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Homework Help: Acceleration of Cylinders

  1. Sep 24, 2010 #1
    [PLAIN]http://img713.imageshack.us/img713/183/334a.jpg [Broken]

    I seem to be having trouble figuring out exactly what forces are going on in this scenario. I know that if the cylinders are to remain in contact, they all have to be accelerating at the same rate, which means the same net horizontal force must be applied to all the cylinders since they are identical.
    What I can figure out is, is that F/3m=a.
    The forces on the bottom left cylinder are the horizontal force to the right, gravity, the normal force from the ground, the normal? force from the right cylinder, mg/2? from the top cylinder, and a force to the left from the top cylinder? This is what I am having trouble with...
    The forces on the top cylinder are mg, the normal forces from the bottom two cylinders, which must be mg/2 for each, a horizontal force coming from the bottom left ball, and a horizontal force to the left coming from the right ball?
    For the bottom right, well mg, a force from the left ball, the normal force from the ground, a force to the right from the top ball, and mg/2 from the top ball.

    As you can see I am having trouble getting my forces right, so I can't begin to calculate anything. Help please :)

    [PLAIN]http://img713.imageshack.us/img713/4729/cylinders.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 25, 2010 #2

    ehild

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    Draw all forces acting on each cylinder, separately. Note that only its own gravity acts on a selected cylinder - there is no term as mg/2. The weight of the upper cylinder is included in the normal force it exerts on a bottom one. And do not forget that the normal force is normal to the tangent plane at the contact.

    Show your work.

    ehild
     
  4. Sep 25, 2010 #3
    But doesn't the weight of the top cylinder act on the bottom cylinders? I put mg/2 because since it is static, there must be a force mg coming from the bottom cylinders.
     
  5. Sep 25, 2010 #4
    [PLAIN]http://img713.imageshack.us/img713/4729/cylinders.jpg [Broken]

    That's the diagram I have now. I thought the question marks would be mg/2 because the top cylinder is resting on the two cylinders. But what is wrong with my diagram?
     
    Last edited by a moderator: May 4, 2017
  6. Sep 25, 2010 #5
    There's something wrong in the direction of some forces....
    You should realize that in a real frictionless world the only direction you can apply a force to a surface is ..... ?

    I think it might help you playing with some real object.
    We don't have real frictionless cylinders, but we can play with some coins on a table.
    Take 4-5 coins of the same type. Try to push a coin by pushing another coins before the first.
    Make a train of 3 coins and try pushing the first to a desired point on a table.
    Try to get a feel of how the first coin can be driven.

    If you want to push a coin from A to B, staying on the AB line, where you have to put the pushing coin behind ?
    Is there a relation betwen the position of the pushhing coin and the line AB ?
    Play, have fun, it's not wasted time.
     
  7. Sep 26, 2010 #6

    ehild

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    What does Newton's third law say about the forces between interacting bodies?

    ehild
     
    Last edited by a moderator: May 4, 2017
  8. Sep 26, 2010 #7
    There is an equal and opposite force. I thought I got them all?
     
  9. Sep 26, 2010 #8

    ehild

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    Look at your figure. The left bottom cylinder acts with N1 normal force at the upper one. What about direction and magnitude of the force the upper cylinder exerts on the bottom one? Should not it be equal in magnitude and of opposite direction as N1?


    ehild
     
  10. Sep 26, 2010 #9
    Yes it should. I think I fixed it up. Right now I have,
    from the bottom left cylinder: N3=mg+N1y
    F3-N1x-N4=ma.

    From the top cylinder: mg=N1y+N2y
    N1x-N2x=ma

    From the bottom right cylinder: N3=mg+N2y
    N4+N2x=ma.

    I have solved some equations, and found
    N1y=N2y=N3/3=mg/2
    2N2x=N1x-N4
    F3=3(N2x+N4)

    I think I have to find N1x and N2x...I think the angle N1 and N2 make is 45o, so I think N1=N1y/cosx=mg/2cos(45), likewise N2=mg/2cos(45), but then this would make N1x=N2x which wouldn't make sense because then the top cylinder wouldn't move.

    I believe I have too many unknowns for the amount of equations I have to solve for acceleration though. Is this rightish though so far?
     
    Last edited: Sep 26, 2010
  11. Sep 26, 2010 #10

    ehild

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    Write all equations for the x components and separately, for the y components. Take care if the signs and notations. Choose the direction of the applied force the positive x direction.

    If you add up all tree equations for the x components, the normal forces cancell and you get F(applied)=3ma.

    You have to exploit the fact that the N-s can not be negative. The cylinders can only push each others and the ground. This provides the conditions for F(applied) so the cylinders remain together.


    ehild
     
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