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Help with moment/force question `

  1. Jun 15, 2010 #1
    Hey guys I was hoping you would be able to solve this question for me with a worked solution :)

    nd1o2r.jpg

    And also this question, thanks a lot :)

    2wefgw7.jpg
     
    Last edited: Jun 15, 2010
  2. jcsd
  3. Jun 15, 2010 #2

    rock.freak667

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    You just need to sum the torques about the point where it makes contact.

    T = r x F

    F points in the opposite direction of the y-axis so F = -Fj.

    To get r just write down how you get from the point to the force.

    Example: If you move along the x-axis 30 mm positively and you reach the force r= 30i mm
     
  4. Jun 15, 2010 #3
    Ok so I r=250i + 75j + 300k

    And F is unkown. Because the moment is around the x axis, I got something like this (75*z)-(300*y)=280Nm

    But wouldn't know where to go from there.
     
  5. Jun 15, 2010 #4

    rock.freak667

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    The unit of r is mm, convert that to m.

    F is perpendicular to the spanner. If the spanner is vertical (parallel to z) then the force is parallel to y. It is pointing in the -j direction so that F= -Fj

    and r x F = 280i
     
  6. Jun 15, 2010 #5
    Ok so, r=0.25i + 0.075j + 0.30k

    because the force points in the -j direction I got, (0.075*0)-(0.30*-Fj) = 280Nm

    which then got F=933.33 ?
     
  7. Jun 15, 2010 #6

    rock.freak667

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    Yes that should be correct.

    Draw the free body diagram for the second problem and use the same set of equilibrium conditions:

    ∑Fx = ∑Fy = ∑Fz = ∑M = 0
     
  8. Jun 17, 2010 #7
    Thanks for your help, lol turns out I should have just used this formula M=F*D

    with distance being (300^2+250^2)^0.5 * F = 280
     
  9. Jun 18, 2010 #8
    Not quite.
    The 250 mm dimension is not relevant. This dim could be made larger or smaller without changing the force (F) required.
    The problem states "Assume that Force (F) acts perpendicular to the spanner". This greatly simplifies the problem.

    Final answer should be about 905 N.
     
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