x m

Force is in Netons

Distance is in meters

T=N x m

Side note: this is not work and we will not use Jules in here but rather we will use

N x m

What are the units of Torque?

In work the Force and distance are parallel to each other. And the object is moving.

In torque the distance and force are perpendicular to one another. And the object is rotating.

So they are totally different things even though the units are the same.

In torque the distance and force are perpendicular to one another. And the object is rotating.

So they are totally different things even though the units are the same.

What is the difference between work and Torque?

-Distance 10m

-Force 5N perpendicular to the distance vector.

T=F x d= 5N x 10m= 50 Nm

-Force 5N perpendicular to the distance vector.

T=F x d= 5N x 10m= 50 Nm

Example # 1 Torque:

-Positive torque is when you rotate counterclockwise.

-Negative torque is when you rotate clockwise.

-Negative torque is when you rotate clockwise.

Torque can be negative or positive:

So in this example you have one force of 5N acting on the ruler over the 10meters distance. On the other side you have distance of 5 meters. Can we find the magnitude of the force so that the ruler doe not move at all? That means that the torque from both sides must be the same.

5N x 10m + (-F) x 5m =0 (why force is negative? since the rotation will be clockwise) all of it must be equal to zero since the net torque is zero since the ruler is not moving.

5N x 10m + (-F) x 5m =0 (why force is negative? since the rotation will be clockwise) all of it must be equal to zero since the net torque is zero since the ruler is not moving.

Lets see how negative and positive torque will work together:

Look at the magnitude of force and the distance relationship.

The longer the distance to the force the smaller the force you need to apply to rotate the object.

-If distance is 5 meters you need 10N force

-If the distance is 10 meters you need 5N force

The longer the distance to the force the smaller the force you need to apply to rotate the object.

-If distance is 5 meters you need 10N force

-If the distance is 10 meters you need 5N force

What can we learn from the example above?

You have a rock in the picture:

-The gravity force will be pulling it downwards lets say 5N

-The rock sits on the ground so according to Newton’s 3rd Law the ground will exert a force back on the rock of 5N that is also called a normal force.

Notice that these forces have the same magnitude but they are going in opposite directions. SO the forces cancel out and balance each other. No net force.

-The gravity force will be pulling it downwards lets say 5N

-The rock sits on the ground so according to Newton’s 3rd Law the ground will exert a force back on the rock of 5N that is also called a normal force.

Notice that these forces have the same magnitude but they are going in opposite directions. SO the forces cancel out and balance each other. No net force.

Lets now look at the different types of forces that will act on objects and see if they are balanced on unbalanced?

You have a rock:

-The force of gravity is still 5N

-Normal force 5N upwards

-But now you are trying to push the rock by applying 2N to the right.

-But there is friction force of 2N to the left.

SO the forces cancel out and balance each other. No net force. The rock will not accelerate.

-The force of gravity is still 5N

-Normal force 5N upwards

-But now you are trying to push the rock by applying 2N to the right.

-But there is friction force of 2N to the left.

SO the forces cancel out and balance each other. No net force. The rock will not accelerate.

Now lets look at another example of the rock being pushed:

You have a rock:

-The force of gravity is still 5N

-Normal force 5N upwards

-But now the guy is pushing the rock 3N to the right

-The friction force is still 2N to the left

The horizontal direction forces in here do not cancel out since you have 1N net force to the right. So the rock will accelerate/move.

-The force of gravity is still 5N

-Normal force 5N upwards

-But now the guy is pushing the rock 3N to the right

-The friction force is still 2N to the left

The horizontal direction forces in here do not cancel out since you have 1N net force to the right. So the rock will accelerate/move.

Now lets look at another example of the rock being pushed but harder:

-Force of gravity is 5N down

-There is air resistance 1N up

In here the force of gravity is greater. There is 4N force downwards.

-There is air resistance 1N up

In here the force of gravity is greater. There is 4N force downwards.

Now imagine that the rock is not on the ground anymore, so the only force that will act on it is force of gravity:

You have a rock on the ground:

-Force of gravity 5N down

-Normal force 5N up

-The guy is pushing with force of 4N to the right

-You have friction force of 2N

-But in this time there is a second person that is pushing the rock to the left with the force of 1N. So you have net 3N to the left .

There is a net force of 1N to the right.

-Force of gravity 5N down

-Normal force 5N up

-The guy is pushing with force of 4N to the right

-You have friction force of 2N

-But in this time there is a second person that is pushing the rock to the left with the force of 1N. So you have net 3N to the left .

There is a net force of 1N to the right.

Now lets look at another example of the rock being pushed by two people:

You have a frozen ice and on that ice you have 2 identical blocks of ice. They each weight 5kg.

The only difference between the two is that:

one is stationary and the other one is moving with constant velocity to the right of 5m/s

Why do you have ice on ice in here? We want to make sure that friction is very small to the point of negligible.

The only difference between the two is that:

one is stationary and the other one is moving with constant velocity to the right of 5m/s

Why do you have ice on ice in here? We want to make sure that friction is very small to the point of negligible.

Lets now study the difference between the normal force and contact force:

Newton’s Law says that they will stay as they are unless unbalanced force acts on them.

However since we live on Earth there is unbalanced force that will act on them and its called gravity that will act downward.

However since we live on Earth there is unbalanced force that will act on them and its called gravity that will act downward.

What is the Newton’s first Law will tell us about these blocks of ice. Where one block is stationary and the other one is moving at constant velocity and there is no friction.

F=g x mass of object

g is gravitational filed that is equal to 9.8m/s²

F=g x m

F=9.8m/s² x 5kg=49N

g is gravitational filed that is equal to 9.8m/s²

F=g x m

F=9.8m/s² x 5kg=49N

Lets now calculate the magnitude of the force of gravity:

The real reason why the ice will not accelerate towards the center of the Earth is because there is a opposing force that acts on the ice blocks and cancels out the gravitational force.

The magnitude of the opposing force is 49N and its direction is upwards.

The magnitude of the opposing force is 49N and its direction is upwards.

So as you can see there is 49N gravitational net force that is pulling that block downwards. But the block will not start accelerating downwards since it is resting on the ice.

Normal force and its magnitude is 49N upwards.

The normal force comes from the ice and the molecular bonds that hold the molecules together. On molecular level normal force is called contact force.

The normal force comes from the ice and the molecular bonds that hold the molecules together. On molecular level normal force is called contact force.

What is the name of the opposing force on the block?

Its the force that is perpendicular to the surface area on which the object is resting.

What is normal force?

Its still perpendicular to the surface.

How about incline plane. How would you draw normal force?

1. The elevator is stationary and velocity is V=0m/s in vertical direction, acceleration= 0m/s²

2. The elevator is going upwards with acceleration of 2m/s²

3. In this situation our acceleration goes back to zero a=0ms/² but now we have velocity of 2m/s in j direction (upward vector)

4. Now we are going down and elevator must slow down (decelerate) to -2m/s²

2. The elevator is going upwards with acceleration of 2m/s²

3. In this situation our acceleration goes back to zero a=0ms/² but now we have velocity of 2m/s in j direction (upward vector)

4. Now we are going down and elevator must slow down (decelerate) to -2m/s²

Lets now look at the Normal Force and elevator problem: where you have 4 situation that we will analyze:

The mass of the person is 10kg

The mass of the person is 10kg

What is the gravitational force in all of the cases?

F=g x m

F=-9.8m/s² x 10kg

F=-98N (why negative? gravity always pulls downwards)

Gravitational force changes as we move towards and away from the Earth center. But when you are on the surface of the Earth is the same in all of the situations.

So when calculating gravitational force F=g x m g will always be constant of 9.8m/s² but the mass of the object changes.

F=g x m

F=-9.8m/s² x 10kg

F=-98N (why negative? gravity always pulls downwards)

Gravitational force changes as we move towards and away from the Earth center. But when you are on the surface of the Earth is the same in all of the situations.

So when calculating gravitational force F=g x m g will always be constant of 9.8m/s² but the mass of the object changes.

Now lets look at what would be the GRAVITATIONAL force in all these 4 cases. The force that will act on us in each of these situations.

The gravity force =-98N and its pointing downwards.

The normal force must be Fn=98N and pointing upwards.

Velocity is zero in here.

The normal force must be Fn=98N and pointing upwards.

Velocity is zero in here.

What is the magnitude and direction of the normal force in case #1:

There is acceleration in here of 2m/s² so there must be a net normal force in here acting on the elevator and a person.

The normal force that is: 98N upwards.

Plus the force due to acceleration:

F=m x a

F=10kg x 2m/s²=20kg m/s²=20N

98N + 20N= 118N of Normal force upwards.

98N gravitational force down

The normal force that is: 98N upwards.

Plus the force due to acceleration:

F=m x a

F=10kg x 2m/s²=20kg m/s²=20N

98N + 20N= 118N of Normal force upwards.

98N gravitational force down

What is the magnitude and direction of the normal force in case #2: The elevator is moving upward with acceleration of 2m/s²

The velocity in here is constant. So remember about Newton’s first law: when you have constant velocity you have no net force on you!!!

Gravitational force=-98N and Normal Force=+98N they cancel each other out.

This elevator is the same as the case #1 and if you sit in that elevator and move at constant velocity you would not be able to tell the difference between case #1 where elevator is stationary and case #3 where elevator moves at constant velocity. Your body can only sense acceleration and not velocity.

Gravitational force=-98N and Normal Force=+98N they cancel each other out.

This elevator is the same as the case #1 and if you sit in that elevator and move at constant velocity you would not be able to tell the difference between case #1 where elevator is stationary and case #3 where elevator moves at constant velocity. Your body can only sense acceleration and not velocity.

What is the magnitude and direction of the normal force in case #3: The velocity is 2m/s but acceleration is 0m/s².

Fnet=m x a

Fnet=10kg x -2m/s²= -20kg m/s²=-20N

Normal force=+78N up since we subtracted the net force.

Gravitational force=-98N down

Fnet=10kg x -2m/s²= -20kg m/s²=-20N

Normal force=+78N up since we subtracted the net force.

Gravitational force=-98N down

What is the magnitude and direction of the normal force in case #4: This is where the elevator is slowing down so you have net acceleration of -2m/s².

When you sit in this elevator the only time you realize that you are moving is in the case # 2 when it is accelerating or in the case #4 that is slowing down. When its standing still or at constant velocity you don’t feel it.

Look at the values of the normal force in each case:

We have incline plane and weight on it. Lets look what forces will act on it?

-Gravity Force pulling it down.

-Normal Force

-Friction Force

Lets go one by one and learn details.

-Gravity Force pulling it down.

-Normal Force

-Friction Force

Lets go one by one and learn details.

Now lest look at incline Plane and try to understand force components: this is very important.

F=m x a (in here you can use g since its gravity)

F=m x 9.8m/s²

As you can see the force of gravity depends on the mass since the gravity is always the same of 9.8m/s²

F=m x 9.8m/s²

As you can see the force of gravity depends on the mass since the gravity is always the same of 9.8m/s²

How do you calculate the force of gravity?

Notation:

Fg⊥ is y component since its perpendicular to the incline plane.

Fg‖ is the x component since its parallel to the incline plane.

Fg⊥ is y component since its perpendicular to the incline plane.

Fg‖ is the x component since its parallel to the incline plane.

Normal force: but remember that normal force acts perpendicular to the surface of the plane. So to find the normal force you must first brake dawn the Gravitational force into x and y competent vectors and the y component will be the normal force.

Angles in right triangle add up to 180°

So if one angle is 90° the other two are:

-60°+ 30°=90°

-45°+45°=90°

So if one angle is 90° the other two are:

-60°+ 30°=90°

-45°+45°=90°

First lets just really fast brush off on the angle in the right triangle:

If you have two parallel lines and there is a third line that is transverse that crosses the 2 parallel lines the angles will be the same in few areas:

Also lets recap on the Congruet angles (angles that have the same angle):

The Force of gravity Fg is parallel to the vertical side of the triangle. The incline place is transverse line.

Lets find out the angles in the small triangle. You have 90° angle, the top angle is ∅ and the left side angle is 90-∅.

Make sure you can see these angles on the big triangle as well.

Lets find out the angles in the small triangle. You have 90° angle, the top angle is ∅ and the left side angle is 90-∅.

Make sure you can see these angles on the big triangle as well.

How can we apply the transverse lines into our triangle?

SOH CAH TOA

Sinθ=O/H

Cosθ=A/H

Tanθ=O/A

Sinθ=O/H

Cosθ=A/H

Tanθ=O/A

No make sure you know the angles:

Using geometry we can find the X and Y components of the gravity vector:

-Y component:

cosθ=A/H

Magnitude of Y component:

‖Fy‖=m × g cosθ

-X component;

sinθ=O/H

Magnitude of X component:

‖Fx‖=m × g sinθ

-Y component:

cosθ=A/H

Magnitude of Y component:

‖Fy‖=m × g cosθ

-X component;

sinθ=O/H

Magnitude of X component:

‖Fx‖=m × g sinθ

Lets find the components of the gravitational force so we can find normal force:

Magnitude of Y component:

‖Fy‖=m × g cosθ

Magnitude of X component:

‖Fx‖=m × g sinθ

‖Fy‖=m × g cosθ

Magnitude of X component:

‖Fx‖=m × g sinθ

Must remember so you can do fast calculation:

You need to remember the angles of 30° and 45° of sine and cosine since these are the ones you will use to find x and y components.

Angles to remember:

In here the friction will be the X component of the gravity force in here.

‖Fx‖=m × g sinθ

‖Fx‖=m × g sinθ

Friction force that keeps that block stationary:

Mass of the block 10kg

Lets do example that will show everything perfectly:

Incline plane 🙂

Incline plane 🙂

Step #1 Calculare the Fg which is gravitational force that pulls block down:

Step #2 Calculate the X component

Step # 3 Calculate the Y component

Step #4 Find the Normal force

So if there would not be any friction in here then there would be no opposing force an the block would move down with some acceleration (ice on ice case). The force that would pull the block down is the X component of the Gravity Force and it is equal to 49N.

Of course this is not the case in normal life we always have some friction. But lets assume there is no friction and lets calculate acceleration of the block down the ramp.

Of course this is not the case in normal life we always have some friction. But lets assume there is no friction and lets calculate acceleration of the block down the ramp.

Step #5 How about Friction in here?

If the block moves and there is acceleration that means that there is a net force in here and its 49N in this case since we assume that there is no friction.

Step #6 Find Accelaration of the block

-What will be the direction of friction?

Opposing to the direction of movement.

-What will be the magnitude of friction in here? 49N

-Since in here we have wood on wood we will use wood on wood coefficient γ=0.72

What is the friction formula?

Ff=γ ×Fn

Opposing to the direction of movement.

-What will be the magnitude of friction in here? 49N

-Since in here we have wood on wood we will use wood on wood coefficient γ=0.72

What is the friction formula?

Ff=γ ×Fn

Step #7 Now lets assume that there is a friction and this is wood on wood: so the block will not move and it will be stationary that means that there is opposing force that balances out the X component of the gravitational force of 49N.

Imagine that you want to move couch. It is very hard at first since but once you get it going its easy. Why?

There are two types of frictions: static and kinetic.

Its harder to push couch at first since standing object with no velocity have static friction that has larger coefficient then moving object that has kinetic friction. Even though you are moving the couch on the carpet the coefficients are different.

There are two types of frictions: static and kinetic.

Its harder to push couch at first since standing object with no velocity have static friction that has larger coefficient then moving object that has kinetic friction. Even though you are moving the couch on the carpet the coefficients are different.

Lets now talk really fast about Static Friction vs Kinetic Friction:

Coefficient of static friction=Magnitude of the “budging” force divided by magnitude of Normal force.

What is budging force? You can see that in here the friction Magnitude is 49N but what is the value of the friction just when the block will start moving? Maybe its 50N (that is usually determined experimentally).

What is budging force? You can see that in here the friction Magnitude is 49N but what is the value of the friction just when the block will start moving? Maybe its 50N (that is usually determined experimentally).

Step #8 Calculate the coefficient of the static Friction of this block?

Angle=30°

Mass of block=10kg

Mass of block=10kg

Lets start another problem which is: Ice block accelerating down on the Ice Ramp:

-Force of Gravity down

-Normal force

-But No friction since its ice on ice

-Normal force

-But No friction since its ice on ice

Step #1 What are the forces that are acting on that ice?

F=m x g

F=10kg x 9.8m/s²

F=98N

F=10kg x 9.8m/s²

F=98N

Step #2 Calculating the Force of Gravity on the block?

‖Fy‖=m × g cosθ

‖Fy‖=10kg x 9.8m/s² cos30°=98Ncos30°=49√3N

Notice a shortcut: 10kg x 9.8m/s²=98N just use this its faster.

‖Fy‖=10kg x 9.8m/s² cos30°=98Ncos30°=49√3N

Notice a shortcut: 10kg x 9.8m/s²=98N just use this its faster.

Step#3 Finding the Y component of the Gravitational force: notice the angle.

‖Fx‖=m x g sinθ

‖Fx‖=10kg x 9.8m/s² sin30°=

98N sin30°=49N

‖Fx‖=10kg x 9.8m/s² sin30°=

98N sin30°=49N

Step #4 Finding the X component of the Gravitational force:

Since the block is not accelerating downwards into the ice (the vertical direction) there must be a normal force that opposes the gravitational force of the same magnitude but different direction.

Normal force=49√3N

Normal force=49√3N

Step#5 Finding the normal force:

We assume that there is no friction. And we do have the X component of the Gravitational Force which will pull the block down the ramp.

Since there is a net force in here of 49N the block will accelerate.

Since there is a net force in here of 49N the block will accelerate.

Step #6 what forces do you have in the horizontal direction?

F=m × a

a=F/m=49N/10kg=4.9m/s²

a=F/m=49N/10kg=4.9m/s²

Step #7 Find the acceleration of the block:

Mass of the block=10kg

Angle 30°

X component of Gravity vector=49N

Y component of Gravity vector=49√3N

Normal Force 49√3 N

Angle 30°

X component of Gravity vector=49N

Y component of Gravity vector=49√3N

Normal Force 49√3 N

Now lets take the same example but we say that the block is made out of wood and it sits on the wooden ramp:

Remember Newton’s first Law: if you have a net force then you will have acceleration! But in here we are moving at constant velocity there is no acceleration. why?

There must be force that acts in opposite direction that keeps this block from accelerating downwards. It must be 49N in opposite direction.

This is the force of FRICTION.

This is different then the other slide since in here the box is already moving. We do not have to overcome static friction.

There must be force that acts in opposite direction that keeps this block from accelerating downwards. It must be 49N in opposite direction.

This is the force of FRICTION.

This is different then the other slide since in here the box is already moving. We do not have to overcome static friction.

Now lets assume that the block is moving down the ramp at constant velocity V=5m/s down the ramp

Coefficient of kinetic friction=magnitude of the Force of Friction divided by Normal force.

Since this box is moving lets calculate the coefficient of kinetic friction:

You have 5kg block that is sitting on the dirt.

You are pushing on the block with 100N force.

Coefficient of static friction=0.60

Coefficient of kinetic friction=0.55

You are pushing on the block with 100N force.

Coefficient of static friction=0.60

Coefficient of kinetic friction=0.55

Lets do one example to illustrate the difference between Static Friction and Kinetic Friction:

Budding force divided by normal force. Where budding force is the enough force that it took to move that block.

Gravitational Force=m x g=5kg x 9.8m/s₂=49N

Normal Force=49N

Gravitational Force=m x g=5kg x 9.8m/s₂=49N

Normal Force=49N

Lets first calculate the Static Friction?

As you can see we calculated the Budding friction. Which is the amount of force that it takes to overcome the static friction.

As you can see we calculated the Budding friction. Which is the amount of force that it takes to overcome the static friction.

100N-29.7N=70.6N that is the net force to the right.

The acceleration at the very moment:

F=m x a

a=F/m

a=70.6N/5kg=14.1m/s²

The acceleration at the very moment:

F=m x a

a=F/m

a=70.6N/5kg=14.1m/s²

What is the acceleration of the object at that vey moment that the block is starting to move?

Its 26.95N

Lets now calculate the Kinetic friction: now the object is moving.

What is the net force?

100N-26.95N=73.05N

What is acceleration?

F=m x a

a=F/m

a=73.05N/7kg=14.61m/s²

100N-26.95N=73.05N

What is acceleration?

F=m x a

a=F/m

a=73.05N/7kg=14.61m/s²

lets calculate the acceleration:

Tension is a force and it usually applies to string or wire when its stretch out.

Lets move on to TENSION:

What is Tension?

You have weight of 100N and its suspended on the wire. Since we are on Earth we know that there is gravitational force that is pulling on the weight downwards 100N.

The weight is not accelerating so the net force on that weight must be zero.

The weight is not accelerating so the net force on that weight must be zero.

Example #1:

There is a force that is in the string that is applied on the weight and its called TENSION.

Its magnitude is 100N and its direction is upwards.

Its magnitude is 100N and its direction is upwards.

What is the counteracting force that prevents the weigh of accelerating downwards towards the center of the earth?

The question is what is the tension in these 2 strings? Find T₁ and T₂

So we know that the block is not moving (the red point is stationary) and that acceleration is zero so there in no net force.

So we know that the block is not moving (the red point is stationary) and that acceleration is zero so there in no net force.

Now lets take the same example and add 2 other strings; 2 green strings. One is attached to sealing and the other one attached to wall.

The force of gravity.

What is the dawn force that is pulling on the weight?

Lets look at the wires:

-T₂ wire only pulls to the left, it does not have y component at all.

-T₁ does the upward lifting it has y and x components

-T₂ wire only pulls to the left, it does not have y component at all.

-T₁ does the upward lifting it has y and x components

What is the upward force in here?

The angle that the wire forms with the sealing is 30° so the inside angle that is parallel is also 30°

So T₁ wire has 2 components:

-T₁y which is the y component

-T₁x which is the x component

So T₁ wire has 2 components:

-T₁y which is the y component

-T₁x which is the x component

Lets look at the vector T₁ and its components:

SOH CAH TOA

Sin 30°=T₁y/T1

T₁ sin 30°=T₁y

All of the lifting of the weigh is done by T₁y component so it must be the same magnitude as the gravity force that is pulling the weight down.

T₁ sin 30°=T₁y=100N

T₁½=100N

T₁=200N This is the tension in the T₁ wire.

Sin 30°=T₁y/T1

T₁ sin 30°=T₁y

All of the lifting of the weigh is done by T₁y component so it must be the same magnitude as the gravity force that is pulling the weight down.

T₁ sin 30°=T₁y=100N

T₁½=100N

T₁=200N This is the tension in the T₁ wire.

What is the T₁y component?

Whatever tension is in the T₂ wire it must be offset by tension in the T₁x component of the T₁ wire.

T₂=T₁x (T₂ equals to the x component of the T₁ wire tension)

200N x cosine30= (200√3)/2=100√3 →this is tension in the T₂ wire

T₂=T₁x (T₂ equals to the x component of the T₁ wire tension)

200N x cosine30= (200√3)/2=100√3 →this is tension in the T₂ wire

Now lets find out what is the tension in the second wire?

When two parallel lines are intersected by the a line, the interior angles opposite of each other are congruent.

Lets just brush off on geometry and angles in here:

Alternate Interior Angles Theorem

Alternate Interior Angles Theorem

When two lines are crossed by another line (called the Transversal): The pairs of angles on opposite sides of the transversal but outside the two lines are called Alternate Exterior Angles.

Alternate Exterior Angles

What is the tension in the bottom wire?

Since the wire is not accelerating in any direction the net forces must be zero.

The gravitational force is 10N down so the tension must be 10N upwards.

Since the wire is not accelerating in any direction the net forces must be zero.

The gravitational force is 10N down so the tension must be 10N upwards.

Problem #2 Lets find out tensions in the wires: as you can see you have 3 wires in here. Lets first start with finding out the tension in the bottom wire and then the two side ones.

-T₁ has vectors T₁x and T₁y

-There is a angle of 30° with the sealing so we assume that the same angle is in the triangle.

What is the value of the T₁x component?

T₁cos30° this is the x component of the T₁

-There is a angle of 30° with the sealing so we assume that the same angle is in the triangle.

What is the value of the T₁x component?

T₁cos30° this is the x component of the T₁

How about the tension in the two side wires?

What you must do is draw the x and y components of each wire.

Lets start with T₁ wire and its components:

What you must do is draw the x and y components of each wire.

Lets start with T₁ wire and its components:

T₂cos60° This is the x component of the T₂

How about the T₂x component?

T₁cos30°=T₂cos60° vectors have the same magnitude.

(√3/2)T₁-½T₂=0

This is equation of the x components with two unknowns. We will use it later.

(√3/2)T₁-½T₂=0

This is equation of the x components with two unknowns. We will use it later.

Now that we found out the x components of the T₁ and T₂ we must remember that their net force must be zero. So they cancel each other out in magnitude but they are opposite in direction.

Finding Y component of T₁:

T₁sin30° +T₂sin60°=10N

T₁½+T₂(√3/2)=10N (multiply the whole thing by 2)

T₁+T₂√3=20N

Now take the equation from previous slide and multiply by 2:

(√3/2)T₁-½T₂=0 /x2

√3T₁-T₂=0

T₁sin30° +T₂sin60°=10N

T₁½+T₂(√3/2)=10N (multiply the whole thing by 2)

T₁+T₂√3=20N

Now take the equation from previous slide and multiply by 2:

(√3/2)T₁-½T₂=0 /x2

√3T₁-T₂=0

Lets now find the y components of the T₁ and T₂: The thing to remember in here is that these two y components should add up to 10N and offset the gravity force.

To do this we must multiply T₁+T₂√3=20N by √3

√3T₁+3T₂=20√3

Now we can subtract:

-T₂-3T₂=0-20√3

T₂=5√3N

Now you can substitute:

√3T₁-T₂ wich we just solved for and its 5√3=0

√3T₁-5√3=0 /divided by √3

T₁=5N

√3T₁+3T₂=20√3

Now we can subtract:

-T₂-3T₂=0-20√3

T₂=5√3N

Now you can substitute:

√3T₁-T₂ wich we just solved for and its 5√3=0

√3T₁-5√3=0 /divided by √3

T₁=5N

Now lets subtract equations:

T₁+T₂√3=20N

√3T₁-T₂=0

T₁+T₂√3=20N

√3T₁-T₂=0

You have incline plane and you have a wight sitting on the incline plane.

Mass of the block is 10kg.

Angle of the incline plane is 35°

Then the weight is attached by wire to a different weight of 20kg by wire.

The whole system will move to the right toward the 20kg weight.

Mass of the block is 10kg.

Angle of the incline plane is 35°

Then the weight is attached by wire to a different weight of 20kg by wire.

The whole system will move to the right toward the 20kg weight.

Problem with in then cline plane: TENSION IN ACCELERATING SYSTEM

-Gravity Force is pulling down on it. And it has x and y components. We need to find the y component with will the equal to normal force. the Y component is perpendicular to the ramp.

10kg x 9.8m/s²=98N→ this is Gravity force.

10kg x 9.8m/s²=98N→ this is Gravity force.

Lets look at the forces that act on the 10kg block:

cosine A/H→98cosine of 35°≈80N

Based on the we can say that normal force will also be 80N.

Based on the we can say that normal force will also be 80N.

Finding the Y component of gravity Force (perpendicular to the ramp)

98 sine of 35°≈56N

Finding the X component of gravity Force:

Ff=coefficient of friction that is specific for each material x normal force

Ff=0.2 x 80N=16N

Ff=0.2 x 80N=16N

Lets now find out the force of friction that is going backwards?

-The x component of gravity force 56N

-Friction force of 16N

The total forces that act on the weight trying to pull it back is 72N (56+16)

-Friction force of 16N

The total forces that act on the weight trying to pull it back is 72N (56+16)

So there are two forces that are pulling on the 10kg object backwards:

The other weight that is attached at the other end.

20kg x 9.8m/s²= 196N Gravitation Force

So you have 72N that want to pull the weight backwards and 196N that want to pull the wight forward.

20kg x 9.8m/s²= 196N Gravitation Force

So you have 72N that want to pull the weight backwards and 196N that want to pull the wight forward.

So what pulls the 10kg object forward?

196N-72N=124N

What is the force net?

F=m x a

Mass=10kg + 20kg=30kg (both weight are moving so we need to add them up)

124N=30kg x a

a=124N/30kg=4.13m/s²

The system will accelerate to the left and down at 4.13m/s²

Mass=10kg + 20kg=30kg (both weight are moving so we need to add them up)

124N=30kg x a

a=124N/30kg=4.13m/s²

The system will accelerate to the left and down at 4.13m/s²

What is the acceleration of the system?

Remember the wire is moving.

The acceleration in this case is 4.13m/s² and not 9.8m/s² which means that there is something that slowed down the acceleration. That must be the tension that slow downs the acceleration.

The acceleration in this case is 4.13m/s² and not 9.8m/s² which means that there is something that slowed down the acceleration. That must be the tension that slow downs the acceleration.

Now lets find a tension in the wire?

F=m x a

F=20kg x 4.13m/s²=83N down

F=20kg x 4.13m/s²=83N down

To find tension just focus on the one object: what is the net force on the 20kg object? we choose this since it easier.

We also know that Tension Force and Force of Gravity will equal to net force.

T+196N=83N

T=-113N

T+196N=83N

T=-113N

Finding Tension:

We have a pie in hand and we want to smash that pie in someones face. Lets find out what acceleration we must have so that the pie will not slide down from our hand downs to the ground.

Another example of Tension in Accelerating System:

The main thing to remember in here is that the force of gravity is pulling pie down but the force of friction opposes force of gravity. For the pie to stay put the force of gravity and force of friction must be equal.

What is the force of gravity that pulls on that pie?

F=m x a

F=m x 9.8m/s²

What is the force of friction? In here we know that force of friction Ff is the same as Force of gravity.

Ff=γ (friction coefficient) x Fn (normal force)

9.8m/s² x m= 0.8 x Fn (we substiuated 9.8m/s² for the Ff since they are equal)

Fn=(9.8m/s² x m)/0.8=12.25m/s² x m

What is the force of gravity that pulls on that pie?

F=m x a

F=m x 9.8m/s²

What is the force of friction? In here we know that force of friction Ff is the same as Force of gravity.

Ff=γ (friction coefficient) x Fn (normal force)

9.8m/s² x m= 0.8 x Fn (we substiuated 9.8m/s² for the Ff since they are equal)

Fn=(9.8m/s² x m)/0.8=12.25m/s² x m

The friction coefficient is: γ=0.8

Given that how fast I need to accelerate it before the force of gravity will pull it down:

Given that how fast I need to accelerate it before the force of gravity will pull it down:

F=m x a

12.25m/s² x m=m x a (cancel mass)

a=12.25m/s²

12.25m/s² x m=m x a (cancel mass)

a=12.25m/s²

So how fast we need to accelerate it?