Question

A uniform beam of length 7.60 m and weight 3.90 102 N is carried by two workers, Sam and Joe, as shown in the figure below.

(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?

Answer 1

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

the torque of Sam = Sam's force * Sam's distance from the doing
the torque of Sam=S*2.8
the torque of Joe = J*1.8

these must be equal in magnitude and opposite in direction so that 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 improve his power 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 two men's torques. In other words, the beam would tip.