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Two blocks on top of each other

  1. Sep 27, 2013 #1
    Hi guys, it's my first time posting here so please forgive me for anything wrong!

    1. The problem statement, all variables and given/known data

    A block of mass m = 0.570 kg rests on top of a block of mass M = 2.30 kg. A string
    attached to the block of mass M is pulled so that its tension is T = 6.20 N at a 20 degree angle to
    the horizontal. The blocks move together. The coefficient of static friction at
    the surface between the blocks is μs = 0.43; there is no friction at the surface between
    block M and the floor.

    a) What is the direction of the frictional force being exerted on the top block of mass m?
    Justify your answer.

    b) The tension T is now increased - what is the maximum tension, Tmax, with which the string can be pulled such that the blocks continue to move together (i.e. that the block of
    mass m does not start to slide on top of the block of mass M)?

    2. Relevant equations

    Fnet = ma
    Fsmax ≤ μFn

    3. The attempt at a solution

    a) (Just getting some clarifications on this one to see if I did it right.)
    I wrote that friction force should be directed to the left, since static friction (and kinetic friction) always opposes the direction of motion/acceleration.

    b) So I'm just trying to draw FBDs for the two blocks and I'm getting confused just doing it. I can't even get the answer if I can't get the FBDs right, so please help me :(

    FBDs:
    Top block (m):
    ƩFx: -Fs = ma
    ƩFy: Fn = mg

    Bottom block (M):
    ƩFx: Tcos20 - Fs = Ma
    ƩFy: Fn + Tsin20 = Mg

    I'm assuming that both blocks have a normal force on them, but I'm not so sure because I'm getting very confused between normal force and contact force... (Newton's Third Law)

    And I'm not sure about the frictional force between the two blocks in the FBDs.. Since it's acting on both blocks (because the friction is between the two blocks), it is included in both of the blocks' FBDs.. right?
     
  2. jcsd
  3. Sep 27, 2013 #2
    Let's tackle a) first. If the static friction opposed the motion of the top block, then what force set it in motion in the first place?
     
  4. Sep 27, 2013 #3
    The tension in the string? I split the tension T into its components, Tx and Ty ... and I believe this force only acts on the bottom block, not the top.
     
  5. Sep 27, 2013 #4
    If that is so, what set the top block in motion?
     
  6. Sep 27, 2013 #5
    The acceleration, since the acceleration is the same for the two blocks.
     
  7. Sep 27, 2013 #6
    I don't understand what you're trying to get at, voko.. >< I'm just trying to get the FBDs..
     
  8. Sep 27, 2013 #7
    Acceleration is a consequence, it is not the cause. When an object begins to move, it will be accelerating just because it begins to move. But what makes an object move, or, more correctly, accelerate?

    A force is required to accelerate an object. What forces act on the top block? Which of them makes it accelerate and stay together with the bottom block??
     
  9. Sep 27, 2013 #8
    I am on this same problem. So let me see if i am on the right track.

    Since the bottom block is pulled by the string it will move, and in order for the top of the mass to move, the top mass will have to "stick" with the bottom mass. Therefore the force that makes the top mass stay still and move with bottom mass involves the static friction. Am i correct on this voko?

    If yes, then would the forces that makes the top block accelerate are the Ty(tension in the y-axis) + Fs or Ty-Fs?
     
    Last edited: Sep 27, 2013
  10. Sep 27, 2013 #9
    Yes.

    How can the force of tension accelerate the top block if it is not applied to it?
     
  11. Sep 27, 2013 #10
    so the static friction is the force that makes the top move with the bottom block. Then is the friction force pointing to the right? Sorry I got the concept but still confuse which direction.

    Also how do you suggest I tackle on the second question?
     
  12. Sep 27, 2013 #11
    According to the rules of the forum, you have to show an attempt.
     
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