1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work problem

  1. Oct 17, 2005 #1
    A 121-kg crate is being pulled across a horizontal floor by a force P that makes an angle of 36.8° above the horizontal. The coefficient of kinetic friction is 0.212. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

    the work of gravity and the normal force are going to be 0 because they are 90 degrees from the distance that the box is travelling. should the work of kinetic friction be 0 because it says "kinetic frictional force is zero"? that would mean that the net work is equal to the force of P * distance * cos 39.9 and we are looking for the force of P, but we don't know the distance... i'm confused can someone help? thanks
     
  2. jcsd
  3. Oct 17, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The problem doesn't say that "kinetic frictional force is zero"; it says that the net work done by the force P and the kinetic friction is zero. What does that tell you about the net horizontal force on the crate?
     
  4. Oct 17, 2005 #3
    it means that the crate is in equilibrium, so would the net horizontal force be zero?
     
  5. Oct 17, 2005 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Right! Now use that fact to figure out what P must be.
     
  6. Oct 17, 2005 #5
    okay... so i figured that the kinetic frictional force + the horizontal component must equal 0. so.... -(.212)*(121)(9.8) + P*cos(36.8) = 0 and i found out that P = 313.95 N but thats the wrong answer so i'm not sure what i did wrong
     
  7. Oct 17, 2005 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Your calculation assumes that the normal force equals the weight of the crate. Not so. The vertical component of P reduces the normal force.
     
  8. Oct 17, 2005 #7
    do i include the vertical component of force P in with the same equation making... the kinetic frictional force + the horizontal component of force P + the vertical component of force P = 0 ??
     
  9. Oct 17, 2005 #8

    Doc Al

    User Avatar

    Staff: Mentor

    To calculate the friction, you need the normal force. That normal force depends on P. Set up force equations for vertical and horizontal components and solve for P.
     
  10. Oct 17, 2005 #9
    ahh i got it thank you sooo much
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work problem
  1. Work problem (Replies: 2)

  2. Problem on Work (Replies: 3)

Loading...