Help with acceleration -- A block on top of a block on a frictionless table

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  • #1
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Homework Statement:

Block A rests on top of block B . The table is frictionless but there is friction (a horizontal force) between blocks A and B. Block Bhas mass 6.00 kg and block A has mass 2.00 kg. If the horizontal pull applied to block B equals 12.0 N, then block B has an acceleration of 1.80 m/s. What is the acceleration of block A?

Relevant Equations:

F=ma
I don't understand the problem. Does not block A and B make a system, so they should have the same velocity and acceleration at all time? If not, why do they have different accelerations? I don't understand this part of the problem either: "pull applied to block B equals 12.0 N, then block B has an acceleration of 1.80 m/s." How do they find this number? And if you apply force to block B, do you not apply force on block A since they lay on top of each other? Sorry, for so many question, I just really want to understand this problem.
 

Answers and Replies

  • #2
TSny
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I don't understand the problem. Does not block A and B make a system, so they should have the same velocity and acceleration at all time?
The two blocks make a "system", but that doesn't mean they have to move together. Think about pulling a tablecloth out from under a set of dishes.

Draw a free-body diagram for each block separately.
 
  • #3
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I have tried but I still dont get it. Maybe I'm doing it wrong? So box B is affected by the weight of box A equal to ##m_A*g## and the normal force from the ground ##m_B*g## and then the pull equal to 12 N.

Box A is only affected by the friction force going in opposite direction of the movement, and by the normalforce from B equal to ##m_A*g##. But I still understand how to calculate the acceleration of A.
 
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  • #4
berkeman
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It would help for you to post a FBD for this, but even without that, it seems like there are two cases to solve for depending on the coefficients of friction between the blocks, no? If they slip that would be one solution, and if they stick that would be another solution, right?
 
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Is this all the problem tells you? It seems almost certain that the problem would also mention a coefficient of friction (probably both static and sliding?)
 
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  • #6
TSny
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Is this all the problem tells you? It seems almost certain that the problem would also mention a coefficient of friction (probably both static and sliding?)
The coefficients of friction are not needed for this problem. From the information given, it is not hard to determine whether or not A slips on B.
 
  • #7
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The coefficients of friction are not needed for this problem. From the information given, it is not hard to determine whether or not A slips on B.
Hmmm ... I’m pretty sure you do need the coefficients. Perhaps we’ll discuss more after the OP takes a swing.
 
  • #8
TSny
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I have tried but I still dont get it. Maybe I'm doing it wrong? So box B is affected by the weight of box A equal to ##m_A*g## and the normal force from the ground ##m_B*g## and then the pull equal to 12 N.
You have to be careful when thinking about the vertical forces acting on B. The weight of box A (##m_A g##) is the gravitational force that the earth exerts on box A. This force does not act on box B. However, there is a normal force that box A exerts on box B. It might turn out that the magnitude of this normal force happens to equal the magnitude of the weight of box A. But, conceptually, you should not think of the weight of box A as a force that acts on box B. Maybe you are actually clear on this point, but from the way in which you phrased it, I am a little concerned.

The important thing for this problem is to make sure that your free-body diagram for B has the correct horizontal forces. If these are depicted correctly, then you should have no trouble finding the value of the friction force that acts between the surfaces of A and B. You won't need to use the vertical forces. However, it is a good exercise to make sure that you can display the vertical forces correctly on each block.

Box A is only affected by the friction force going in opposite direction of the movement,
Think carefully about the direction of the friction force on A. Don't fall into the trap of always thinking that friction acts in the opposite direction of the movement.

and by the normal force from B equal to ##m_A*g##.
Good. The normal force from B acting on A does have the magnitude ##m_A g##. This follows from a correct drawing of the free-body diagram of A.

But I still understand how to calculate the acceleration of A.
You should be able to determine the acceleration of A from your free-body diagram for A after you find the friction force from the free-body diagram for B.
 
  • #9
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The coefficients of friction are not needed for this problem. From the information given, it is not hard to determine whether or not A slips on B.
Oh, I get it. Never mind. Sorry.
 

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