- #1

zhaos

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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.

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?

Sum of forces = ma

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.

## 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.

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