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How to calculate the angle using work, displacement and 2 forces?

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Given:
    Work = 12000 J
    Weight: 800 N
    displacement: 200 m
    Force: 120 N
    angle = ?

    2. Relevant equations
    W = Fd

    W = Fd cos θ

    cos -1 (W/Fd) = θ


    3. The attempt at a solution

    First I added the 2 forces together.

    800 + 120 = 920 N

    calculated work; 200 m * 920 N = 184000 J

    then used the second formula;

    cos -1 (12000 J /184000 J) = 86.3°

    but the answer in my book says the answer is supposed to be 60°....

    help me please.
     
  2. jcsd
  3. Feb 10, 2014 #2
    You're going to have to clarify the situation. What is going on in the problem? A block sliding, it looks like? Also, what do mean by the angle? The angle between what?
     
  4. Feb 10, 2014 #3
    The question says;

    a person uses a cord to pull a boat with a mass of 1000 kg by 50.0 m along the board walk. The cord makes a 45 degrees angle with the horizontal force. If a force of 40 N is applied to the boat, how much work is realized?
     
  5. Feb 10, 2014 #4
    I think you have the wrong problem -- in that one you're given the angle, but in this one you're trying to find it.
     
  6. Feb 10, 2014 #5
    You're right, I got them mixed up.

    The real one says,

    12000 J of work is required to pull a wagon that weighs 800 N, in a distance of 200 m. If the applied force over the cord is 120 N, what is the angle?

    Sorry for that.
     
  7. Feb 10, 2014 #6
    No problem.

    You have the same problem as before -- adding the forces to find the work, when they need to be done separately. So first find [itex]W_{weight}[/itex] and then [itex]W_{app}[/itex].
     
  8. Feb 10, 2014 #7

    BvU

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    @Jacko: Please stop trying to lift boats and wagons. They stay level. No Wweight.
     
  9. Feb 10, 2014 #8
    The point was to figure out that the work done by weight is zero.
     
  10. Feb 10, 2014 #9

    BvU

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    Fixo: You're doing the right thing. W is work in your second relevant formula, not Weight. Don't let the weight or Jacko distract you. What is W/Fd ? right, no need for a calculator!

    The first formula doesn't apply: F isn't in the direction of d. That's where the cos comes in: to calculate the component of F that is in the direction of d.
     
  11. Feb 11, 2014 #10
    I would have never figured that out. But with this new found information I will try to do it again.
     
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