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A uniform beam of length 7.60 m and weight 4.50 102 N is carried by two workers, as shown in the figure below?

A uniform beam with a length of 7.60 m and a weight of 450 N is being carried by two workers, as illustrated in the figure linked here: http://www.webassign.net/sercp8/p8-8.gif.

(a) Determine the forces exerted by each worker on the beam. Let ( F_{\text{Sam}} = F_{\text{Joe}} = ) N.

(b) Qualitatively, how would the forces change if Sam moved closer to the midpoint of the beam?

(c) What would happen to the forces if Sam moved beyond the midpoint?

2 Answers

A
Anonymous

Feb 22, 2025

we consider both vertical and rotation equilibriumthe total vertical forces of Sam and Joe (S, J) must equal the weight of the beam, so we haveS+J=450the beam is uniform, so its midpoint and center of mass are at the 3.8m markwe can sum torques around any convenient point, let’s choose the center of gravity since the torque due to the weight around that point is zeroso we have that the torque of Sam + torque of Joe =0torque of Sam = Sam’s force * Sam’s distance from the midoingtorque of Sam=S*2.8torque of Joe = J*1.8these must be equal in magnitude and opposite in direction, so we can writeJ=2.8S/1.8 = 1.56Ssubstitute into vertical equilibrium:S+1.56S=450S=176J=274If Sam moved closer to the midpoint, the force he would have to exert would increase; since he would have a shorter moment (lever arm), he would need to increase his force so that his torque would balance Joe’s torque.

If he walked past the midpoint, the beam could no longer stay in equilibrium since there would be no torques on the other side of the center of mass to balance the torques of the two men. In other words, the beam would tip.

A
Anonymous

Feb 21, 2025

F_Sam * 2.8 = F_Joe*a million+(14/9).8 Taking moments approximately CG of the beam. or F_Joe = bcb02c4ee68ff576f7ab94edc169abdbcb02c4ee… .additionally anybody recognize F_Sam a million+(14/9) F_Joe = 3bcb02c4ee68ff576f7ab94edc169abd0 N or [bcb02c4ee68ff576f7ab94edc169abdbcb02c4e… – or F_Sam = bcb02c4ee68ff576f7ab94edc169abd3bcb02c4e… = bcb02c4ee68ff576f7ab94edc169abd52.6 N and F_Joe = 237.a million+(14/9) N

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