Friction

A surface can always supply a normal force, perpendicular to the surface. However, a surface quite often also supplies a fiction force parallel to the plane. Friction forces always oppose the motion -- or prevent the motion.

Think of pulling on a block to the right with an external force F as shown here:

We know gravity pulls down with a force w, the weight of the block. From the y-components of F = m a, we have seen that the plane responds by exerting a normal force n . But the surface also responds by exerting a parallel force, f_s; this is the force of static friction. When we first push on this block it does not move; it is held in place by this force of static friction. No matter how smooth the surfaces of the block and the plane appear at first glance, if we look at them under a microscope, we find they are quite rough.

But there is some maximum value of this force of static friction. If we increase the external force F, the block finally breaks loose and starts to move to the right. Now the forces on it are as shown in this sketch:

The surface exerts a force of kinetic friction ( or sliding friction) that is labeled f_k. "kinetic" simply means "moving"; this is the friction force once the block is in motion. This force of kinetic friction is less than the maximum value of the force of static friction; that is

f_k < f_s

This behavior can be summarized in this graph,

Upon closer investigation, we find that the maximum value of static friction and the force of kinetic friction are each proportional to the normal force; that is,

f_s,max = _s n

and

f_k = _k n

These 's are called the coefficients of friction. _s is the coefficient of static friction and _k is the coefficient of kinetic friction. Since f_s,max > f_k, this means _s > _k. If it is clear from context, it is common to say simply the "coefficient of friction" and to label it merely as .

Now let us return to earlier examples:

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Example Once again, we have a man pulling a crate along a concrete floor. This time, let's be specific. The crate has a mass of 500 kg and the man pulls with a force of 1 250 N. The coefficient of friction between the crate and the floor is 0.2. What is the acceleration of the crate? For this example, take g = 10 m/s² for arithmetic convenience.

The free-body diagram looks about as it did earlier -- except there is an additonal force now, the force of kinetic friction, f_k.

Applying F = m a to the y-component forces, we find

n = w = m g = (500 kg) (10 m/s²)

n = 5 000 N

f_k = n

f_k = (0.2) (5 000 N)

f_k = 1 000 N

Now we know the values of all the forces involved and we can proceed

F_net = F - f_k

F_net = 1 250 N - 1 000 N

F_net = 250 N

F_net = 250 N = m a

250 N = (500 kg) a

a = 250 N / 500 kg

a = 0.5 [ kg / N ] [ N / (kg m/s²)]

a = 0.5 m/s²

General Intro			Strings and Springs
Return to ToC, Ch 6, Application of Newton's Laws

(c) 2005, Doug Davis; all rights reserved