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

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In summary: You are welcome. Keep up the good work!In summary, the conversation discusses a practice problem involving two blocks on a frictionless surface and the maximum horizontal force that can be applied without causing the top block to slip. The solution involves considering the forces on each block and using equations to solve for the maximum force, with the added equation serving as a check. It is also mentioned that if the top block were to slip, the value of friction used would be mu k.
  • #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.

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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  • #2


zhaos said:
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!
 
  • #3


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.
 
  • #4


Yes, that looks very good.
 
  • #5


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!
 

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

1. What are frictional forces?

Frictional forces are forces that oppose motion between two surfaces that are in contact with each other. They are caused by the microscopic irregularities on the surfaces of the objects and can act in both directions.

2. What is the significance of finding the maximum F for no slippage?

When two objects are in contact with each other and one is moving relative to the other, there is a possibility of slipping or sliding between the two surfaces. By finding the maximum F for no slippage, we can determine the maximum force that can be applied to prevent slipping from occurring.

3. How is the maximum F for no slippage calculated?

The maximum F for no slippage is calculated by using the equation F = µN, where µ is the coefficient of friction and N is the normal force acting between the two surfaces.

4. What factors affect the maximum F for no slippage?

The maximum F for no slippage is affected by the coefficient of friction between the two surfaces, the normal force acting between them, and the roughness of the surfaces.

5. Why is understanding frictional forces important?

Frictional forces play a crucial role in everyday life and have a significant impact on the design and functioning of many objects. Understanding frictional forces can help in improving the efficiency of machines, preventing accidents, and developing new technologies.

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