- #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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