# Mechanics 5: Friction and Uniform Circular Motion

Friction is a force that always is directed so that it opposes motion. It is also referred to as a “contact force.” The force of friction depends on the normal force, which we learned about last chapter, and we call it \(\mathbf{F_n}.\)

The force of friction is a vector, but its direction is defined as opposing motion. So in some problems, you must first figure out the direction of motion to know which direction to assign to \(\mathbf{F.}\) The force of friction is defined as

$$\mathbf F = \mu \mathbf{F_n}$$

The coefficient of friction is written as the greek letter \(\mu\) (pronounced “moo”) and it comes in two versions. We’ll get to that later.

Many books call the normal force \(\mathbf{N},\) so you may see it written as

$$\mathbf{F} = \mu \mathbf N$$

which leads to the physics joke: *This equation proves that Physics is Fun!* But, I think it is confusing to call the normal force \(\mathbf N\) and the unit of force N, so we’ll stick with the notation, \(\mathbf{F_n}.\)

### Frictional Forces

Static friction is a force that must be overcome in order to make something move. In other words, if you apply a force to an object that is equal to the force of static friction, it will just begin to move. It is noted as \(\mu_s.\) So, the force of static friction is calculated by

$$\mathbf{F_s} = \mu_s \mathbf{F_n}$$

and depends on the normal force, \(\mathbf{F_n}.\) Its direction is defined such that it opposes motion.

Kinetic friction is the force that you must overcome in order to keep an object moving that is already in motion. It is noted as \(\mu_k.\) So, the force of static friction is calculated as

$$\mathbf{F_k} = \mu_k \mathbf{F_n}$$

and depends on the normal force, \(\mathbf{F_n}.\) Its direction is defined such that it opposes motion. Let’s look at some examples.

### Examples

- Say we have a block sitting on the floor. How much force do we need to apply to just get it to move?
Here is the free body diagram

So the pushing force needs to equal the force of static friction, or \(F_p = F_s.\) We know how to calculate the force of friction, so this becomes

$$F_p = \mu_s F_n$$

We can see that the normal force is equal to the force of gravity (\(mg\)), so the final answer is

$$F_p = \mu_s mg$$**(With Numbers)**

Say that the block has a mass of 5 kg, and the coefficient of static friction is 0.4. How hard do we need to push to get the block to move? From above, we know

$$F_p = \mu_s mg = (0.4) (5\,{\rm kg}) (9.81\,{\rm m/s^2}),$$

which equals 19.6 N. So, to get this block to move, we would have to apply a force of (at least) 19.6 N. - Let’s work the problem again, this time with a given \(F_p > F_s.\) So, we know that it will move — now can we calculate the acceleration of the block?
As long as \(F_p > F_s\) (given) we know that the block will move. Once it starts moving, the friction force changes (lessens) from static friction \((F_s)\) to kinetic friction \((F_k).\) If \(F_p > F_s,\) it must also be that \(F_p > F_k,\) so Newton’s law tells us that the sum of the forces equals mass times acceleration, or \((F_p – F_k) = ma\) giving us

$$a = (F_p – F_k) / m \; .$$**(With Numbers)**

For a mass of \(m=6\)kg; between the block and the floor, a coefficient of static friction \(\mu_s = 0.5\) and kinetic friction \(\mu_k = 0.2.\) For a pushing force of \(F_p = 30\) N, calculate the acceleration of the block.Check: force of friction is \(F = \mu F_n,\) and \(F_n = mg,\) giving

$$F_s = \mu_s mg = (0.5)(6\,{\rm kg})(9.81\,{\rm m/s^2}) = 29.4\,{\rm N}$$

So, yes, a pushing force of \(F_p = 30\)$ N is \(> F_s,\) and the block moves. Next, what the force of kinetic friction?

$$F_k = \mu_k mg = (0.2)(6\,{\rm kg})(9.81\,{\rm m/s^2}) = 11.8\,{\rm N}$$

So, as mentioned above, the sum of forces equals mass times acceleration, so the acceleration of the block is

$$a = (F_p – F_k)/m = 3\,{\rm m/s^2}$$

Newton’s law tells us that any force in excess of \(F_k\) goes into accelerating the block. - Here we have two blocks, of masses \(m_1\) and \(m_2.\) They have different coefficients of friction. For now, we’ll call those coefficients: \(\mu_{s_1},\) \(\mu_{k_1},\) \(\mu_{s_2},\) and \(\mu_{k_2}.\)

- How much force, \(F_p,\) must be applied to the bottom block to move the stack?
Calculate the normal force from the floor: \(F_n = F_g = (m_1 + m_2) g.\) Notice that we have to sum the masses to get the total mass. Now, the force of static friction opposes \(F_p,\) so to get the stack to move, the pushing force must be greater than \(F_s\) which equals \(\mu_s F_n,\) so we get \(F_p > \mu_{s_2} (m_1 + m_2) g.\)

- How hard can we push on the bottom block without the top block sliding?
If the whole stack is moving with acceleration \(a,\) then the sum of the forces applied to the top block equals \(m_1a.\) If it is not moving relative to \(m_2,\) then the only force applied to it is the force of static friction, or \(F_s = \mu_{s_1} m_1 g.\) So Newton’s law tells us that \(F_s = m_1 a,\) or

$$m_1 a = \mu_{s_1} m_1 g$$

This tells us that as long as the pushing force, \(F_p < \mu_{s_1} m_1 g,[/latex] then the top block will not slide on the bottom block. - Can we push on the top block and get whole stack to move?
In part (a) we figured out how much force, [latex]F_p,\) applied to the bottom block, is needed to get the stack to move. Now, we can calculate that if we push on

the top block, that push gets translated to the bottom block through the force of static friction from the top block, which we saw above was \(F_s = \mu_{s_1} m_1 g.\) So we now have bounds for the pushing force to move the stack from the top block, and we know that the pushing force needs to be

$$ \mu_{s_2} (m_1 + m_2) g < F_p < \mu_{s_1} m_1 g$$

**(With Numbers)**

Say we have \(m_1 = 4\) kg, \(m_2 = 5\) kg, \(\mu_{s_1} = 0.6\) and \(\mu_{s_2} = 0.25.\) This would represent the top of \(m_2\) is rough and the floor is smooth.- In order to get the stack to move,

$$ F_p > \mu_{s_2} (m_1 + m_2) g = 0.25 (4\,{\rm kg} + 5\,{\rm kg}) (9.81\,{\rm m/s^2}) = 22.1\,{\rm N}$$

- For the top block not to move, we need

$$F_p < \mu_{s_1} m_1 g = 0.6 (4\,{\rm kg}) (9.81\,{\rm m/s^2}) = 23.5\,{\rm N}$$ - Combining these results, the whole stack can be pushed from the top block as long as

$$ 22.1\,{\rm N} < F_p < 23.5\,{\rm N}$$

- How much force, \(F_p,\) must be applied to the bottom block to move the stack?
- In this problem we have a block sitting on an inclined plane

This is similar to Example 3 in the previous chapter, but now we are accounting for the friction force. Note that gravity (\(F_{g_x}\)) is trying to pull the

block down the plane, and friction (\(F_s\)) is trying to hold it in place (opposing the possible motion).So the question is: how can we tell if the mass, \(m,\) will slide down the plane?

We know from last chapter that \(F_{g_x} = m g \sin\theta\) and \(F_{g_y} = m g \cos\theta.\) From this chapter, we know that \(F_s = \mu_s F_n.\) For the no motion case, \(F_s = F_{g_x}\) and \(F_n = F_{g_y}.\)

Putting these together, we get

$$F_n = m g \cos\theta,\ \ \ {\rm so}$$

$$F_s = \mu_s F_n = \mu_s m g \cos\theta$$

and for the no motion case, \(F_{g_x} = F_s,\) giving

$$m g \sin\theta = \mu_s m g \cos\theta,$$

which simplifies to \(\sin\theta = \mu_s \cos\theta\) and further simplifies to \(\tan\theta = \mu_S.\) This tells us that whether or not the block slides on its own*does not*depend on mass, but only on the angle of incline (\(\theta\)) and the coefficient of static friction for the surface (\(\mu_s\)).

- Here, we have two blocks tied together with a rope. One block is on the inclined plane, the other is hanging from the rope.

What is the mass of \(m_2\) for no motion in the system?Here are the free-body diagrams for these two blocks. Note that \(m_1\) has been rotated for clarity in this diagram.

We have drawn \(F_s\) so that it opposes \(T,\) the direction that \(m_2\) is trying to pull \(m_1.\) From this diagram (and from before) we know that

\(F_{g_x} = m_1 g \sin\theta\) \(F_{g_y} = m_1 g \cos\theta\) \(F_n = F_{g_y}\) \(F_s = \mu_s F_n\) \(F_{g_2} = m_2 g\) \(T = F_{g_2}\)

So the force of friction can be calculated like so

$$F_s = \mu_s F_n = \mu_s F_{g_y} = \mu_s m_1 g \cos\theta$$For the no motion case (forces are balanced), the diagram tells us that \(T – F_{g_x} – F_s = 0,\) or \(T = F_{g_x} + F_s.\) Substituting for these three quantities, we get

$$ m_2 g = m_1 g \sin\theta + \mu_s m_1 g \cos\theta$$

We can cancel a \(g\) from each term, leaving the answer for \(m_2\):

$$ m_2 = m_1 \sin\theta + \mu_s m_1 \cos\theta$$**(With Numbers)**

Say we have \(m_1 = 7\) kg, \(\mu_s = 0.3\) and an incline angle of \(\theta = 25^\circ.\) This means that we need a mass \(m_2\) equal to

$$m_2 = m_1 \sin\theta + \mu_s m_1 \cos\theta = (7\,{\rm kg}) \sin 25^\circ + 0.3 (7\,{\rm kg}) \cos 25^\circ = 4.86\,{\rm kg}$$

for no motion to occur.

### Uniform Circular Motion

Recall that velocity is a

*vector.*That means that it has a

*magnitude*and a

*direction.*Also, recall that acceleration is a change in velocity divided by a change in time. That means there are three ways to accelerate (change the velocity). You can speed up (increase magnitude), slow down (decrease magnitude) or turn a corner (change direction). We’ve talked about the first two — now, let’s talk about the third one.

When an object moves at a constant (magnitude) velocity in a circle, it is constantly changing direction. The acceleration is towards the center of the circle. Think of it this way — if an object is trying to travel in a straight line, and is being constantly pushed (accelerated) toward the center of the circle, it will bend around and travel in a circle. A standard illustration of this is a mass on a string. If you twirl it in a circle, the mass wants to travel in a straight line, but the string is constantly pulling it towards the center of the circle. In fact, if while you were whirling the mass, the string broke, the mass would fly off in a straight line, absent the pull towards the center of the circle.

This type of acceleration is called *centripetal* acceleration. This is a latin word that means “center seeking” — it is always being pulled to the center. This acceleration is calculated by \(v^2/r,\) where \(r\) is the radius of rotation. The centripetal force on the mass is mass times acceleration, which is calculated as \(mv^2/r.\)

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