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Find the acceleration of blocks - Friction problem

  1. Jul 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Refer to diagram. Find acceleration of each block:
    n1,n2, n3 are coefficients of friction.

    if a force of 10 N acts on the middle block, find the acceleration of each.

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    a1,a2,a3 are accelerations of bodies, taken from top to bottom.
    Fr1= Friction force for n1
    Fr2= Friction force for n2
    fr1+fr2=10
    7*a1=fr2
    2*a2=fr1
    7*a1+2*a2=10
    the middle block will be stationary with respect to ground.

    I cant understand where am I going wrong?????
     

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    Last edited: Jul 30, 2008
  2. jcsd
  3. Jul 30, 2008 #2

    tiny-tim

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    Hi ritwik06! :smile:

    I'm not really following your equations … your a's seem to be attached to the wrong ones.

    Two points:

    i] it helps if you state clearly which body (or combination of bodies) you are examining.

    so, for example, if you're examining all three together, then the only force you need is the (only) external force, but you must include all three mass x accelerations.

    ii] for the top or bottom block, the acceleration in the equation will be relative acceleration. :smile:
     
  4. Jul 30, 2008 #3
    Re: Friction

    from the top, I count the bodies as 1,2,3. fr1 is the friction acting beteen body 1 and body 2. fr2 is friction between body 2 and body 3.

    when body 2 tends to move ue to the 10 N force being applied on it. Body 1 and body 3 apply a frictional force to oppose its motion. The maximum value of the combined friction can be (.2*20+50*.3)=19 N. Thus only 10 N will be applied.
    fr1+fr2=10N
    Body 2 will remain at rest.
    A reaction of fr1 will act on body causing it to accelerate in opposite direction.
    2*a1=fr1
    similarly reaction of fr2 will act on the bottom body.
    fr2=7*a3
    a2=0

    are these things correct?
     
  5. Jul 30, 2008 #4

    tiny-tim

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    Hi ritwik06! :smile:

    Sorry, still not understanding that.

    You have three unknowns, a1 a2 and a3.

    The three blocks as a whole will slide with a total m1a1 + m2a2 + m3a3 which depends on the applied force.

    The middle block presumably accelerates fastest, and tries to pull the other blocks with it, but it can only impart the appropriate friction force on each, which gives you the relative accelerations, (a1 - a2) and (a3 - a2).

    So that give you three equations for three unknowns, which is ok. :smile:
     
  6. Jul 31, 2008 #5
    Re: Friction

    Ok, :smile:
    I will try to explain what I mean to say. I want to do this by using a free body diagram.
    See the attached free body diagrams carefully. The normal reaction and the weights acting downwards are not shown as the blocks are in equilibrium along the vertical axis.
    only the horizontal forces are shown.
    Fr means friction
    The maximum value of Fr1 can be = 0.2*20 =4N
    The maximum value of Fr2 can be = 0.3*(20+30) =15N

    As only a force of 10 N is applied on Block 2:
    it will not move because a frictional force of 10N will act in the opposite direction.
    Fr1+Fr2=10 N
    a2=0;

    For Body 1
    2*a1=Fr1

    For Body 3

    7*a3=Fr2

    2*a1+7*a3=Fr1+Fr2=10

    Now whats the mistake in this process?
     

    Attached Files:

  7. Jul 31, 2008 #6

    Doc Al

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    Re: Friction

    OK (assuming those are the coefficients of static friction). The actual value of the static friction force can be anything from 0 to the maximum value as required to prevent slipping.

    How did you arrive at that conclusion?

    (That would make sense if blocks 1 and 3 were fixed, but they are free to move.)
     
  8. Aug 1, 2008 #7
    Re: Friction

    Thats exactly what I have assumed. thats why instad of writing th original values of friction in each case I have assumed Fr1 and Fr2 whose sum can safely be said to be 10, isnt it? Can their individual values be determined?
    I do understand that
    but only 3 forces are acting on block B, which add upto 0. so the block will not move. doesnt the FBD show that?

    Please dont go roundabout with the explanation. I really dont have time to stick to one question for days together. I dont ask for the complete solutions. But please give concrete hints. Tell me where I am wrong. Or if u feel , some of my concept is not clear, then do give me a link where I can read about the topic. I will gladly do that. But going on with a trivial problem as this, is unaffordable. Dont u think so?

    Dont feel I am being rude or something. The thing is that I am preparing for JEE. People do a lot of coaching for that. I cannot afford it. Whatever I do, is done by myself from the sources at my disposal. My teachers are of no use. They refuse to discuss topics that does not belong to the school syllabus. How very miserable!!
     
  9. Aug 1, 2008 #8

    Doc Al

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    Re: Friction

    No. Why do you think that?
    Again, why do you think that?

    I've done so about 3 times now. As I said before, if blocks 1 and 3 were fixed (could not move) then the applied force of 10 N would easily be balanced by static friction and no blocks would move: acceleration of all blocks would equal zero. But blocks 1 and 3 are not fixed, they can move.

    What would it take for all three blocks to move together?
     
  10. Aug 1, 2008 #9

    Doc Al

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    Re: Friction

    Another way to think of things that might help you: Realize that the three blocks are resting on a frictionless table. Exert any kind of horizontal force on them and something must surely move!
     
  11. Aug 1, 2008 #10
    Re: Friction

    The friction forces will try to stop the middle block. So th net force they apply will definitely be 10N. If it is anything greater, the middle block will accelerate, this is not possible. the total magnitude of friction will never exceed 10 N.

    right. i agree. blocks 1 and 3 will move. But I am unable to find thier acceleration. Only block 2 will not move. as the net force on it is 0. on block 1 and 3, th reaction of frictional force will make them move.

    A general question. What general approach should I follow in these types of questions? FBD's dont seem to work here???
     
  12. Aug 1, 2008 #11

    tiny-tim

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    Oh, ritwik06, ritwik06, ritwik06! :smile:

    You've got this weird idea in your head, and you won't let go of it.

    The middle block will accelerate.

    FBD's do work here.

    Stop trying to use reason, as if you were some ancient Greek philosopher.

    This is physics … it's all equations.

    In this case, good ol' Newton's second law.
    hmm … don't think I'm being rude :wink:, but the reason I gave up several posts ago is because you weren't following concrete hints.

    As I said before:
    Just draw the FBD's, write the equations out (Newton's second law), and solve them … and stop reasoning! :smile:

    Start with the FBD for the three blocks as-a-whole. :smile:
     
  13. Aug 1, 2008 #12
    Re: Friction

    Correct me if I am wrong guys.

    Wouldn't you treat this problem into different cases? I would split this problem into 3 different cases; A system as a whole, the first and second block as one object and the second and third block as one object.

    From what I see, the second and third block would have the same acceleration and the first block would be slightly less (due to it sliding). (First block being the top)

    Again, this is just me thinking out loud. I only did some rough quick calculations. Point out any errors in my logic.
     
  14. Aug 1, 2008 #13

    Doc Al

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    Re: Friction

    Friction forces will act to prevent slipping between surfaces.
    Nope. Again, you are confusing this with a situation (for example) where you have a block resting on a table with friction. If the maximum static friction between block and table is 15N, then a 10N applied force will produce a 10N static friction and the block won't move. The key point is that the table doesn't move. (But blocks can move!)

    Maybe solving a problem with only two blocks will make it clearer. Imagine just two blocks on top of each other on a frictionless table. Say the maximum static friction between the two blocks is 15 N (perhaps the top block has a mass of 3 kg and the coefficient of friction is 0.5). If you push the bottom block with a force of 10 N, what happens? (Say the bottom block is also 3 kg.)

    Hint: The blocks will only start to separate (slip) if the required force between them is greater than static friction can provide.

    So you think that you can push on the middle block and have it remain at rest while the other blocks move? I'd pay cash money to see that trick! :smile:

    FBD's work just fine. Your problem is assuming (from experience with different situations) that the net force on the middle block must be zero.
     
  15. Aug 1, 2008 #14
    Re: Friction

    That seems the best hint .
    We are sure that block B can't move independently(with respect to the system) because the friction forces are enough to oppose 10 N. Hence B will move with the system , ie the system as a whole will move with an acceleration of 10/(2+3+7) =5/6 m/s^2
     
  16. Aug 2, 2008 #15
    Re: Friction

    Yeah, I understand it now. both will move with an accleration of 10/6 ms^-2. right?

    I shall be very grateful if you could correct the FBD I made.
     
  17. Aug 2, 2008 #16

    Doc Al

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    Re: Friction

    Right.
    There's nothing wrong with the FBD that you provided in post #5. It's perfect. You made an error in your analysis when you assumed that Fr1 + Fr2 = 10 N. The key is understanding how friction works.

    The way to analyze these static friction problems is to first consider the situation as if no slipping occurs. What would the acceleration of each block be? Then, using your FBD and Newton's laws, figure out what the friction force must be to support such an acceleration. As long as the needed force is less than (or equal to) the maximum static friction there will be no slipping--the blocks move as one. That's what friction does--it resists slipping between surfaces.
     
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