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Newton's Laws Of Motion

  1. Oct 20, 2006 #1

    app

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    INTRODUCTION: The following is a problem of "Newton's Laws Of Motion". It is a typical problem of this chapter. I am having problems with some parts.
    THE PROBLEM IN BLACK AND WHITE: "Find the acceleration [tex] a_{A}, a_{B} and a_{C} [/tex] of the three blocks shown in the diagram (not included here), if a horizontal force F= 10 N acts on block A. Mass of the blocks A,B and C are 2kg, 5kg and 8kg. The coefficient of friction between the blocks is 0.2 and the table is smooth."
    THE DIAGRAM IN WORDS: There is a frictionless table, on which three blocks are kept one on top of the other. The block A is at the top, followed by B and the block C is at the bottom.
    WHAT I KNOW: In these types of problems we need to do the following: (i) Identify the forces on each of the blocks.
    (ii) Form the equations of motion.
    (iii) Solve them to get the final answer.
    WHAT MY BRAIN SUGGESTED: Let's find the forces on each of the blocks.
    (A) Forces on Block A:
    (i) Its own weight [tex] m_{A}g [/tex] acting vertically downwards.
    (ii) The normal reaction force [tex] N_{A} [/tex] acting vertically upwards.
    (iii) The horizontal force F(=10 N) acting towards right.
    (iv) The frictional force between A and B, [tex] f_{AB} [/tex]. It acts towards left on A.
    (B) Forces on Block B:
    (i) Its own weight [tex] m_{B}g [/tex] acting vertically downwards.
    (ii) The normal reaction force [tex] N_{B} [/tex] acting vertically upwards.
    (iii) The normal force due to A, [tex] N_{BA} [/tex] acting downwards.
    (iv) The frictional force between A and B, [tex] f_{AB} [/tex]. It acts towards right on B.
    (v) *The frictional force between B and C, [tex] f_{BC} [/tex]. I dont know about the direction.
    (C) Forces on Block C:
    (i) Its own weight [tex] m_{C}g [/tex] acting vertically downwards.
    (ii) The normal reaction force [tex] N_{C} [/tex] acting vertically upwards.
    (iii) *The normal force acting downwards due to the blocks kept on C. I don't know about how much this force would be. Should it be just due to the block C kept above or would it also include A on top?
    (iv) *The frictional force between B and C, [tex] f_{BC} [/tex]. Once again i'm not sure about the direction. But it would certainly act in a direction opposite to the frictional force on B.
    WHAT I CANNOT UNDERSTAND: I cannot understand the points marked with a star (*) above. I have explained the difficulty in italics beside that particular point.
    WHAT MY BOOK SAYS: In my book it is given that the blocks B and C will move together (i.e. they will have a common acceleration). I wanna know why and also how to identify that those two blocks will move together?
    CONCLUSION: I will be highly grateful if someone kindly explains the points marked with a (*) above. These are common problems in "Newton's Laws Of Motion". So, it would be nice if someone can also give detail reasons to his answers. That would help me solve similar problems in future. Thanks a lot for your help.
     
    Last edited: Oct 20, 2006
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  3. Oct 20, 2006 #2

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    The friction force acts always in the direction against movement. Since the blocks are moving right, the friction forces must all be directed left.
    About the normal force acting on C. Hold an object in your hand. Now put another object over the first. Will your hand sense only the weight of the first object or of both?
     
  4. Oct 20, 2006 #3

    app

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    Oh, ok. I got it. But what about the blocks B and C moving together? why will they move together? I mean why will B and C have the same acceleration?
    but i all the frictional forces will not act towards left. the frictional force on B due to A and on top of C due to B, will be directed towards right (newton's 3rd law), right? But the frictional forces at the bases of A and B will obviously be directed towards left, as u say.
    Anyway, thanks a lot for helping. and please do tell me about the common acceleration of B and C.
     
  5. Oct 20, 2006 #4

    OlderDan

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    You are correct about the friction forces and Newton's third law. If the top block A is pulled to the right, the friction on Block A from block B will be to the left, and the friction on block B from block A will be to the right.

    The problem is telling you that B and C move together is just to help you approach the problem. B and C will move together with the right combination of masses and coefficients of friction, but for some other combination they might not move together. You could have all three blocks moving together, or you could have any two adjacent blocks moving together with the third moving differently, or you could have all three moving differently. You really cannot know for sure which situation is the correct one until you find a solution to the problem that is self consitent.

    Suppose you assume that all blocks move together with a common acceleration. You do the problem and find that acceleration. Then you look at the forces acting on block B and discover that the net force required to give the block that much acceleration exceeds the net force that is acting. That tells you your assumption was wrong, so now you have to look at the problem again assuming block A is sliding relative to block B and B and C move together (as given in this problem). You find an acceleration for A and another acceleration for B and C. Then you should check to see if the forces on block C are sufficient to give block C that acceleration. If not, again you must revise your assumption and do the problem assuming all blocks move separately.

    The difficulty is that there is no formula for the frictional force that acts when the blocks are not slipping. When there is no slipping, all you know is the maximum possible friction force based on the coefficient of static friction and the normal force. You cannot assume the actual force that is acting is this maximum force.
     
    Last edited: Oct 20, 2006
  6. Oct 21, 2006 #5

    app

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    The problem does not state that the two blocks, B and C move together (i.e. that they have the same acceleration). It was just a hint given at the back of the book. So, u say that there is no direct method to identify which blocks will move together, or whether they all wud move with different accelerations or not?
    Dont u think that instead of taking their accelerations to be same and then change them according to the forces(as u say), it wud be better to assume all the accelerations to be different, and then if some of them share the same acceleration(i.e. move together), then its ok?
    Actually we have to do these types of problems in not more than 1 minute each. We have to attempt 75 questions in 75 minutes. That's why i always have to ask u'll for the shorter mothods for any problem.
    Anyway, thanks a lot for helping.
     
  7. Oct 21, 2006 #6

    OlderDan

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    You can start with whatever assumption you want about the relative motion of the blocks. You still have to check the answer for consistency. If you assume all blocks are moving relative to one another you will be using the maximum possible friction forces in your calculation. There is no guarantee that the maximum possible forces are acting. If your assumption was wrong, your results will be wrong, and inconsistent.

    The most efficient way to solve the problem is to make the right assumption the first time. In this particular problem, the fact that mass A is small compared to the other masses suggests that it will probably slip on block B. In addition, that the table is frictionless, the combined mass of blocks A and B is comparable to the mass of C, and the applied force is small compared to the weight of the blocks is suggestive that block C will move with block B, but all you need to do is make C more massive or add some friction at the table and that will change the problem to one of slipping between B and C.

    I am sorry if somebody thinks you should be able to do a problem like this in 1 minute. It is totally unrealistic. I can only assume that some of the 75 problems you have to do are a lot simpler than this one.
     
    Last edited: Oct 21, 2006
  8. Oct 21, 2006 #7

    app

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    I cant do anythng abt that, thats our exam rules here, its tough competition. And ya, ofcourse, some others out of that 75 are simpler. But if one can do this problem in the least possible time,(i dont mean 1 minute for this problem), then he can save time for the other problems. And it is not expected that one will ever complete 75 in 75 mins. Its as many as u can. our score doesnt depend on how many we can do, it depend on the overall performance. doesnt matter if u get a 90%. the people who r getting more than u will get chance into universities.
    Anyway, thanks for guiding.
     
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