## Physics: Principles with Applications (7th Edition)

Part A) To find the forces at the supports we will sum the torques about support B and then sum the forces vertically to find both reactions. $(+ \circlearrowleft) \sum\tau=0$ $1m * F_A-52kg*9.8m/s^2*3m=0$ $F_A=\frac{52kg*9.8m/s^2*3m}{1m}$ $F_A\approx1500N$ The reaction at support A is 1500N downward. $(+ \uparrow) \sum F_y=0$ $-52kg*9.8m/s^s-F_A+F_B=0$ $F_B=52kg*9.8m/s^s+F_A$ $F_B=52kg*9.8m/s^s+1528.8N$ $F_B\approx 2000N$ The reaction at support B is 2000N upward. Part B) The same as part A except we will take the weight of the board into account. $(+ \circlearrowleft) \sum\tau=0$ $1m * F_A-52kg*9.8m/s^2*3m-28kg*9.8m/s^2*3m=0$ $F_A=\frac{52kg*9.8m/s^2*3m+28kg*9.8m/s^2*3m}{1m}$ $F_A\approx1800N$ The reaction at support A is 1800N downward. $(+ \uparrow) \sum F_y=0$ $-52kg*9.8m/s^s-F_A+F_B-28kg*9.8m/s^2=0$ $F_B=52kg*9.8m/s^s+F_A+28kg*9.8m/s^2$ $F_B=52kg*9.8m/s^s+1528.8N+28kg*9.8m/s^2$ $F_B\approx 2600N$ The reaction at support B is 2600N upward.