Block on top of a Block

  • Thread starter phil ess
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  • #1
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Homework Statement



Consider two blocks of mass M1 and M2. Mass M1 is resting on top of M2, which rests on a horizontal surface. The coefficient of static friction between the two blocks is u1 and the coefficient of kinetic friction between the bottom block and the ground is u2. A force F is applied to the bottom block at an angle R to the horizontal. The blocks move together in the horizontal direction. Draw a FBD for each block separately.

Untitled.jpg


Homework Equations



None yet.

The Attempt at a Solution



Untitled2.jpg
Untitled3.jpg


This is what I have so far, but I don't understand the frictional forces acting between the blocks very well. I know static friction is keeping m1 on top of m2, but then it would be going to the right to oppose the motion of m1, but what would I balance that force with then? Any help is much appreciated.
 

Answers and Replies

  • #2
LowlyPion
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Looks to me like you need to resolve the pulling force (which you have labeled as kinetic force) into its components x,y as a function of its angle R and properly reflect them in your drawings..
 
  • #3
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Actually the Fkinetic is meant to represent the frictional force of the bottom block trying to slip under the top block, which is countered by Fstatic between the two blocks, though I'm not sure that makes sense. Also, you're saying that I need a force in the y direction representing the y-component of the applied force being transferred to the top block from the bottom?
 
  • #4
LowlyPion
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Actually the Fkinetic is meant to represent the frictional force of the bottom block trying to slip under the top block, which is countered by Fstatic between the two blocks, though I'm not sure that makes sense. Also, you're saying that I need a force in the y direction representing the y-component of the applied force being transferred to the top block from the bottom?
Be a little more careful then. The top block only has the backward inertial force balanced by the forward effect of friction holding the block on top of the second block. Otherwise, it would be in motion with respect to the bottom block.

Your Normal force from M2 and M1*g are both at the bottom surface in opposite directions.

On the bottom block the frictional force opposes the forward motive force. Moreover there are 2 components of the motive force. One horizontal as a function of R and one up as a function of R that opposes the combined (M1+M2)*g down. The force normal from the ground makes up the difference.
 
  • #5
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Ok if I'm understanding this correctly then this is what I have for the bottom block:

Untitled3-1.jpg


With F kf (force of kinetic friction) opposing the horizontal component of the applied force, and (m1+m2)g, along with Fn, opposing the remaining vertical component?


Then for the top block:

Untitled2-1.jpg


You said that there is an inertial force trying to push the block left relative to the bottom block, which is being countered by the force of static friction? I'm still bothered by this FBD for the second block because it's accelerating via the bottom block, and I don't have an unbalanced force, so I must be missing something.
 
  • #6
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Ok I have figured it out! There was a good explanation of this kind of situation that I found in one of my textbooks. Thanks for the help Pion.
 

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