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Turning effect of forces

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data

    physics.jpg

    Find the smallest horizontal force required to push the barrel over the step.



    2. Relevant equations

    n/a



    3. The attempt at a solution

    By the principle of moments,
    F x 0.2 = 1500 x 0.5 .
    Thus, F = 3750N.

    I think that theres a smaller force, right??
     
  2. jcsd
  3. Aug 5, 2010 #2

    Doc Al

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    What's the height of the force F? Where's your pivot?
     
  4. Aug 5, 2010 #3
    the height is 0.2 + 0.5 = 0.7m.

    The pivot , i suppose is the edge of the step ??
     
  5. Aug 5, 2010 #4
    Last edited by a moderator: May 4, 2017
  6. Aug 5, 2010 #5
    Last edited by a moderator: May 4, 2017
  7. Aug 5, 2010 #6

    Doc Al

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    Your diagram shows the force level with the center of the barrel. I guess that's wrong? And the height of the step is labeled as 0.2, but it's drawn as being at the height of the force.

    What's the height of the step? Where is the force applied? At h = .5? or .7?

    Right. So what's the moment arm (distance) for computing the torque due to F?
     
  8. Aug 5, 2010 #7
    My apologies. the height of the step is 0.2m . And 0.5 represents the radius of the circle.
    The force is applied at the height of 0.7m
     
  9. Aug 5, 2010 #8

    Doc Al

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    Excellent. So what's the torque due to that force about the pivot? And what's the torque due to the weight of the barrel?
     
  10. Aug 5, 2010 #9
    It should be, F x D = 1500N x 0.7m = 1050Nm
     
  11. Aug 5, 2010 #10

    Doc Al

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    What's the value of D? I assume the barrel weighs 1500 N? How far is the center of the barrel from the edge of the step?
     
  12. Aug 5, 2010 #11
    Distance of Barrel from the centre to the edge of the step
    = 0.5m

    so is it F x D = 1500N x 0.5m
    = 750Nm ??
     
  13. Aug 5, 2010 #12
    the weight (1500N) is not perpendicular to the distance so it's not 1500N x 0.5m.

    you need to use some trigo here, i think (unless there's a shortcut method).
     
  14. Aug 5, 2010 #13
    But i thought its a circle ??
     
  15. Aug 5, 2010 #14
    I think i got it.

    F x D , = 1500N x ( 0.5m - 0.2m )
    = 450Nm
     
  16. Aug 5, 2010 #15

    Doc Al

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    No. That's the radius of the barrel. (Use a bit of trig/geometry.)

    Again, D is the distance between the line of the applied force F and the pivot. So what is D?
     
  17. Aug 5, 2010 #16
    0.5m - 0.2m = 0.3m
     
  18. Aug 5, 2010 #17
    actually the distance of Barrel from the centre to the edge of the step is the radius of the circle :tongue:
     
  19. Aug 5, 2010 #18

    Doc Al

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    LOL... yeah, I should have said "What is the horizontal distance from barrel center to the edge of the step". :redface:
     
  20. Aug 5, 2010 #19
    Sorry .. ;(
     
  21. Aug 5, 2010 #20

    Doc Al

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    What's the height of F? What's the height of the step? The vertical distance between them is D.
     
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