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?

5 answers

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
its cool the answer is 732.57 N

i got it =]
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.
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
83.7(-9.8)(3.2)+30.5(-9.8)(2.25)-Fa(4.5)

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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