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Pulley problem

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    The mass of the two objects on both sides if the pulley is given. So is the acceleration to start moving. It's asking to find the applied force. How do I do that?
    Mass of 1- 4.5kg
    Mass if 2- 3.0 kg
    How much force must u exert to get M1 to accelerate at .25m/s2
    2. Relevant equations
    Fnet= ma
    -F1g+ f2g= (m1+m2)a


    3. The attempt at a solution
    I tried putting Fa into the fnet equation and solving for Fa but. It's not working??
     
    Last edited: Sep 27, 2013
  2. jcsd
  3. Sep 27, 2013 #2
    4.5(a) = 4.5(9.8) - 3(9.8) - F
    if a = .25

    F = 13.575 Newtons added to mass 2 in order to slow the decent of mass 1 to .25m/s^2. Assuming frictionless pulley.
     
  4. Sep 27, 2013 #3
    The answer is supposed to be 17N
     
  5. Sep 27, 2013 #4
    It's just asking the applied force needed to open the window which is M1
     
  6. Sep 27, 2013 #5
    The other person who replied said that you need to add what they got (13.575) to M2, which is 3
    Rounding that gives you 17
     
  7. Sep 27, 2013 #6
    -F
    -3(9.8) + T = 3a
    4.5(9.8) - T = 4.5a
    -F -3(9.8) +4.5(9.8) = 7.5a

    solve for F the force.
    12.825 Newtons
     
  8. Sep 27, 2013 #7

    PhanthomJay

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    Solutions should be attempted by the requestor, not the responder.
    Draw free body diagrams of each mass, and apply Newton's Laws to each mass.
    The book answer looks correct.
     
  9. Sep 27, 2013 #8
    Just wondering why is the force F- and why the 4.5(9.8) isn't negative since that's on the left side of the pulley?? The applied force is going up the pulley on the left so..
     
  10. Sep 27, 2013 #9

    PhanthomJay

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    Ok so m1 is the 4.5 Kg window on the left, its weight acts down on it, you apply a force F upward on it to open it, and together with the cord tension acting up on it, it accelerates up at 0.25 m/s^2. Apply Newton's 2nd law, and continue on , next identifying the net force acting on m2 which accelerates m2 downward at 0.25 m/s^2.....
     
  11. Sep 28, 2013 #10
    Ok I understand it now however why do u have to add the mass to the 12.825N ? How can u add mass to a N force?
     
  12. Sep 28, 2013 #11

    PhanthomJay

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    You can't. You should attempt the problem yourself rather than follow someone elses work.

    Let's look again at the window, which has a mass m1 = 4.5 kg. Draw a free body diagram. The windows weight, m1(g) acts down. You apply a Force F, up. The tension, T, in the cord acts up (tension forces always pull away from the objects on which they act).

    Now apply Newton's 2nd law, F_net = ma, thus

    F + T -(m1)g = m1(a)

    That is your first equation. You know m1, g, and a, but you still have 2 unknowns, F and T, so you need a second equation. You can get that second equation by drawing a free body diagram of m2, the 3 kg mass, and identifying the forces acting on it, and applying Newton's 2nd law. What do you get?
     
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