## Friction

A surface can always supply a

normalforce, perpendicular to the surface. However, a surface quite oftenalsosupplies afiction forceparallel 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

Fas shown here:We know gravity pulls down with a force

w, the weight of the block. From the y-components ofF= ma, we have seen that the plane responds by exerting anormalforcen. But the surfacealsoresponds by exerting aparallelforce,f_{s}; this is the force ofstatic 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 frictionthat is labeledf_{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 isf _{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}nand

f _{k}=_{k}nThese '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:

ExampleOnce 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 frictionbetween the crate and the floor is 0.2. What is the acceleration of the crate? For this example, take g = 10 m/s^{2}for arithmetic convenience.The free-body diagram looks about as it did earlier -- except there is an

additonalforce now, the force of kinetic friction, f_{k}.Applying

F= mato the y-component forces, we findn = w = m g = (500 kg) (10 m/s ^{2})n = 5 000 N

f

_{k}= nf

_{k}= (0.2) (5 000 N)f

_{k}= 1 000 NNow 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 NF

_{net}= 250 NF

_{net}= 250 N = m a250 N = (500 kg) a

a = 250 N / 500 kg

a = 0.50 [ N / kg ] [ (kg m/s

^{2}) / N]a = 0.50 m/s

^{2}

ExampleFind the acceleration of an inclined Atwoods machine with a hanging mass of m_{1}= 1 kg and a mass of m_{2}= 5 kg sitting on an inclined plane which is inclined at 30^{o}from the horizontal. The coefficient of kinetic friction between this mass and the plane is 0.25.The forces on the hanging mass, m

_{1}, are just as they were before:But the forces on the other mass, m

_{2}, which sits on the plane now have africtionforce to be included:Now we apply Newton's Second Law to these forces acting on mass m

_{2}.F _{y,net }= 0F

_{y,ne}= 0 because there is no motion -- and, certainly, no acceleration -- in the y-direction.F _{y,net }= n - m_{2}g cos 30^{o}= 0n = m

_{2}g cos 30^{o}n = (5 kg) (10 m/s

^{2}) (0.866)n = 43.3 N

Notice that the normal force is

notequal to the weight!This is important.Now that we know the normal force, we can immediately calculate the kinetic friction force,f _{k}= nf

_{k}= (0.25) (43.3 N)f

_{k}= 10.8 NNow we can apply

F= mato the x-component forces to findF _{x,net}= m_{2}g sin 30^{o}- T - 10.8 N = m_{2}a(5 kg) (10 m/s

^{2}) (0.5) - T - 10.8 N = (5 kg) a25 N - T - 10.8 N = (5 kg) a

14.2 N - T = (5 kg) a

We still have

one equationwithtwo unknowns. But from the forces on the hanging mass, m_{1}, we knowT - m _{1}g = m_{1}aT = m

_{1}g + m_{1}aT = (1 kg) (10 m/s

^{2}) + ( 1 kg) aT = 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/s

^{2}

ApplicationsSummaryReturn to ToC, Ch5, Newton's Laws of Motion(c) Doug Davis, 2001; all rights reserved