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Homework Help: What is the resultant torque about the hinge?

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data
    What is the resutant torque about the hinge?

    known data: CCW force 280N, CW force 400, see diagram attached for all data. All red drawings are done by me. Black is the original problem.

    2. Relevant equations

    M = F * perpendicular distance D.

    3. The attempt at a solution
    Ok. This was my method.
    For the right side of the hinge I drew the equivalent angle of 40 deg to the 30 cm bar.
    30cm cos40deg gave me the perpendicular distance of 22.96 cm

    I performed the same thing on the left side of the hinge. I cut the 90deg in half and drew a line to the line of sight of the hinge. I assumed it was half of 90.
    60cos45deg gave me 42.426cm

    so now Mhinge = 400N*0.2296m - 280N*0.4243m
    = -26.964N
    So the resultant torque is a CCW force of 26.964 N.
    My buddy got an answer of -90.87 N.

    Thanks in advance.

    Attached Files:

    • s4.JPG
      File size:
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  2. jcsd
  3. Oct 3, 2007 #2

    Chi Meson

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    Homework Helper

    I can't see the drawing yet, but I can say that it doesn't seem that the net torque can be as much as 90 N-m. (notice the correct unit for torque).

    Other than that we have to wait for the picture to be approved.
  4. Oct 3, 2007 #3
    here is a link incase the attachment doesnt work. http://img402.imageshack.us/img402/1069/s4bv6.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  5. Oct 3, 2007 #4
    So you want the angle between the force and the lever arm. (Defining ccw positive)

    M_A = F*d*sin90 = F*d

    M_B = -F*d*sinĂ˜

    what is the angle for the clockwise torque?
  6. Oct 3, 2007 #5
    i dont follow what you have done.
    can you see my diagram?
  7. Oct 3, 2007 #6
    Yeah I can see the url one. Your 280N torque is perpendicular to your lever arm, so the cross product will simply give M_a = F_a*d_a since the sine of 90 degrees is 1.

    The other one, the one that you should give a shot at is opposite to the 280N torque (what I have called M_a), but the force is not perpendicular to the lever arm, so you have to calculate the sine of whatever that angle is.

    Also, be careful to get the right units, you have cm in your drawing, but you will want meters for your torque.
  8. Oct 3, 2007 #7
    ok i am getting now -90.87 N-m. same as my buddy.

    For clarification, a force must be perpendicular to the point you are analzying and the distance between the force and point is the perpendicular distance you need when doing the M calculations?
  9. Oct 3, 2007 #8
    Yes, that is correct.
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