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Force required to pull an object with 2 different methods

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    You are given a choice of two methods of pulling a 2 kg sled for a distance of 7 meters at a constant speed. If the coefficient of friction is 0.1, which method do you select as involving the least amount of pull on your part?

    Option 1 is a picture depicting the sled on a horizontally flat surface with the pull being in the direction of 30 degrees above the surface,

    Option 2 depicts the sled being pulled up an incline of 30 degrees above the relative horizontal with the pull being in the same direction.

    Please help, I determined an answer using correct work, however my instructor insists I am missing a step somewhere.

    2. Relevant equations
    Ff=μ*Fn
    Fg=m*9.8
    Ft=?

    3. The attempt at a solution
    For option 1, I calculated the force of friction using the coefficient given so that Ff=0.1*(2kg*9.8)=1.96
    Using this and the angle of the pull (30 degrees) I drew a triangle for the force that must be applied to move the sled so that cos(30)=1.96/x, x=2.236N applied in the direction of the pull in order to overcome friction. I multiplied this by the 7 meters it needs to travel to get the total work necessary for option 1. W=7*2.236= 15.652

    for option 2, I set the axis so that the x-axis was parallel to the table and the y-axis was perpendicular and solved for each component of gravity: Fg-x=19.6sin(30)=9.8 Fg-y=19.6cos(30)=16.974. Again using the coefficient of friction given, I solved for friction by setting the normal force equal to the y-component of gravity so that Ff=0.1*16.974=1.697. I then added this the the x-component of gravity and I multiplied the sum (11.497) by 7 to get the work (80.47)

    The problem is that my instructor informed me that there is an error in my work. Something regarding the normal force in option 2 I believe but i cant seem to figure what it is. Please help.
     
  2. jcsd
  3. Mar 9, 2016 #2
    Why do keep finding work? You want the case with minimum force required - no relation to work, here. Also, please post a diagram. It is difficult to follow all that you have written.

    This is one place where you've made a mistake. Is Fn equal to gravity, here?
     
  4. Mar 9, 2016 #3

    Charles Link

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    In your method 1, since you are applying an upward pull, your frictional force is less than .1*2*9.8 because the downward force of the sled (with the upward pull) is less than 2*9.8. Your method 2 looks right to me. Just an additional item-a more interesting question with the second method is to pull it horizontally and not up a 30 degree incline. Are you sure the second method asked to pull it up a 30 degree incline? Pulling it with a horizontal pull on a level surface might take more force and more work than pulling it with a slightly upward pull to reduce the friction.
     
    Last edited: Mar 9, 2016
  5. Mar 9, 2016 #4

    haruspex

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    This is a strange question. Are you sure you have the right two pictures for it? They do not represent two different methods of pulling the sled, they represent two different trajectories for it. Given the wording of the question, I would have expected the sled to be going on the same path in each, the only difference being the angle of the rope to the horizontal.
     
  6. Mar 9, 2016 #5
    image.jpeg
     
  7. Mar 10, 2016 #6
    Ok, so first off - Do you understand the question? If so, could you explain the reasoning behind your working?
     
  8. Mar 10, 2016 #7

    CWatters

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    Draw the two free body diagrams.
     
  9. Mar 10, 2016 #8

    Charles Link

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    I also asked the same thing in post #3. It is possible the instructor gave the problem in this manner, but the more interesting physics is for the same trajectory.
     
  10. Mar 10, 2016 #9
    And it would've helped the OP realise the mistake he was making in the first case.
     
  11. Mar 10, 2016 #10
    I believe I am trying to solve for the amount of force necessary to move the sled in each situation.
     
  12. Mar 10, 2016 #11
    2
    I figured the normal force would be equal to the y-component of gravity, yes.
     
  13. Mar 10, 2016 #12

    haruspex

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    The normal force is the (minimum) force required that there is no acceleration into the surface. Add up all the forces on the sled normal to the surface in each case.
     
  14. Mar 10, 2016 #13
    What are the forces acting on the mass? As suggested earlier, draw a free body diagram. Which forces/ their components act along the vertical direction? Is it only Fg and Fn?
     
  15. Mar 12, 2016 #14
    This is what my problem is I guess
     
  16. Mar 12, 2016 #15
    Ok; do you know how to free body diagrams? Please draw them, and only then, can we proceed.
     
  17. Mar 12, 2016 #16
    Before doing anything I would advise drawing two free body diagrams including at least the force of gravity, force of friction, and the force exerted on the sled by the rope that is being pulled
     
  18. Mar 15, 2016 #17
     

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  19. Mar 15, 2016 #18

    haruspex

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    Yes, those are the two diagrams, except that you have left out friction.
     
  20. Mar 16, 2016 #19
    Ok, so I was told that my error was that I set the normal force in the second method equal to the y-component of gravity, relative to the table. So, this is as far as I get image.jpg
     
  21. Mar 16, 2016 #20

    haruspex

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    And you are still making that error in the horizontal motion case. How can you determine the correct value of FN?
     
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