A uniform aluminum beam 9.00 m long , weighing 273 N , rests symmetrically on tw

Question

A uniform aluminum beam 9.00 m long, weighing 273 N, rests symmetrically on two supports 4.55 m apart (see figure). A boy weighing 624 N starts at point A and walks toward the right.

(a) In the same diagram construct two graphs showing the upward forces FA andFB exerted on the beam at points A and B as functions of the coordinate x of the boy. Let 1 cm = 100 N vertically, and let 1 cm = 1.00 m horizontally.

(b) From your diagram, how far beyond point B can the boy walk before the beam tips? _________ M

THE ANSWER IS NOT 1.1 M OR 6.1 M OR 5 M

Explanation / Answer

net torque about the poin B = 0

weight of the beam ats at a point r = (4.5-2.225) = 2.275 from B

torque due to weight of the beam = 273*2.275 = 621.1 Nm

the boy is at a distance of x from B

torque due to weight of the boy = 624*x

net torque = 0

624*x = 621.1

x = 0.99 m   <----answer

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