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

(a) Let's consider an axis of rotation at the left end of the board. Note that the force at the left-hand end exerts zero torque about this axis. Let $F_s$ be the force of the support. $\sum \tau = 0$ $(F_s)(1.00~m) - (280~N)(1.50~m)-(500~N)(3.00~m) = 0$ $(F_s)(1.00~m) = (280~N)(1.50~m)+(500~N)(3.00~m)$ $F_s = \frac{(280~N)(1.50~m)+(500~N)(3.00~m)}{1.00~m}$ $F_s = 1920~N$ The force at the support point is 1920 N. (b) The sum of the vertical forces must equal zero. Let $F_L$ be the force at the left-hand end of the board. $\sum F = 0$ $F_L+280~N+500~N-1920~N=0$ $F_L = 1920~N-280~N-500~N$ $F_L = 1140~N$ The force at the left-hand end is 1140 N.