Frictional Forces: Find Max F for No Slippage of m1 on m2

[zhaos]

Hi all, I am studying for an exam and changed a practice problem by putting the force on the bottom block (you'll see). I just wanted to check if my thinking is correct.

Homework Statement

Two blocks sit on top of each other, block 1 (m1) on top of block 2 (m2), on a frictionless surface. There is friction between the two blocks, coefficient of static friction mu. If a horizontal force, F, is applied to the block 2, what is the greatest magnitude of F that can be applied such that block 1 does not slip?

Homework Equations

Sum of forces = ma

The Attempt at a Solution

We mainly need to concern ourselves with the horizontal physics.
Is it valid that I consider the forces on block 2 and the forces on block 1?
Block 2: F - m1 * g * mu = m2 * a
Block 1: m1 * g * mu = m1 * a
Also, F = (m1 + m2)a

So then, I used equations for block 1 and block 2, to solve for F. I got F = g * mu (m1 + m2). Looks like F = (m1 + m2)a isn't very useful.

Is this the right reasoning and the right answer?

Perhaps the weird thing for me is that let's say I use the equation for block 2 and the equation F = (m1 + m2)a. If I substitute in (m1 + m2)a for F, then I can solve for a and get g * mu. If I substitute F / (m1+m2) for a, and solve for F, I get a much more complex expression [m1 * g * mu (m1 + m2)] / (m1 + 2*m2). I wonder if this is the same? It works out dimensionally, but not numerically if I choose random values for m1 and m2. What did I do wrong?

update again: oh oh. no I guess it's right, I did the math wrong. The third way solving for F does give the same answer. Never mind.

so i think what i did is right..

But what about this complexity? Let's say F were great enough for the top block to slip. How would I consider the forces in such a situation, given mu k and mu s?

Thanks.

 

[PhanthomJay]

Hi all, I am studying for an exam and changed a practice problem by putting the force on the bottom block (you'll see). I just wanted to check if my thinking is correct.

Homework Statement

Two blocks sit on top of each other, block 1 (m1) on top of block 2 (m2), on a frictionless surface. There is friction between the two blocks, coefficient of static friction mu. If a horizontal force, F, is applied to the block 2, what is the greatest magnitude of F that can be applied such that block 1 does not slip?

Homework Equations

Sum of forces = ma

The Attempt at a Solution

We mainly need to concern ourselves with the horizontal physics.
Is it valid that I consider the forces on block 2 and the forces on block 1?
Block 2: F - m1 * g * mu = m2 * a
Block 1: m1 * g * mu = m1 * a
Also, F = (m1 + m2)a

So then, I used equations for block 1 and block 2, to solve for F. I got F = g * mu (m1 + m2). Looks like F = (m1 + m2)a isn't very useful.

Is this the right reasoning and the right answer?

Perhaps the weird thing for me is that let's say I use the equation for block 2 and the equation F = (m1 + m2)a. If I substitute in (m1 + m2)a for F, then I can solve for a and get g * mu. If I substitute F / (m1+m2) for a, and solve for F, I get a much more complex expression [m1 * g * mu (m1 + m2)] / (m1 + 2*m2). I wonder if this is the same? It works out dimensionally, but not numerically if I choose random values for m1 and m2. What did I do wrong?

update again: oh oh. no I guess it's right, I did the math wrong. The third way solving for F does give the same answer. Never mind.

so i think what i did is right..

Yes, good. The extra equation serves as a good check that you did your work correctly when using the other 2 equations.

But what about this complexity? Let's say F were great enough for the top block to slip. How would I consider the forces in such a situation, given mu k and mu s?

Thanks.

In that case, the top block slips, so it will have a different acceleration with respect to the ground than the bottom block's acceleration with respect to the ground. The bottom block's acceleration with respect to the ground will depend on what the value of F is...the top block's acceleration with respect to the ground is independent of the value of F. What value would you use for friction, mu k or mu s?
Welcome to PF, zhaos!

 

[zhaos]

Hi PhantomJay

Thanks for your reply. Since the block is slipping we would use mu k.
I think the situation is now..
F - muk * m1 g = m2 * a [for block 2]
and
muk * m1 g = m1 * a' [for block 1]
(a' not the same as a)
Ah so I see what you mean about a' being independent of F. Thanks.

 

[PhanthomJay]

Yes, that looks very good.

 

[rwisz]

I am happy to see that you are actively thinking and problem-solving for your exam. Your reasoning and approach to the problem seem correct. However, I would like to clarify a few things.

Firstly, when dealing with friction, it is important to consider the maximum static friction force before considering the kinetic friction force. This is because the block will not slip until the static friction force is exceeded. Therefore, in your equations, it would be more accurate to use the maximum static friction force (mu_s) rather than the coefficient of static friction (mu).

Secondly, when substituting F for (m1+m2)a, you should get F = (m1+m2)g*mu_s, which is the same as your first answer. The more complex expression you got is incorrect, as it does not take into account the maximum static friction force.

Lastly, if the force F were great enough for the top block to slip, then the equation would be F - (m1+m2)g*mu_k = (m1+m2)a, where mu_k is the coefficient of kinetic friction. This is because once the block starts slipping, the kinetic friction force comes into play. So, in this case, the maximum force F would be (m1+m2)g*mu_k.

I hope this helps clarify any confusion and good luck on your exam!