Questioned Attached. The drawing shows a horizontal 5.0-m, 290 kg beam just afte

Question

Questioned Attached.

The drawing shows a horizontal 5.0-m, 290 kg beam just after one of its supports has collapsed, leaving it resting on a single support with an 83-kg worker standing on the left end. A cable attached to its right end is anchored in the ground, but will pull free if the tension exceeds 4.0 kN = 4000 N. Draw a FBD of the beam shown in the drawing. (Start fresh, don't simply add forces to the drawing.) Assume that the beam's center of gravity is at its center. Will the beam remain balanced, or will it fall? Hint: if the cable holds, the net torque on the beam about an axis through the point of contact with the remaining support is zero.

Explanation / Answer

The torque going counterclockwise (due to forces on the left side of the rod) has to be balanced out by torque going clockwise (due to forces on the right side of the rod--ie the tension in the cable) in order for the rod to remain balanced. The forces producing torques are the weight of the person, the weight of the rod and the cable. The torque due to the weight of the man is

( au_{man} = rW_{man} = (3.5m)(83kg)(9.81m/s^2) = 2850Nm)

The weight of the rod acts at it's center, which is 2.5m from the left end of the rod (which is 5m long), so it is 3.5m - 2.5m = 1m from the pivot. Thus, the torque due to the weight of the rod is

( au_{rod} = rW_{rod} = (1m)(290kg)(9.81m/s^2) = 2845 Nm)

Finally, the cable makes an angle of 67 degrees from the rod (using alternating interior angles), so the torque due to it would be

( au_{cable} = rTsin heta = (1.5m)Tsin(67))

The total torque counterclockwise balances out the torque clockwise, so

((1.5m)Tsin(67) = 2850Nm + 2845Nm = 5695 Nm)

So,

(T = 5695Nm/(1.5m*sin(67)) = 4124 N)

This is above the maximum tension the cable can handle, thus it would snap.

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