————————————————- INTERPRETATION OF RESULTS In the last experiment, relation between mass of an object or the net force with acceleration was defined. Another type of force acting on a system is to be discussed in this experiment. When two objects interact by touching or contacting each other, interaction is said to be subjected to contact forces. Among those contact forces, we have normal force and frictional force. Normal forces are forces acting perpendicularly towards the object. It is due to molecules of the surface is resisting the molecules of the object squeezed on it.
Principal concern in this experiment is the frictional force. The use of oil in a cart engine, the use of right lubrication in skating as well as in basketball, the screeching sound of a sudden stopping car are all occurrences which is exploited by friction in the reality. Meanwhile, frictional force is component acting on them. It acts parallel to the interface or the surface of contact. Furthermore, this force acts to oppose any relative motion between surfaces. Thus, it is a force which resists the relative motion of one surface in contact with another. This force is not only exerted by one object but between the contacting surfaces.
One good example of this is the air drag – the frictional force exerted by the air on body moving through it. Frictional force is occurring in the same way as the normal force. When an object is laid on a surface, molecules of the object will form very little bonds with the molecules of the surface. First part of the experiment deals with the determination of coefficient of friction of motion of a moving object. Tension force pulling the wooden block is associated by mass carried by pan and the gravity. It is gradually increased until object moves with constant velocity.
At those times that the object is not yet moving, static frictional force is present. It is not constant. We added some load but still, it is not yet moving. By that, static frictional force is equal to the force to the tension force applied thus, capable of contradicting the force. The static frictional force will continually increase as load is added until it reaches the maximum force that the static frictional force can exert. It is also known as (fs)max. When a frictional force is at its maximum, the body in question will either be moving or will be on the verge of moving.
At this point, when the object starts to move with constant velocity, the frictional force will be present is now kinetic frictional force, f k. It is easier to move an object in motion rather than starting with stationary object. It is because fs > fk ,usually. In connect with this, we have coefficient of friction is represented by symbol µ which is a dimensionless scalar value. The coefficient of friction (static or kinetic) is a measure of how difficult it is to slide a material of one kind over another; the coefficient of friction applies to a pair of materials and not simply to one object by itself.
The coefficient of kinetic friction is usually less than compared to static frictional force. It is the reason why fs > fk because µs is usually smaller than µk. On static frictional forces, its coefficient starts at zero, increasing depending on the force applied until it reaches it maximum value. Thus it is correct to interpret that the static frictional force is given by fs ?? µsN and the kinetic frictional force is given by f k ?? µkN. In the experiment, the kinetic coefficients (since friction is measured when object starts to move constantly) we had calculated are: 0. 562, 0. 502, 0. 493, 0. 19, and 0. 515 giving an average of 0. 5062. This coefficient is extremely independent with the weight of the object and the normal force. It is dependent with two contacting materials because of their different attraction and repulsion with each other. By increasing the surface area starting on the third trial, we could get the same frictional force. Thus, coefficient of frictional force is roughly independent with the surface area, unlike in some misconceptions. Note that we use a horizontal track. Thus, normal force is just equal to the force exerted by the weight of the object times the gravity, N=Wg.
Moreover, the frictional force present is proportional to the normal force. This equation is given to be f = µN. So, as we make the wooden block heavier, the greater frictional force is exerted with its kinetic coefficient remained constant. In non horizontal surfaces, an object will begin to slide down on a specific angle. This is also called as the angle of repose at which the object at a certain surface would be in the verge of sliding. It was determined on the second part. For certain applications it is more useful to define static friction in terms of this maximum angle.
It is defined as: where µ is the maximum static coefficient of friction between the objects. It is in accordance with Newton’s second law of motion where a body’s total force is zero. By solving for the coefficient, we could found that coefficient is just equal to the tangent of the angle of inclination. Consequently, at horizontal objects static friction coefficient is just zero assuming that no external force such as tension is applied. By analysis, if we use angle greater than angle of repose, the object will tend to accelerate, on the other hand, it would stay if the angle is lower than the angle of repose.
Adjustment with the height is made until the block starts to slide. Therefore, at that height (with constant hypotenuse), the angle made by the ground and the wooden beam is equal to the angle of repose. In the experiment, angles computed are, 27. 420, 25. 850, 26. 700, 26. 800 and 27. 870. This phenomenon occurs because the component of gravitational force along the wooden platform just overcomes the frictional force. Remember that at greater angle, gravitational force is more influential which is discussed on experiment 2. It is easier and more convenient to determine the static coefficient friction in laboratories using this method.
By analyzing the result, the pair of materials with higher angle of repose is the one who have greater frictional force and frictional coefficient. This principle is highly utilized at recreational parks and resorts, particularly on their slides. At the last part of the experiment, maximum force which causes an object into uniform motion was obtained. The maximum force is the force needed to overcome the static frictional force and to move an object at constant velocity at a specific angle. This maximum force is co-occurring with (fs)max. At different angles, (fs)max are also different.
The higher the angle, the higher the maximum force needed if the orientation of the tension is going upward. Greater maximum force means greater mass of pan needed. It is also because of higher influence of gravity at larger angle. By variation of weights of load which brings the object tension, the maximum force was achieved. The calculated weights of the pan are: 215. 6255, 249. 2073, 324. 9080, 364. 5126 and 494. 7816 all in grams, g. On the other hand, experimental values are: 258. 7, 288. 7, 398. 7, 448. 7 and 478. 7 also in g. Small discrepancy has been committed.
Percentage difference is around 14% From the obtained experimental values, comparison has been made with the theoretical value. Certain errors have been committed. It may be rooted from misapprehension of the motion at constant velocity. This would bring us great discrepancy at the end. ————————————————- PROBLEMS 1. A 60 kg package rests on the level floor of a warehouse. If the coefficients of static and kinetic friction are 0. 56 and 0. 37 respectively, what horizontal pushing force is required to (a) start the package’s motion (b) slide the package across the floor at constant velocity? Given:
Forces| x-component| y-component| W| 0| -60kg9. 8ms2| N T W F N| 0| + |N|| T| + |T|| 0| Frictional Forces| (Fs)max| -0. 56|N|| 0| | Fk| -0. 37| N|| 0| a. ?Fx=0= 0+0 + |T| -0. 56|N| T=0. 56N let this be the equation 1 ?Fy=0= -60kg9. 8ms2+N+ 0+0 N=60kg9. 8ms2=588 N Substituting N from equation 1: T=0. 56588 N=329. 28 N b. ?Fx=0= 0+0 + |T| -0. 37|N| T=0. 37N Again, by substituting N from equation 1: T=0. 37N = 0. 37588 N T=217. 56 N 2. The coefficient of friction between the sole of a shoe and a tile span roof is 0. 36. What minimum slope of the roof can start slipping? ? (Fs)max= fN 0. 36=µ=tan? ?= tan-1(0. 36) ?= 19. 80 3.
A 1-kg block of wood is pressed against a vertical wall (? =0. 25) by a force inclined 40o upward with respect to the vertical axis. Determine the maximum force which can start the motion of the block. Force| x-component| y-component| W| 0| -1kg 9. 8 ms2| N| -|N|| 0| (Fs)max| 0| -0. 25|N|| FApplied| |Fapplied|sin 40| |Fapplied|cos 40| N 400 W (Fs)max Type equation here. ?Fx=0= 0-N+ 0+|Fapplied|sin 40 N=Fappliedsin 40 let this be the equation 1 ?Fy=0= -1kg 9. 8 ms2+0-0. 25|N|+|Fapplied|cos 40 Fappliedcos40=+1kg 9. 8 ms2+0. 25N SubstitutingN from equation 1: Fappliedcos40=+1kg 9. 8 ms2+0. 25Fappliedsin 40 Fapplied(cos40-0. 5sin40)= +1kg 9. 8 ms2 Fapplied=1kg 9. 8 ms2cos40-0. 25sin40 Fapplied=16. 19 N ————————————————- GRAPHS What is the slope of the line? The curve is linear. In the graph, it is computed that the slope is 0. 5263. This can be computed using: slope, m=ymax-yminxmax-xmin m= 382. 7g-182. 7g743. 7 g-363. 7 g=0. 5263 What can you say about the slope of the line and the average value of ?? The slope of the line is simply the average coefficient friction, µ. The two values are relatively near to each other. Obtained experimental value is 0. 5062. Errors have been committed leading to a semi line graph.