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Moments metre stick question

  1. Dec 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A metre stick is pivoted at 50cm and is in equilibrium.
    There is a 2.0N upwards force at the 5.0cm mark, and a 12N downwards force at the 40cm mark. Force F acts at the 75cm mark. Find force F. Assume the upwards direction is positive.

    2. Relevant equations
    3. The attempt at a solution

    The leftmost upwards force: 2 x 0.45 = 0.9 Nm (as the force would be 45cm from the pivot)
    The leftmost downwards force: 12 x 0.1 = 1.2 Nm
    Resultant leftmost force: 0.9 - 1.2 = -0.3 Nm

    Force F acts 0.15 m from the pivot, therefore: F x 0.15 m = 0.3 Nm. (resultant would be positive in order to counterbalance).
    Rearranging for F, 0.3/0.15 = 2 N. The programme marks this as incorrect.
  2. jcsd
  3. Dec 20, 2015 #2
    See anything strange going on there?
  4. Dec 20, 2015 #3
    Nope. The stick is pivoted at 0.5m. If the force acts at the 0.75m mark, it is acting 0.15m away from the pivot.

    EDIT: Ugh, I'm an idiot. Thanks!
  5. Dec 20, 2015 #4
    Haha, when I saw your reply earlier without the edit, I was like, "huh"? xD
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