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, fs; 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 that is labeled fk. "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
fk < fs 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,
fs,max = s n
and
fk = These
Now let us return to earlier examples:
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 100 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, fk.
Applying F = m a to the y-component forces, we find
n = w = m g = (500 kg) (10 m/s2) n = 5 000 N
fk =
fk = (0.2) (5 000 N)
fk = 1 000 N
Now we know the values of all the forces involved and we can proceed
Fnet = F - fk Fnet = 1 250 N - 1 000 N
Fnet = 250 N
Fnet = 250 N = m a
250 N = (500 kg) a
a = 250 N / 500 kg
a = 0.50 [ N / kg ] [ (kg m/s2) / N]
a = 0.50 m/s2
Example Find the acceleration of an inclined Atwoods machine with a hanging mass of m1 = 1 kg and a mass of m2 = 5 kg sitting on an inclined plane which is inclined at 30o from the horizontal. The coefficient of kinetic friction between this mass and the plane is 0.25.The forces on the hanging mass, m1, are just as they were before:
But the forces on the other mass, m2, which sits on the plane now have a friction force to be included:
Now we apply Newton's Second Law to these forces acting on mass m2.
Fy,net = 0 Fy,ne = 0 because there is no motion -- and, certainly, no acceleration -- in the y-direction.
Fy,net = n - m2 g cos 30o = 0 n = m2 g cos 30o
n = (5 kg) (10 m/s2) (0.866)
n = 43.3 N
Notice that the normal force is not equal to the weight! This is important. Now that we know the normal force, we can immediately calculate the kinetic friction force,
fk = fk = (0.25) (43.3 N)
fk = 10.8 N
Now we can apply F = m a to the x-component forces to find
Fx,net = m2 g sin 30o - T - 10.8 N = m2 a (5 kg) (10 m/s2) (0.5) - T - 10.8 N = (5 kg) a
25 N - T - 10.8 N = (5 kg) a
14.2 N - T = (5 kg) a
We still have one equation with two unknowns. But from the forces on the hanging mass, m1, we know
T - m1 g = m1 a T = m1 g + m1 a
T = (1 kg) (10 m/s2) + ( 1 kg) a
T = 10 N + (1 kg) a
Now we substitute that to find
14.2 N - [10 N + (1 kg) a] = (5 kg) a 14.2 N - 10 N - (1 kg) a = (5 kg) a
4.2 N - (1 kg) a = (5 kg) a + (1 kg) a = (6 kg) a
a = 6 kg / 4.2 N
a = 1.43 m/s2
Applications Summary Return to ToC, Ch5, Newton's Laws of Motion (c) Doug Davis, 2001; all rights reserved
