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Work and Force Problem

  1. Mar 24, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    known date
    1. angle= 29 degrees
    2. frictional force = 48 N
    3. mass of cart = 16 kg
    4. displacement = 22 meters
    5. velocity = constant

    2. Relevant equations
    F=ma
    F(s)=mas
    W=F cos theta (s)


    3. The attempt at a solution

    this problem gave me a headache when i attempted to solve it. first i resolved that since velocity is constant, then a=0 but that would mean that the force = 0 and that would say that the person is not exerting any force on the cart. that's all i have down.

    v=constant
    a=0
    F(s)=mas
    0=0
    whats the force shes exerting then? i know how to figure out work, thats easy. just use the work formula.
     
  2. jcsd
  3. Mar 24, 2007 #2
    so you know a). for b),c) and d) just draw a FBD for each.
    hint: d) should be just mg.
     
  4. Mar 24, 2007 #3
    i dont know a! my guess is a = 48 N but my textbook says 54.9. how the heck did it get that answer?
     
  5. Mar 24, 2007 #4
    48 N is the magnitude of the frictional force. Don't guess! Draw a force-body diagram and figure out how much force the person is exerting on the cart.
     
  6. Mar 24, 2007 #5

    Mentz114

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    Gold Member

    The ratio of 48/54.9 is cos( 29 degrees). Does that give you a clue ?
     
  7. Mar 24, 2007 #6
    wait i dont get it, doesnt the force of friction act opposite of the direction of the displacement and force that the person exerts? if thats the case then why am i using this ratio?
     
    Last edited: Mar 24, 2007
  8. Mar 24, 2007 #7
    hey guys i drew a free body diagram, can you tell me whats wrong

    [​IMG]
     
  9. Mar 24, 2007 #8
    The direction of the frictional force is wrong.
    You are missing two other forces.
    If a is zero, then the NET force is zero, that is to say that when you add up all the forces (vectorially) the sum(net) is zero.
     
  10. Mar 24, 2007 #9
    i really have no clue as to the direction of the frictional force

    are the two other forces im missing the normal force and w=mg? i dont i need those in this problem.
     
  11. Mar 24, 2007 #10
    Yes, the other two forces I am referring to are the normal and gravity.
    The path of the cart is on a horizontal surface. The frictional force opposes the motion.
     
  12. Mar 24, 2007 #11
    it says that the woman is pushing the cart in a direction 29 degrees below the horizontal
     
  13. Mar 25, 2007 #12
    I see, I have assumed that the motion of the cart is in the horizontal direction only. The problem does not seem to state that though, hmmm. Is there a picture that goes with the problem?
     
    Last edited: Mar 25, 2007
  14. Mar 25, 2007 #13
    Yea I was thinking the woman was foolishly pushing down when the cart was moving horizontally as well, but if she is on a hill, answer to a of 54.9 doesn't seem to fit, ie if she is pushing harder than the frictional force, the cart should accelerate, even gravity is helping her???
     
  15. Mar 25, 2007 #14
    no picture to go with the problem.
     
  16. Mar 25, 2007 #15
    well robb is offline, so I'll butt in and suggest that the situation per my last post is the cart is on a horizontal surface.

    We know first that in order for there to be no acceleration, her force must exactly oppose the frictional force. If she were pushing horizontally, then 48N is the right answer. But she is pushing at an angle and only some of her force is used to overcome friction, and no we needn't worry yet about mg/normal forces, ie since we are given the total frictional force we needn't worry about its components. So redo the FBD only assume its moving horizontally. The rest should be fairly straightforward as there is no acceleration. Just need to consider force times displacement I think.
     
  17. Mar 25, 2007 #16
    oooooh thanks a ton denverdoc, i guess i needed to read the problem more carefully. it was just confusing as rather the cart she was pushing was 29 degrees below the horizontal or the force she exerts on the cart is 29 degrees below....
     
  18. Mar 25, 2007 #17
    Hey, lots of these guys who word these problems need to be put in a small room and forced to solve each others problems. :tongue:

    Physics is tuff enuf w/o ambiguous wording of questions.
     
  19. Mar 25, 2007 #18
    Lol! :biggrin:
     
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