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Crate being pushed varying force

  1. Oct 11, 2008 #1
    given/known data
    A 170 kg crate hangs from end of rope length L = 15 m. You push horizontally on crate with varying force F to move distance d = 5 m sideways. (a) What magnitude be of F when crate is in final position? During crate's displacement, what are (b) the total work done on it, and (c) the work done by the pull on the crate from the rope?

    Relevant equations

    attempt at a solution
    (a) F-(Tension in x component)/170 = 0 ---> F=1666/tan(cos[tex]^{}-1[/tex](5/15))=589

    (b)0=Wa+Wg+Wt --->? Wa, i am not sure how to calculate as it is a varying force

    (c)Wt= [tex]\Delta[/tex]K=0
    how would this be proved numerically?

    Basically what i need help with is calculating Wa. Any help is appreciated. Thanks.
    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 11, 2008 #2


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    For a) why isn't the component of Force along x = M*g*Sinθ ?

    For b) what do you know about the relationship between potential energy and work.

    For c) Work is given by distance times the force acting along the path of that distance. Since at all points any Force from Tension is at right angles ...
  4. Oct 11, 2008 #3
    for (a) gravity does not have any components. i think you misread the question. ive attached a diagram.

    for (b) since the change in kinetic energy is zero, mgh is equal to the negative of work

    for (c) i dont quite understand

    Attached Files:

  5. Oct 11, 2008 #4
    (a) I can't view the attachment till it's approved, but I'm pretty sure that we can safely say that when the mass is pulled sideways ie. the string is not vertical, then force due to will have an x and y component.

    (b) I agree there.

    (c) I think he's saying that you can calculate the work done in each plane, and because they are at right angles to each other you can use pythagoras to find the work done by the rope (when the rope moves from vertical it has components in both directions).
  6. Oct 11, 2008 #5
    so for a, are the calculations correct? tension will have its corresponding components, yes.
    and for c work due to tension is always 0 then (cos90)

    "why isn't the component of Force along x = M*g*Sinθ" how did you come to that conclusion???

    EDIT: how would one calculate the Wa without setting it to negative work of gravity? it is a varying force
    Last edited: Oct 11, 2008
  7. Oct 11, 2008 #6
    For A, you got 589 right? This is the x component of tension only. You need to calculate the net (pythagoras)

    I'll look at C in a sec

    EDIT: Hang on, sorry, you are right.
  8. Oct 11, 2008 #7
    What are you defining Wa?
    When LowlyPion said why isn't it mgsin(theta) he was simply taking theta to be on the other corner of the triangle.

    Why is the work due to tension always 0?
  9. Oct 11, 2008 #8
    Wa is the work done by the applied force.

    since at any point on the rope it is right angles to displacement cos90=0 so then tcos90=0. the answer in the book is 0 too.

    thats what i think.
  10. Oct 11, 2008 #9


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    That is "Wa is the work done by the applied force ...and its magnitude is the scalar product of that force along the direction of its application.

    Cos90 = 0 => Work from T is 0 You are correct.

    For a) my apologies. I was thinking about another part of the problem and hurriedly proceeded to mislead by writing Sin instead of Tan. Sorry.
  11. Oct 11, 2008 #10
    but since it is a varying force, how would i use the scalar product, it isnt a constant force. should i only apply it to the final position where. what is the formula for Wa? since its the negative of Wg= mghcos180=-(9.8)(170)(0.86)=-1429 J. Wa should be +1429J but how can i calculate it. Thanks for the help.
  12. Oct 11, 2008 #11


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    For b) what is the change in potential energy?
  13. Oct 11, 2008 #12
    mg(df-di)=(9.8)(170)(14.14-15)=-1432.76J ? same as above i already calculated it?
  14. Oct 11, 2008 #13


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    How did you calculate the 14.14?
  15. Oct 11, 2008 #14
    since it gets pushed sideways, the height changes vertically but the length of the rope remains the same. sqrt(15^2-5^2)=14.14
  16. Oct 11, 2008 #15


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    Cos19.47 was what I was expecting to see. .86 is fine.
  17. Oct 11, 2008 #16
    ok so we have the change in potentila enrgy = Wg but what are the calculations or formula for calculating Wa. that is what i need to understand. thnks
  18. Oct 11, 2008 #17


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    It is the change in potential energy.
  19. Oct 11, 2008 #18
    it all makes sense now. thanks for the help.
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