Calculating torque when the lever arm has mass

  1. 1. The problem statement, all variables and given/known data
    A board is set on top of a scaffolding, with dimensions as shown in the diagram. The board weighs 15 kg. A man weighing 70 kg stands on the board as shown. How far out can he stand before the board falls?


    2. Relevant equations
    Torque: [itex]\tau = rFsin(\theta)[/itex]
    Force due to gravity: [itex]F = mg[/itex]

    3. The attempt at a solution
    I decided to use calculus, and then combine with the torque equation for the torque the man applies. I am assuming the leftmost point where the scaffolding touches the board is the fulcrum, and the rightmost point only serves to keep the board from rotating clockwise.

    First, since gravity is our only force and it points downwards, we have [itex] \tau=rmg[/itex].
    Now we use calculus:
    [itex] d\tau=rg\cdot dm[/itex], and [itex] dm=\frac{15}{5.5}\cdot dr [/itex], so
    [itex] d\tau=\frac{15g}{5.5}r\cdot dr[/itex].

    Now I assume the first 1.5m of board on each side of the fulcrum cancel out, so the torque on the right of the fulcrum is [itex]\tau=\int_{1.5}^{4} \frac{15g}{5.5}r\cdot dr=...=18.75g[/itex]. I leave the g because I am about to cancel it out.

    Now for the lefthand side of my equation, i just need the torque from the man, which will be at a distance of [itex] x [/itex] out, which we must solve for. So
    [itex] x\cdot 70\cdot g = 18g[/itex], and so [itex] x\cong 0.27m[/itex].

    Okay, that's it for my solution. The book says the answer should be [itex] 1.2m [/itex]. I haven't done physics in a while but I have done a lot of math, which is why my solution is overly mathy. I assume there's an easier thing to do here that doesn't require calculus like I did...
  2. jcsd
  3. PhanthomJay

    PhanthomJay 6,298
    Science Advisor
    Homework Helper
    Gold Member

    Your answer is correct, but try to avoid using calculus when algebra will suffice. The 15 kg mass of the board can be represented by a 15g force acting at the center of mass of the board, that is, acting dead center at 1.25 m from the fulcrum. Thus, 70x = 15(1.25), solve x = .27 m.

    I am not sure if you realized that tipping occurs when there is no reaction force at the right support as the board starts to lift off from it.

    Looks like the book messed up big time on this one, maybe a calculus error:biggrin:

    Wecome to PF!
  4. Thank you so much, PhanthomJay! Yeah, wrong answers in a book can really make you question yourself... and yes, center of mass was the magic thing I needed, thanks! One question, what do you mean by:

    ...And thanks again!
  5. PhanthomJay

    PhanthomJay 6,298
    Science Advisor
    Homework Helper
    Gold Member

    Supposing the guy walks less than 0.27 m from the left support, say 0.1 m. The system is still in equilibrium, so when you sum moments about the left support equals 0, you will find there must be an upward reaction force at the right support in order to satisfy this equilibrium condition. As the guy moves further away, this reaction force at the right support gets lower and lower, becoming 0
    when the guy has moved .27
    m. That's the tip point. If he were to continue walking a hair further away, the system could no longer be in equilibrium, since the support reaction at the right can't be less than zero, and the plank would fall over, taking Johnny with it. Unless the plank was nailed down to the support, but it is stated that this is not so.
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