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What force does Pete exert on the rope?

  1. Jan 10, 2005 #1
    How would I calculate this
    A horizontal force of 809 N is needed to drag a crate across a horizontal floor with a constant speed. Pete drags the crate using a rope held at an angle of 33°. What force does Pete exert on the rope?
     
  2. jcsd
  3. Jan 10, 2005 #2
    How about show some working, so we can help you on where you're stuck?

    Remember this forum does not do your homework :smile:
     
  4. Jan 11, 2005 #3
    I was wondering what the formula was?
     
  5. Jan 11, 2005 #4
    i think u missed some ideas in ur homework. pls clarify it. i think it is work.
     
  6. Jan 11, 2005 #5
    Use a bit of trigonometry. I'm sure you've learned trigonometry in class.
     
  7. Jan 11, 2005 #6
    Just think of it as a triangle.

    The floor is one length (A) and the rope it the other(H), in this case we make pete the other(O). since we have the angle and we want the length of H we use the following formula. sin33=A/H therefore A/sin33=H just carry on from here.
     
  8. Jan 11, 2005 #7
    Draw a diagram of th situation. Draw the forces acting on the crate. in this case, since it's moving at constant speed, there is no acceleration.

    since the rope is at 33 degrees, it would be a right angled triangle with the adjacent of the 33-degree angle being of magnitude 809

    from there, u can use the cosine formula to calculate the hypotenuse of the triangle, which will give u the force exerted by Pete!
     
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