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Dynamics: Time taken for Block A to slide on Block B

  1. Feb 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the time it takes for block A to slide 0.5m on block B.
    Mass A=10 kg
    Mass B=35 kg
    I included a screen shot from my online homework.
    2. Relevant equations
    FR=FNμk


    3. The attempt at a solution
    For Block A:
    mAa=T-FRA-WAsin(30)
    T=10a+68.67

    For Block B:
    FRC=(mAg+mBg)μBC
    FRA=mAAB
    mBa=T+FRC-FRA-WAsin(30)-WBsin(30)

    After substituting in all my values:
    35a=T-196.2
    T=35a+196.2

    Setting my Tensions equal to each other:
    10a+68.67=35a+196.2

    magnitude of a=5.1012

    Finding time:
    s-s0=0.5at2+v0t
    s0=0
    v0t=0

    0.5=.5(5.1012)t2

    t=0.443s

    Which is wrong... I can't figure out what exactly I'm doing wrong.
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2016 #2

    Doc Al

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    Staff: Mentor

    (1) Careful with signs: What are the directions of those frictional forces on block B?
    (2) Careful with signs: What is the direction of the acceleration of block B? (Since you use "a" as the magnitude of the acceleration, apply the signs correctly.)
    (3) Why does the weight of A enter into an equation for B?
     
  4. Feb 12, 2016 #3
    Ok, I see what you mean with the friction forces.
    Does the weight of block A not put a force on block B?
     
  5. Feb 12, 2016 #4

    Doc Al

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    Staff: Mentor

    Certainly block A exerts both a normal force and a friction force on block B. But the weight of block A is a force that acts on block A, not B.
     
  6. Feb 12, 2016 #5
    Ok! So my new equation for B is this:
    mBa=T+mBAB+mBBC-mBgsin(30)

    Plugging in values...
    T=35a+68.67

    I still think I messed up, when I set these equations equal to each other, my acceleration ends up being 0 :(
     
  7. Feb 12, 2016 #6

    Doc Al

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    Staff: Mentor

    Note my previous item (2): Take care with the sign of the acceleration. There is a constraint: If block A moves up the incline, how must block B move?
     
  8. Feb 13, 2016 #7
    I agree that block B should move down the incline. I'm not sure which part of my equation you're saying needs the sign change.
     
  9. Feb 13, 2016 #8

    Doc Al

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    Staff: Mentor

    Block A has acceleration of magnitude "a" going up the incline; Block B has the same magnitude of acceleration going down the incline. Adjust your signs accordingly. (Compare the direction of acceleration with the directions of the forces.)
     
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