Jim and Mary are carrying Bob on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 80 N. Bob weighs 600 N. Bob's center of gravity is 80 cm from Mary. Jim and Mary are at the ends of the stretcher. The upward force that Jim is exerting to support the stretcher, with Bob on it, is
A) 280 N.
B) 320 N.
C) 380 N.
D) 400 N.
E) 520 N.

Question

Quizplus · Accepted Answer

To find the force that Jim is exerting, we can use the principle of moments (torque) about Mary's end of the stretcher. Let $ F_J $ be the force exerted by Jim. The total weight of Bob and the stretcher is $ 600\,N + 80\,N = 680\,N $. The center of gravity of the system can be found by considering the moments about Mary's end. The stretcher's weight acts at its center, 1.00 m from Mary, and Bob's weight acts 0.80 m from Mary.The moment due to the stretcher's weight is $ 80\,N \times 1.00\,m = 80\,Nm $, and the moment due to Bob's weight is $ 600\,N \times 0.80\,m = 480\,Nm $. The total moment about Mary's end is $ 80\,Nm + 480\,Nm = 560\,Nm $.For equilibrium, the moment about Mary due to Jim's force must equal the total moment due to the weights, so:$$ F_J \times 2.00\,m = 560\,Nm $$$$ F_J = \frac{560\,Nm}{2.00\,m} = 280\,N $$Therefore, Jim is exerting a force of 280 N.

Jim and Mary Are Carrying Bob on a Horizontal Stretcher