Answer to Question #188428 in Physics for erika

Question

A 500-kg load is suspended from the end of a horizontal boom, as in the figure. The angle between the boom and the cable supporting its end is 45 degree. Assuming that the boom’s mass can be neglected compared with that of the load, find (a) the tension in the cable and, (b) the inward force the boom exerts on the wall.

Answer 1

a)

$$T=\frac{mg}{\cos{45}}\\T=\frac{500(9.8)}{\cos{45}}=6930\ N$$

b)

$$F=mg\tan{45}\\F=(500)(9.8)\tan{45}=4900\ N$$