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IIT Physics Concepts: Understanding the difference between Static and Kinetic friction

Newton’s law problems involving friction forces are common on IIT-JEE and AIEEE exam but if you are not careful you will make a classic mistake which will cost you some easy marks.

Calculating the Friction force – Yes more precisely the Kinetic Friction Force

In this article, I will like to dwell in more detail on the calculation of the friction force in various problems. Though calculation of friction force is really simple, many mistakes made in examinations are due to the inability to calculate the friction force. Consider the example illustrated in the example below. A block of weight w is being pulled with a force F applied to a rope which makes an angle a with the horizontal; the coefficient of friction is k. Find the force of sliding friction. How will you go about it?

Friction free body - IIT PhysicsIIT Physics Friction

It’s really simple the friction force is = kN, where k is the friction co-efficient and N is the normal force. Please note N = mg= w in this case.

Some student directly write the expression for friction force as = kw

It is worth noting that, In a particular case, the weight and the normal reaction may be equal to each other, but, in genera!, they are entirely different forces. Consider the example I proposed. The forces applied to the body (the block) are the weight w = mg, normal reaction N, force of sliding friction and the tension F of the rope (see Fig. 2). We resolve force F into its vertical (F sin a) and horizontal (F cos a) components. All forces acting in the vertical direction counterbalance one another. This enables us to find the normal reaction:

N = w -F sina

As you can see, this force is not equal to the weight of the block, but is less by the amount F sin a. physically; this is what should be expected, because the taut rope, being pulled at an angle upwards seems to “raise” the sled somewhat. This reduces the force with which the block bears on the surface and thereby the normal reaction as well. So, in this case,

Friction force =k (w- Fsina)

If the rope were horizontal (a =0) , then instead of equation we would have N=w, from which it follows that Fr=kw.

“This is quite a common error of examinees who attempt to treat the force of sliding friction as the product of the coefficient of friction by the weight and not by the normal reaction. Try to avoid such mistakes in the exam.”

Difference between Kinetic and Static Friction

So far we have been dealing with the force of kinetic friction. Now let us consider static friction. This has certain specific features to which students do not always pay sufficient attention. Take the following example. A body is at rest on a horizontal surface and is acted on by a horizontal force F which tends to move the body. How great do you think the friction force will be in this case?

If the body rests on a horizontal plane and force is horizontal, then N=P. Then it follows that the friction force equals kw. If you thought so, then you have made a typical mistake by confusing the forces of kinetic and static friction. If the body was sliding al<mg the plane, your answer would be correct. But here the body is at rest. Hence it is necessary that all forces applied to the body counterbalance one another. Four forces act on the body: the weight P, normal reaction N, force F and the force of static friction Fr (see diagram). The vertical forces w and N counterbalance each other. So should the horizontal forces F and Friction force Therefore

Frictional Force = F

It follows that the force of static friction depends on the external force tending to move the body, and the force of static friction increases with the force F. It does not increase infinitely, however. The force of static friction reaches a maximum value:

Friction force (Static) Max = Ko (N)

Coefficient ko slightly exceeds coefficient k which characterizes, according to equation the force of sliding friction. As soon as the external force F reaches the value koN, the body begins to slide. At this value, coefficient ko becomes equal to

k, and so the friction force is reduced somewhat. Upon further increase of force F, the friction force (now the force of kinetic friction) ceases to increase further (until very high velocities are attained), and the body travels with gradually increasing acceleration. The inability of many examinees to determine the friction force is disclosed by the following rather simple question: what is the friction force when a body of weight w is at rest on an ‘inclined plane with an angle of inclination a? One hears a variety of incorrect answers. Some say that the friction force equals kw, and others that’ it equals kN=kw cos(a)

Since the body is at rest, we have to deal with the force of static friction. It should be found from the condition of equilibrium of forces acting along the inclined plane. There are two such forces in our case: the friction force FIr and the sliding force Wsin(a) acting downward along the plane. Therefore, the correct frictional force is Wsin(a)

Let us now consider another problem: A load of mass m lies on a body of mass M; the maximum force of static friction between the two is characterized by the coefficient ko and there is no friction between the body and the earth. Find the minimum force F applied to the body at which the load will begin to slide along it.

static friction force

First I shall assume that force F is sufficiently small, so that the load will not slide along the body. Then the two bodies will acquire the acceleration:

A = F / (m+M)

Now, what force will this acceleration impart to the load?

It will be subjected to the force of static friction Fr by the acceleration. Thus

Fr= ma= Fm/ (m+M)

It follows that with an increase in force F, the force of static friction Fr also increases. It cannot, however, increase infinitely.

Its maximum value is Fr Max = koN = komg

Consequently, the maximum value of force F. at which the two bodies can still travel together as an integral unit is determined from the condition

Komg = Fm/ (M+m)

From which

F = (M + m) kog

This, then, is the minimum force at which the load begins to slide along the body.

Summary: We take friction force Fr to behave as follows:

When bodies are not in relative motion,

0 <= Fr <=kN

Fr opposes the motion that would occur in its absence.

For bodies in relative motion,

Fr = kN

Fr is directed opposite to the relative velocity.

I will post problems testing this concept later.

Thanks for reading and do comment.

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2 Awesome Responses so far

  1. mentor,
    this is an amazing article which will help us students understand the concepts very well.
    The best part about your articles is that they are very student friendly i.e. easy to read and understand.

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