2 blocks, frictions, and no numbers

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Homework Help Overview

The discussion revolves around a physics problem involving two blocks, Block A and Block B, with a focus on static friction and forces acting on the blocks. The original poster is attempting to determine the force required to initiate sliding between the two blocks, given various options related to their masses and the coefficient of static friction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the force needed for Block A to slide out from under Block B using equations related to friction and forces. Participants question the normal force acting on Block A and discuss the implications of Block B's weight on the normal force.

Discussion Status

Participants are actively engaging in clarifying concepts related to normal force and friction. Some have provided insights into the relationship between the normal force and the frictional force, while others are still grappling with the implications of these relationships in the context of the problem.

Contextual Notes

There is some confusion regarding the normal forces acting on each block and how they relate to the frictional forces. The original poster expresses uncertainty about the correct approach and the implications of the equations they have derived.

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Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is initially at rest on the horizontal top surface of A. The coefficient of static friction between the two blocks is μs. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:

A. mAg
B. mBg
C. μsmAg
D. μsmBg
E. μs(mA+mB)g

I came up with a couple of equations (that may not be right).

f = μmBg
f = F - mAaA


So I tried substitution and solving for F, but I can't get the right answer (E).
So what's the right way to do this problem?
 
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The equation for frictional force is the coefficient of friction times the normal force. What is the normal force on block A?
 
It would be mBg, I guess.
 
Really? Draw a free body diagram.

Don't forget that block B is sitting on top of block A.
 
I did.

I have force F acting on mA.
mAg is directed down
mBg is directed down, and then up as a normal force
friction is in opposite direction of the 2 blocks

What am I doing wrong here? =(
 
The problem is that block A has block B on top of it. That means that the weight of block A is (Ma + Mb)g. What that means is that since block A is not accelerating down, it's normal force must be the same as the weight. Now do you understand?
 
Ok, I see now. So E says that F = μNA
But what does that mean? I thought friction is supposed to be the same for both blocks in this problem. This is getting more confusing.
 
The coefficient of friction is the same. The frictional force depends upon the normal force though. Because A has a greater normal force than B, it would produce a greater frictional force.
 
I see. Thanks for the explanations. Now I really need to study hard for that test.
 
  • #10
No problem :) Glad I could help. Good luck!
 

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