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Another Work Problem

  1. Dec 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A 35kg box needs to be lifted to the top of a loading dock, which is also accessibly by ramp. The ramp is 5.0m long and has a vertical height of 1.7m.
    a) What minimum force is required to lift the box straight up onto the loading dock?
    b) What minimum amount of work is required to lift the crate straight up onto the loading dock?
    c) What force is required to push the crate up the ramp such that the amount of work is the same as in b)? Assume no friction.

    [​IMG]

    2. Relevant equations
    E = W
    Ep = mgh
    W = f x d


    3. The attempt at a solution
    I solved all the problems:
    First solved to get the angle of the ramp using sin law. It was 19.88, rounded to 20.
    a) Ep = mgh
    = (35)(9.8)(1.7)
    = 583.1N

    W = f x d
    583.1 = F x 5(sin20)
    F = 340.97N

    b) solved in a (Ep) = 583.1N

    c) W = F x d
    583.1 = F x 5(cos20)
    F = 124.1N


    My question is in parts b) and c) where I used sin/cos. I knew those had to be used because it was at an angle, but only knew which to use because I compared my answers with those given in the answer section of the textbook. So how do I know when to use sin or cos, is one of the lengths of the diagram equal to work or force?

    Thank you!
     
  2. jcsd
  3. Dec 13, 2007 #2

    Doc Al

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    Staff: Mentor

    Realize that 5(sin20) is just the height, which you already know is 1.7 m.

    But for part a, since you're lifting straight up, the force you need to exert is the full weight (since the net force on the box must be zero). So you shouldn't have to calculate anything to find the force in this case.

    This time you made an error. Since your pushing the box up the ramp, the distance you must push equals the length of the ramp, which is 5 m.

    You can also figure out the force by realizing that you have to push the box with a force that just equals the component of the box's weight parallel to the ramp. That component is mg(sin20).

    Solve it both ways and compare.
     
  4. Dec 13, 2007 #3
    The unit for work or energy is joules not newtons.

    In part A you didnt to do 5sin20 because you already knew the vertical distance of 1.7m.

    As for part C, are you sure that is the confirmed answer? It only looks like the x-component of the force vector to push the block up.

    edit: yes you have to take into account the force of gravity bringing the box back down as doc al has said. you should the be able to solve it now.
     
    Last edited: Dec 13, 2007
  5. Dec 13, 2007 #4
    For part C am positive that's the answer, according to the answer section of our textbooks.
     
  6. Dec 13, 2007 #5

    Doc Al

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    Staff: Mentor

    Are you saying that the book gave the answer to part C as 124.1N? If so, the book is wrong. :wink: The correct answer is close to that number though. The important point is that your method for C was incorrect.
     
  7. Dec 13, 2007 #6
    These are the book's answers:
    a) 3.4 x 10^2N
    b) 5.8 x 10^2N
    c) 1.2 x 10^2N
     
  8. Dec 13, 2007 #7

    Doc Al

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    Staff: Mentor

    Good. Note that the book rounds off the answer to 2 digits. But your method for part C is still incorrect. (You're just lucky that cos20 is close enough to 1 that it doesn't matter in this case.)

    Please redo it correctly and confirm that you still get an answer that matches the book.
     
  9. Dec 13, 2007 #8
    Okay, so then:
    a) Ep = mgh
    = (35)(9.8)(1.7)
    = 583.1J

    W = f x d
    583.1 = F x 1.7
    F = 343N

    b) solved in a (Ep) = 583.1J

    c) W = F x d
    583.1 = F x 5(sin20)
    F = 340.98N

    ? Still unsure about C.
     
    Last edited: Dec 13, 2007
  10. Dec 13, 2007 #9

    Doc Al

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    Staff: Mentor

    Force should be in Newtons, not Joules.

    C is still wrong. See my comments earlier.
     
  11. Dec 13, 2007 #10
    W = F x d
    583.1 = F x 5
    F = 116.2N

    F = mg(sin20)
    = (35)(9.8)(sin20)
    = 117.3N


    Is this right?
     
  12. Dec 13, 2007 #11

    Doc Al

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    Staff: Mentor

    Exactly right. The reason that your second answer is slightly different is because you rounded off the angle to 20 degrees.
     
  13. Dec 13, 2007 #12
    Thank you! But one more question:
    How did you know to use F = mg(sin). I understand the force = component part.
    But, when I tried to draw it out.. it doesn't seem to make sense.
    Sin = opposite/hypotenuse .. so sin20.. then the opposite of that would be the Fg (mg).
    So sin20 = fg/h .. h is the force? Then rearranged, wouldn't it be h (force) = mg /sin20?
     
  14. Dec 13, 2007 #13

    Doc Al

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    Staff: Mentor

    Careful with your triangles. Whenever you use a triangle to find the components of a vector, the full vector is the hypotenuse. So you need to make a new triangle with the weight as the hypotenuse (but the angles will be the same as the angles on the ramp). The other sides of the triangle will be parallel and perpendicular to the surface of the ramp.

    Read this: Incline Planes - Force Vector Components

    And this: Inclined Planes
     
  15. Dec 13, 2007 #14
    Oh, okay. I get it. Didn't know that before.
    Thanks again!

    Sorry for all the questions- we have a test tomorrow, and our teacher isn't the best for explaining things fully.
     
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