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Homework Help: Varying Force

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A 230kg crate hangs from the end of a 12.0 m rope. You push horizontally on the crate with a varying force F to move it 4.00m to the side.

    What is the magnitude of F when the crate in the final position?

    During the displacement, what are the work done on it, the work done by the weight of the crate, and the work done by the pull on the crate from the rope?

    Knowing that the crate is motionless before and after displacement, use the answers to find the work your force does on the crate.

    Why is the work of your force not equal to the product of the horizontal displacement and the initial magnitude of F?

    2. Relevant equations
    f= ma
    w = fdcos(0)

    3. The attempt at a solution
    Magnitude of F is Fx = Tsin(Θ) = Fp
    Work done by tension on the crate changes with (theta)
    Work by weight = 0 because F is parallel to displacement
    Work = ∫ from 0 to 4 (Tsin(Θ)) dΘ How do i do this? is this correct
    The work of your force, is not equal to the product of horizontal displacement and the initial magnitude of F because F is varying. Also It is the sum of all the F*d from 0 to 4 m
  2. jcsd
  3. Mar 15, 2015 #2

    Suraj M

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    When you are integrating for work, your saying that it's from 0 to 4 which is the displacement but you have a ##d\theta## you can't integrate that, you should either get ##\theta## in terms of the displacements or easier thing to do is to find the angle of displacement for 4 m then integrate from 0 to that angle.
  4. Mar 15, 2015 #3
    Thank you, that makes sense. So ∫ from o to 18.4 degrees
    One extra question, is it right for me to assume the function for force is t(sinθ) ?
  5. Mar 17, 2015 #4

    Suraj M

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    Im not a 100% sure. It should work, but whats your T(tension)? I mean, value. is that constant? I doubt it.
  6. Mar 17, 2015 #5


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    I don't think you calculated the angle correctly. When the crate is displaced 4 m horizontally, the cord length is still 12 m.
    There is more than one force acting on the crate.
    Which of those forces do work?
    What is the total net work done by all forces? (HINT: use work-energy theorem.)
    You don't have to use calculus.
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