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Static Equilibrium homework

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A painter (mass 61 kg) is walking along a trestle, consisting of a uniform plank (mass 20.0 kg, length 6.00 m) balanced on two sawhorses. Each sawhorse is placed 1.40 m from an end of the plank. A paint bucket (mass 4.0 kg, diameter 28 cm) is placed as close as possible to the right-hand edge of the plank while still having the whole bucket in contact with the plank.

    (a) How close to the right-hand edge of the plank can the painter walk before tipping the plank and spilling the paint?
    (b) How close to the left-hand edge can the same painter walk before causing the plank to tip?

    (Hint: As the painter walks toward the right-hand edge of the plank and the plank starts to tip clockwise, consider the force acting upward on the plank from the left-hand sawhorse support.)

    2. Relevant equations

    Summation of forces

    Summation of torques , tao=Fr

    3. The attempt at a solution



    Wp=weight of the plank, wg= weight of the guy, wb weight of the bucket, nr=normal of the right hand fulcrom

    Wp=196N, Wg=588N, Wb=39.2N

    Fy: -Wp-Wg-Wb+Nr
    Nr=823.2 newtons

    Torque: N(1.6m)-Wg*x

    x=2.24 meters.

    I did my torque about the center of the plank. Wp

    this is for part a. i am still working on part b and i will post it up in a bit. So basically in the torque you exclude the normal on the left hand fulcrum and the weight of the paint. is this right?

    [url=http://www.freeimagehosting.net/pq8zg][PLAIN]http://www.freeimagehosting.net/t/pq8zg.gif[/url][/PLAIN]
     
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2
    drawing is now up.forgot to post it up so sorry.
     
  4. Nov 19, 2012 #3

    PhanthomJay

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    Why would you exclude the weight of the paint? Why do you exclude the left hand support reaction? And from what point is 'x' being measured?
     
  5. Nov 19, 2012 #4

    gneill

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    Take your torques around the point that you expect the plank to pivot. Don't ignore any of the masses on either side of the pivot point; As far as torques are concerned, the weights act through the center of mass for each object (or portion of the object producing a given torque) in question.
     
  6. Nov 19, 2012 #5
    my origin is all the way to the left end of the plank, sorry. I also thought that the sentence meant that when the normal on the left is zero and the paint has fallen off. It is implying only when the normal on the left is zero only though right?
     
  7. Nov 19, 2012 #6
    I redid it and got this:

    Wp=weight of the plank, wg= weight of the guy, wb weight of the bucket, nr=normal of the right hand fulcrom

    Wp=196N, Wg=588N, Wb=39.2N

    Fy: -Wp-Wg-Wb+Nr
    Nr=823.2 newtons

    Torque(from Nr): Wp(1.6m)-Wg(x)-Wb(2.86m)

    588N(x)=313.6N*m-112.112N*m
    x=0.343m from Nr
     
  8. Nov 20, 2012 #7

    PhanthomJay

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    Yes, correct.

    The distance from the bucket to the right pivot is not 2.86m.
    make correction for moment caused by bucket about right pivot
    when you correct this value for x, it is measured from the right pivot. The problem asks for the distance from the right edge of the plank.
     
  9. Nov 20, 2012 #8
    I messed uo again lol. Here is my fix:

    Torque: -Wg*x-Wb(1.26m)+Wp(1.6m)

    -588N*x-39.2N(1.26m)+196N(1.6m)
    x=.449m

    So from Nr to him is 0.449m But from the right end to him is 1.4m-0.449m. Would this be correct?
     
    Last edited: Nov 20, 2012
  10. Nov 20, 2012 #9

    PhanthomJay

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    You can sum moments about any point you want, and the distance x is the distance from the point about which you sum moments. So your last approach summing momens about the right edge is OK, once you make the correction I noted in red above. But as gneill pointed out, it is better to sum the moments about the right support, since this avoids the need to calculate the value of N_r. If you were to do it this way, you would get a value for x measured from the right support. Then to get the value from the right edge, it would be (1.4 - x).
     
  11. Nov 20, 2012 #10
    Hey i just fixed it!! I thought you could only do torques about a force. I did not know that you could do it on the right end and that is why I changed it.
     
  12. Nov 20, 2012 #11
    So it is .951m then. Alright thanks. Let me post up the next one. :)
     
  13. Nov 20, 2012 #12
    Hey sup. Here is part b:


    Same y forces.

    Torque(about Nl) nl is left fulcrum.

    Torque: Wg(x)-Wp(1.6m)-Wb(4.46m)

    -Wg*x=-Wp(1.6m)-Wb(4.46m)

    -588N*x=-190N(1.6m)-39.2N(4.46m)

    x=0.814 meters. from fulcrum

    from left end. 1.4m-0.814m=0.586 meters.
    Is this correct?
     
  14. Nov 20, 2012 #13

    PhanthomJay

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    More or less, with math/typo corrections, then this should be OK, good!
     
  15. Nov 20, 2012 #14
    Is each flucrum equidistant from the center of the plank?. I keep getting 4.46m.

    From Left fulcrum to center is 1.6m, then from center to right fulcrum is 1.6m too, then add 1.4m-.14m. That is what I did. Is that wrong?
     
  16. Nov 20, 2012 #15

    haruspex

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    4.46m looks right to me, but Wp is indeed 196N. Btw, it would have been slightly simpler to work in terms of multiples of g rather than converting to Newtons. All the g's would have cancelled.
     
  17. Nov 20, 2012 #16
    Alright thanks to both of you. :)
     
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