## University Physics with Modern Physics (14th Edition)

(a) If the rod balances, then the net torque about the fulcrum is zero. $\sum \tau = 0$ $W~(0.75~m)+(255~N)(0.25~m) - (225~N)(0.75~m) = 0$ $W~(0.75~m) = (225~N)(0.75~m)-(255~N)(0.25~m)$ $W = \frac{(225~N)(0.75~m)-(255~N)(0.25~m)}{0.75~m}$ $W = 140~N$ (b) Let the new position of the fulcrum be a distance $x$ from the left end of the rod. If the rod balances, then the net torque about the fulcrum is zero. $\sum \tau = 0$ $(140~N)~(x-0.75~m)+(255~N)(x-1.00~m) - (225~N)(2.00~m-x) = 0$ $(140~N)~x+(255~N)~x + (225~N)~x = (140~N)(0.75~m)+(255~N)(1.00~m) + (225~N)(2.00~m)$ $(620~N)~x = 810~N~m$ $x = \frac{810~N~m}{620~N}$ $x = 1.31~m$ The new position of the fulcrum should be 1.31 from the left end of the rod. Since the original position of the fulcrum was 1.25 m from the left end of the rod, the fulcrum should be moved 6.0 cm to the right.