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Turning Effects of Forces ( principle of moments )

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data
    a uniform rule has a mass of 90g and its centre of mass is at the 50cm mark. the ruler is suspended by a thread at the 70cm mark, balanced by a mass of weight W hanging from the 80cm mark. Calculate the moment of the weight of the ruler about the 70cm mark. What is the moment of weight W about the 70m mark ??

    2. Relevant equations

    Principles of moment = Sum of Clockwise motion = sum of anti clockwise motion.

    3. The attempt at a solution

    Calculation of moment of weight of ruler above the 70cm mark.
    sum of anticlockwise moment = sum of clockwise moment
    Weight of rule x 20 = (0.090)(10) x 10
    => weight of rule = 0.45 N
    => moment of weight of ruler about 70 cm mark = 9 Nm

    I know that this is the actual solution to the question but can somebody explain why the distance of the anti clockwise moment is 20cm and the distance of the clockwise moment is 10cm ?

    As far as possible, can somebody give me reasons to derive at this solution ? As for the moment of the weight W about the 70cm mark, can somebody help ?? I dont really understand this well ..
  2. jcsd
  3. Aug 5, 2010 #2
    Last edited by a moderator: May 4, 2017
  4. Aug 5, 2010 #3
    Based on this, im required to find the moment of the weight W about the 70cm mark. can i know what does this mean ??
  5. Aug 5, 2010 #4
    this just means that now we take the 70cm mark as the pivot point/fulcrum, and the ruler will rotate about that point.

    just take the perpendicular distances from the forces applied at the 50cm mark and the 80cm mark to the fulcrum point at 70cm.
  6. Aug 5, 2010 #5
    So the answer is, also 9Nm?
  7. Aug 5, 2010 #6
    actually that's wrong. a correction to what i said earlier: the ruler will not rotate because the net moment is zero.

    and since the net moment is zero:

    anticlockwise moment (due to weight of ruler) = clockwise moment (due to weight W)
    0.09(9.81) x 20 = moment of weight W

    moment of weight W = 17.7Nm
  8. Aug 5, 2010 #7
    Thanks :D 1 more thing, is the value of W = 0.45N ??
  9. Aug 5, 2010 #8
    nope. it's 1.80N, twice of that of the weight of the ruler (taking g = 10m/s). The closer you get to the fulcrum, the more force you need to exert to produce the same moment (think of how it's harder to open the door near the hinge than using the knob).
  10. Aug 5, 2010 #9
    O.O why must it be twice of the weight of the ruler ?? Although i understand you, that the force exerted must be larger
  11. Aug 5, 2010 #10
    It's twice by virtue of this equation:

    [tex](0.09)(10) \times (20) = W \times (10)[/tex]

    Solve for W and you will get 1.8N. It doesn't always have to be 1.8N, just for this particular problem.
  12. Aug 5, 2010 #11
    Thanks , i have learnt to solve another type of question ! May god bless you =x
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