Chapter 11 - Equilibrium and Elasticity - Problems - Exercises - Page 357: 11.7

(a) The motor's weight is 1000 N and its center of gravity is located a distance of 0.80 meters from the end of the board where the 600 N force is applied. (b) The motor's weight is 800 N and its center of gravity is located a distance of 0.75 meters from the end of the board where the 600 N force is applied.

(a) Since the sum of the two upward forces must equal the weight of the motor, the motor must weigh 1000 N. Let's suppose the 400 N force is applied at the left end of the board and the 600 N force is applied at the right end. Let's consider an axis of rotation about the center of the board. Since the magnitude of the torque from the 600 N force is larger than the torque from the 400 N force, in order for the net torque to equal zero, the motor's center of gravity must be to the right of the center of the board. Let $r$ be the distance from the center of the board that the motor's center of gravity is located. $\tau = 0$ $(400~N)(1.00~m) - (600~N)(1.00~m)+(1000~N)~r = 0$ $(1000~N)~r = (600~N)(1.00~m)-(400~N)(1.00~m)$ $r = \frac{(600~N)(1.00~m)-(400~N)(1.00~m)}{1000~N}$ $r = 0.2~m$ The motor's center of gravity must be 0.2 meters from the center of the board. Note that this is a distance of 0.80 m from the end of the board where the 600 N force is applied. The motor's weight is 1000 N and its center of gravity is located a distance of 0.80 meters from the end of the board where the 600 N force is applied. (b) Since the sum of the two upward forces must equal the sum of the weights (board and motor), the motor must weigh 800 N. Let's suppose the 400 N force is applied at the left end of the board and the 600 N force is applied at the right end. Let's consider an axis of rotation about the center of the board. Since the magnitude of the torque from the 600 N force is larger than the torque from the 400 N force, in order for the net torque to equal zero, the motor's center of gravity must be to the right of the center of the board. Let $r$ be the distance from the center of the board that the motor's center of gravity is located. Note that the board's weight exerts zero torque about this axis of rotation. $\tau = 0$ $(400~N)(1.00~m) - (600~N)(1.00~m)+(800~N)~r = 0$ $(800~N)~r = (600~N)(1.00~m)-(400~N)(1.00~m)$ $r = \frac{(600~N)(1.00~m)-(400~N)(1.00~m)}{800~N}$ $r = 0.25~m$ The motor's center of gravity must be 0.25 meters from the center of the board. Note that this is a distance of 0.75 m from the end of the board where the 600 N force is applied. The motor's weight is 800 N and its center of gravity is located a distance of 0.75 meters from the end of the board where the 600 N force is applied.