figure link http://www.webassign.net/sercp8/p8-8.gif (a) Determine the forces that each person exerts on the beam. F_Sam = N F_Joe = N (b) Qualitatively, how would the answers change if Sam moved closer to the midpoint? (c) What would happen if Sam moved beyond the midpoint?

### 2 Answers

we consider both vertical and rotation equilibrium the total vertical forces of Sam and Joe (S, J) must equal the weight of the beam, so we have S+J=450 the beam is uniform, so its midpoint and center of mass are at the 3.8m mark we 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 zero so we have that the torque of Sam + torque of Joe =0 torque of Sam = Sam’s force * Sam’s distance from the midoing torque of Sam=S*2.8 torque of Joe = J*1.8 these must be equal in magnitude and opposite in direction, so we can write J=2.8S/1.8 = 1.56S substitute into vertical equilibrium: S+1.56S=450 S=176 J=274 If 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.

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