# Friction between two moving blocks

1. Sep 11, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
Mass MA = 4 kg rests on top of mass MB = 5 kg that rests on a frictionless table. The coefficient of friction between the two blocks is such that the blocks just start to slip when the horizontal force F applied to the lower block is 27 N. Suppose that now a horizontal force is applied to the upper block. What is the forces maximum value for the blocks to slide without slipping relative to each other?

2. Relevant equations
Friction equation
Newton's second law

3. The attempt at a solution

So essentially, I got the problem right, but by only making a correction to an earlier mistake. I am trying to figure out why I made that mistake in the first place.

The mistake was that when using Newton's second law on the lower block, I wrote that $F_{net} = (m_A + m_B)a$. I found that the correct way is to write $F_{net} = m_B a$. Why is this the correct way? Isn't is so that block B consists of the masses of both of the blocks?

2. Sep 11, 2016

### BvU

There's two parts to this: 1. $f$ the friction factor and then 2. The maximum force $F$ on the upper block.

3. Sep 11, 2016

### Mr Davis 97

Well let's just talk about finding the coefficient for now. For the first body, in the x-direction I get $f = m_A a \implies a = \mu N_{AB}$, which is the maximum acceleration that the friction can take.

Now, for the second body, in the x-direction, I get $F_A - f = (m_A + m_B)a$. Combing this with the previous result we get $\displaystyle \mu = \frac{F_A}{(2m_A + m_B)g}$ for $\mu$. However, it turns out that the correct expression for mu is $\displaystyle \frac{F_A}{(m_A + m_B)g}$, which is only the case when, for the second body, we have that $F_A - f = (m_B)a$. rather than $F_A - f = (m_A + m_B)a$. I'm not sure my latter is wrong while the former is right.

4. Sep 11, 2016

### andrewkirk

Look at it from the point of view of the upper block A when the lower block B is pushed by F. Block B is applying a horizontal force to A that accelerates it at rate $a=F/(m_A+m_B)$. Hence that force is $m_Aa=m_AF/(m_A+m_B)$. So block A exerts that force backwards on block B. The max horizontal force of the blocks on one another without slipping is $\mu m_A$. So, if $F1$ is the max force on the lower block without slipping (27N) we have $\mu gm_A=m_AF1/(m_A+m_B)$.

Now consider when the upper block is pushed with force $F$. By similar reasoning, we see that the horizontal force between the blocks is $m_BF/(m_A+m_B)$. Let the max force on the upper block without slipping be $F2$. Then we have
$\mu m_A=m_BF2/(m_A+m_B)$

Hence we have
$$m_AF1/(m_A+m_B)=\mu gm_A=m_BF2/(m_A+m_B)$$
whence
$$m_AF1=m_BF2$$
so that $F2=F1m_A/m_B$.

It turns out that one doesn't need to calculate $\mu$ in order to solve the problem.

5. Sep 11, 2016

### Mr Davis 97

I am not seeing exactly how that applies to my specific problem. What I'm not understanding is that when using Newton's 2nd law on the lower block, the mass is $m_B$ rather than $m_A + m_B$. Also, note that I already solved the problem using $m_B$. The thing is on my first try I used $m_A + m_B$, and I'm wondering why this is wrong.

6. Sep 11, 2016

### andrewkirk

My understanding was that you did not see a justification for the formula for $\mu$.

My post derives two such formulas. They are obtained by dividing the three-part equation
$$m_AF1/(m_A+m_B)=\mu gm_A=m_BF2/(m_A+m_B)$$
by $gm_A$ to obtain
$$\frac{F1}{(m_A+m_B)g}=\mu =\frac{F2(m_B/m_A)}{(m_A+m_B)g}$$
Switching to your notation, with $F1,F2$ corresponding to $F_B,F_A$ respectively (assuming that by $F_A,F_B$ you mean the maximum possible forces applied to blocks A and B respectively, without slipping), this is
$$\frac{F_B}{(m_A+m_B)g}=\mu =\frac{F_A(m_B/m_A)}{(m_A+m_B)g}$$

The formula you have quotes for $\mu$, in post 3, is
$$\mu=\frac{F_A}{(m_A+m_B)g}$$
I don't think that formula is correct. Where did you get it? Are you sure you transcribed it correctly?

7. Sep 11, 2016

### Mr Davis 97

By $F_A$ I just mean force applied. It does not correspond to the force on a particular block, which is probably a mistake in naming on my part. And by $F_A$ I am referring to the force that is 27 N. So I think my formula is correct. My question still stands from my last post (as I don't want to copied it all again here).

8. Sep 11, 2016

### andrewkirk

I'm afraid I can't understand your second post. It uses a symbol $f$ that has not been defined. There is another undefined term $F_{net}$ in the first post. Given that you have used $F_A$ for the 27N force that is applied to block B - a force that convention would be labelled $F_B$, it is not possible to reliably guess what your undefined terms mean, using convention as a guide.

I suggest you write out exactly what your question is, making sure that the meaning of every symbol you use is clearly defined. Bear in mind that there are two physical scenarios discussed - one where block A is pushed and one where block B is pushed - so it needs to be clear at all times which scenario you are referring to.

9. Sep 11, 2016

### Mr Davis 97

Okay, so first of all, I am just talking about the first part of the problem, where we need to find the a coefficient of friction from the given data.

So for body A, in the x-direction I get $f = m_A a \implies a = \mu N_{AB}$, where $f$ is the force of friction. To get the maximum $a$, I subsitute $f = \mu m_A g$ and solve for $a$.

Now, for block B, in the x-direction, (using $F_B$ for the applied force on B) ,I get $F_B - f = (m_A + m_B)a$, where once again $f$ denotes the friction acting on block B due to block A. I have $m_A + m_B$ because it seems that block B must account for both its own mass and the mass on top of it, mass A. Combing this equation with the previous result, (by plugging in $a$ from the first equation to the second), we get $\displaystyle \mu = \frac{F_B}{(2m_A + m_B)g}$. However, it turns out that the correct expression for mu is $\displaystyle \frac{F_B}{(m_A + m_B)g}$, which is only the case when, for the second body, we have that $F_B - f = (m_B)a$. rather than $F_B - f = (m_A + m_B)a$. My question is,

10. Sep 11, 2016

### andrewkirk

Thank you, that is much clearer.
I think where it runs into trouble is here:
$m_A+m_B$ should not be used because the effect of the top block's mass is already taken account of by deducting $f$. So by using $m_A+m_B$ instead of $m_B$ the effect of the top block has been double counted.
The formula should either be, this, as Newton's 3rd law applied to block B alone
$$F_B-f=m_Ba$$
or
this, as Newton's 3rd law applied to blocks A and B combined
$$F_B=(m_B+m_A)a$$

11. Sep 11, 2016

### Mr Davis 97

Okay, that exactly answers my question. So in any scenario would we have Newton's law where we sum multiple masses on the right hand side? What about if we had two bodies, one on top of the other, different masses as in this example, except there was no friction, and we used to different forces to accelerate them at the same rate. Then wouldn't Newton's law for the bottom body require that we take the aggregate mass into account in order to accelerate it?

12. Sep 11, 2016

### andrewkirk

If there's no friction between the two bodies, it's not possible to apply a horizontal force to the top body by pushing on the bottom one, or vice versa. Viewed from the perspective of horizontal motion, the two bodies are unrelated systems and must be treated separately, with two separate instances of Newton's 3rd law.

We can only treat multiple bodies as a single system, with a single application of Newton's 3rd Law, if a force applied to one of the bodies is transmitted to all of the bodies in proportion to their mass. In the problem given that condition holds, but only up to a limit that depends on the relative masses and the coefficient of friction between the bodies. If the coefficient is zero, the condition never holds, no matter how small the applied force.