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Torque problems

  1. Nov 12, 2004 #1


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    Im having some difficulty with this problem..i know im getting the wrong answer but not sure why.

    Question: A uniform plank of length 5.0m and weight 225N rests horizontally on two supports, with 1.1m of the plank hanging over the right support. To what distance x can a person who weighs 450N walk on the overhanging part of the plank before it just begins to tip?

    i i

    i - supports
    p - person
    ----- - plank
    the distance between the rightmost i and p is (x)
    and the distance between the rightmost i and the edge of the plank is 1.1m

    My (wrong) solution:

    Sum of the forces in the y direction = 0
    0 = N(Normal from first support) + N(normal from second support) - Ww(Weight of the wood) - Wp(Weight of the person)

    2N = Ww + Wp
    N = (Ww +Wp)/2

    Sum of the torques = 0

    0 = Wp(x) - Ww(1.4) - N(3.9)
    Wp(x) = Ww(1.4) + [(Ww +Wp)/2)](3.9)
    subbing in the numerical values...

    (450)x = (225)(1.4) + [(225 +450)/2](3.9)
    x = 3.625

    but this answer is wrong because x has to be less than 1.1 m!! :frown:

    thanks for your help :smile:
  2. jcsd
  3. Nov 12, 2004 #2


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    oh oh the diagram didnt work hmmm...
    the plank is supported by two supports, one at the very left edge and the other 1.1 m from the right edge, the person is a distance x from the right support on the way to the right edge
  4. Nov 12, 2004 #3
    Well, you're confused on a couple of points, even though you're getting the main idea. First of all, of course x won't be smaller than 1.1, since you're measuring x _from the left support_, not from the right support - the only condition is that it has to be between 4 and 5, therefore. Second, N should be two times bigger, because in equilibrium the weight of the plank and the man are not held up by two supports, it's held up only by the letmost one (think about it - would it matter in equilibrium if you just took the leftmost pillar away?). So you don't divide by two. Third, if you're calculating torques around the leftmost point, the lever arm for the weight of the plank is 2.5 m, not 1.4, and it should have the same sign as the torque contribution of the person (and the opposite of the contribution of the normal). Think about it - it's an important point.

    These problems are much easier to solve if you set up the right equilibrium point. This time, the right pillar would've been a much cleaner choice (since, in equilibrium, the left pillar is supporting no weight so the contribution from its normal is zero - make sure you understand this point).
  5. Nov 12, 2004 #4


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    i did use the right pillar for the equilibirum choice...the N i was using in the torque equation refers to the torque for the left pillar and thats why its (5- 1.1) 3.9 m away.. i think i might be doing something wrong for calculating the torque for the weight of the wood ..is it supposed to be at the centre of the plank so the distance would be (2.5 - 1.1) 1.4m away?
  6. Nov 12, 2004 #5
    yes. but, as i pointed out, there is no force being exerted on/by the left pillar during equilibrium, so the contribution from it is zero (in equilibrium, the man being at the farthest possible distance on the right, the plank is just about to lift from the left pillar, so it exerts no force on the pillar and, therefore, there is no normal force)
  7. Nov 12, 2004 #6


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    O i understand now...just because the question says just before the plank begins to tip, it doesn't necessarily mean that the plank is not raised above the support
    I get 0.7m if i dont use normal force for the left pillar which is alot more reasonable..and the question becomes much more simple
    thanks so much duarh :)
  8. Nov 12, 2004 #7
    that's what i got too. incidentally, it doesn't have to be raised above the support - in fact, it better not be, or you're starting to lose equilibrium. the condition is one of 'light touching'. :D like peaceful coexistence
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