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Two blocks on top of eachother with pulleys

  1. Jul 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F. The coefficient of static friction between all surfaces is 0.60 and the kinetic coefficient is 0.36.

    a)What is the minimum value of F needed to move the two blocks?
    b)If the force is 10% greater than your answer for (a), what is the acceleration of each block?

    2. Relevant equations

    Sum of forces = ma

    3. The attempt at a solution

    Forces acting on the bottom box are normal force up the y axis, normal force from the top box and mg going down the y axis. On the x axis, the F we need to find is positive and the friction of 1 on 2, the kinetic friction on the ground and the tension of the rope is going the opposite way. For the box on top, there is normal force up y axis, mg down, friction of 2 on 1 going the positive x direction and tension going the other way.

    I know for the top box that n=mg so n =3(9.8) regarding the y axis
    for the x axis:
    F=ma
    Friction of 2 on 1 = u*n = .6(3*9.8) = 17.64 so
    17.64 - T = -3a


    Then for the bottom box I know that :
    n2-mg-n1= 0
    n2 = 5(9.8)+3(9.8)
    so n2=78.4

    for the x direction:
    F-T-kinetic friction-Friction1on2 = ma
    so F-T-28.224-17.64 = 5a

    And now I'm stuck!
     
  2. jcsd
  3. Jul 30, 2009 #2

    alphysicist

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    Hi harrinj4,

    The minimum force that just starts the blocks moving, is also the maximum force that you can apply and not have the blocks move. So what is the acceleration a for the x-equations of your two blocks?
     
  4. Jul 30, 2009 #3
    Zero?

    So from there could I find T, then plug it in the other equation to find F, make acceleration zero again?
     
  5. Jul 30, 2009 #4

    alphysicist

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    That's right; the acceleration is zero for both blocks in part a, so you have two equations with two unknowns (F and T) and so can solve for F.
     
  6. Jul 30, 2009 #5

    See I thought that, but I get 64 as the answer and it's wrong!

    Hmm, see any other things I did wrong? something seems wrong...
     
  7. Jul 30, 2009 #6
    This is due at five! SHITT....
     
  8. Jul 30, 2009 #7

    alphysicist

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    In your equation for the bottom block, you are using kinetic friction for the friction from the ground. However, it is not moving (yet), so this should be static friction.
     
  9. Jul 30, 2009 #8
    ha just figured that out a minute ago and got it right. So now for the second part:


    Take the force and multiply it by 1.1 (adding ten percent), change all the frictions to kinetic frictions, and then???
     
  10. Jul 30, 2009 #9

    Did the work and found tension then put tension in to find acceleration and it worked. Thanks for help before tho!!!
    !!
     
  11. Jul 30, 2009 #10

    alphysicist

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    Perfect! (and I'm glad to help!)
     
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