Question

A uniform horizontal beam with a length of 8.00 man and a weight of 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle 53.0 with the horizontal.If a 600-N person stands 2.00 m from the wall, find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam

Answer 1

$\Sigma M=0: -2P-4F+8T\sin53°=0$

$T=\frac{2P+4F}{8\sin53°}=\frac{2×600+4×200}{8×\sin53°}=313N$

$R_x=T\cos53°=313×\cos53°=188N$

$R_y=P+F-T\sin53°=600+200-313×\sin53°=550N$

$R=\sqrt{R_x^2+R_y^2}=\sqrt{188^2+550^2}=581N$

$\alpha=\tan^{-1}(\frac{R_x}{R_y})=\tan^{-1}(\frac{550}{188})=71°$