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Introductory physics force problems

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    1) http://i.imgur.com/JbK13.gif
    The figure below depicts the forces acting on a safe being pushed up an ramp by movers, Fa. The mass of the safe is 500 kg, the coefficient of static friction along the incline is μs = 0.41 and the coefficient of kinetic friction along the incline is μk = 0.34. The ramp forms an angle θ = 23° above the horizontal. The top of the wall is h = 1.0 m off the floor. What force needs to be applied to hold the safe in place?

    2) http://i.imgur.com/ZTYbR.gif
    A young boy with a broken leg is undergoing traction. Find the magnitude of the total force of the traction apparatus applied to the leg, assuming the weight of the leg is 75 N and the weight hanging from the traction apparatus is also 75 N, θ = 28°.


    2. Relevant equations
    None were given to me, but this is what I think they are:
    1)Fs = μs * N
    w = mg

    2) horizontal component of the traction force = 2 * 75 * cos (28) = 132 N


    3. The attempt at a solution

    1) static friction cannot hold the safe in place without having it slide back down

    x components: -mg sin(theta) + Ffr = 0 = mg sin(theta) + uN
    y components: N - mg cos(theta) = 0
    N = mg cos(theta)

    therefore F = -mg sin(theta) + u(mg cos(theta))
    I don't think this is right...

    2) horizontal component of the traction force = 2 * 75 * cos (28) = 132 N
    magnitude of the force exerted on the femur by the lower leg = 132 N

    I really appreciate any help you can give me! Thank you so much!
     
    Last edited: Sep 10, 2011
  2. jcsd
  3. Sep 10, 2011 #2

    kuruman

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    Problem 2 is fine.

    For problem 1: Assume that the safe is standing on the ramp without anyone touching it. As you say, static friction is not enough to keep it at rest. OK, but what is the net force acting on it if this were true? That is the force that needs to be applied up the ramp to keep it at rest.
     
  4. Sep 10, 2011 #3
    For problem 2, the horizontal component isn't enough apparently. I don't know what the vertical component is though.

    For problem 1, I have F = mg sin(theta) + u(mg cos(theta)). But I don't think that gets me the right answer. I don't know if I'm setting up my x and y components correctly.
     
  5. Sep 10, 2011 #4

    kuruman

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    The vertical component is such that the leg is at rest. What forces act on the leg in the vertical direction?
    The two force components are in opposite directions; if one is positive, the other must be negative.
     
  6. Sep 10, 2011 #5
    Problem 2. The forces that act on the leg at rest are the weight (mg).

    Problem 1. Would it be F = mg sin(theta) - u(mg cos(theta))?
     
  7. Sep 11, 2011 #6

    kuruman

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    That's only the vertical component exerted by the traction apparatus. To find the total or net force exerted by the apparatus you also need to consider its horizontal component.

    It would be that.
     
  8. Sep 11, 2011 #7
    For problem 2, the horizontal component of the traction force = 2 * 75 * cos (28) = 132 N. Would it just be 132 N + mg?
     
  9. Sep 11, 2011 #8

    kuruman

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    It would not. The two components are at right angles to each other. How do you find the resultant in this case?
     
  10. Sep 11, 2011 #9
    Would it be sqrt (75^2 + 132^2) using the Pythagorean theorem?
     
  11. Sep 11, 2011 #10

    kuruman

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    It would be that.
     
  12. Sep 11, 2011 #11
    Thank you so much! For the first problem would it be:

    F = mg sin(theta) - u(mg cos(theta)) = 500*9.8 sin 23 - 0.41(500*9.8 cos 23) = 65.3 N?
     
  13. Sep 11, 2011 #12

    kuruman

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