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A construction supervisor of mass M = 83.7 kg is standing on a board of mass m = 30.5 kg. The board is supported by two saw horses that are a distance of 4.50 m apart. If the man stands a distance x1 = 1.30 m away from saw horse 1 as shown, what is the force that the board exerts on that saw horse?

  • physics -

    Take moments about the other saw horse (#2) and set the total moment equal to zero. Let F be the upward force exerted by saw horse 1.

    83.7g * 3.20 + 30.5g * 2.25 - F*4.50 = 0

    Solve for F, in Newtons. g is the acceleration of gravity, 9.8 m/s^2
    3.2 m is how far the man is from sawhorse 1 and 2.25 m is the distance of the center of mass of the board from sawhorse 1

  • physics -

    its cool the answer is 732.57 N

    i got it =]

  • physics (correction) -

    I should have written "3.2 m is how far the man is from sawhorse 2 and 2.25 m is the distance of the center of mass of the board from sawhorse 2" I believe the equation is correct anyway.

  • physics -

    1. Suppose a freely orienting chain with 1000 segments each of length 7 AO is subjected to a force on its ends of 10-5 N. What will be the average separation of the chain ends?
    2. How large a force is needed to elongate the following piece of polymer to a length of 2.54 cm?
    Original length = 10 cm.
    Area of cross section = 0.1 cm2
    Mn = 30,000, Mc = 6000
    Density = 0.90 gm/cc
    3. Consider a polymer for which the potential energy resulting from attractive forces between the chains decreases markedly as the chains are oriented. What can one say about the effect of these forces upon the equilibrium stress-strain curve for that polymer rubber?

  • physics -

    this was what i did and it worked for me

    ok but the Fb*d the d=0 so that cancels out so i have

    where the guy weighs 83.7 Kg
    the board weights 30.5 kg
    the distance between the saw horses is 4.5 m

    and Fb of the second saw horse cancels out to 0 because the distance from the pivot point is non-existent

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