A uniform 200-Ib boom, 24 ft long, is supported by a cable as shown in Figure (a

Question

A uniform 200-Ib boom, 24 ft long, is supported by a cable as shown in Figure (a) at right. The boom is hinged at the wall, and the cable makes a 30 degree angle with the boom, which is horizontal. A load of 500 lb is hung from the right end. What is the tension, T, in the cable? What are the horizontal and vertical components of the force exerted by the hinge? Note the components of the tension, T, displayed in Figure (b) above, and those of the force exerted by the hinge, F, shown in Figure (c). Remember that the system must be in both translational and rotational equilibrium. You must complete this problem using the English units as given; do not convert to Metric units (you will not receive any credit for this problem if you convert to Metric units). Recall that in the English System of units, force is measured in pounds (abbreviated "lb") and distances are determined in feet (abbreviated "ft"), so torque is expressed in "foot-pounds" (abbreviated "ft-lb"), the product of a force and a distance in these units.

Explanation / Answer

given,

weight of load = 500 lb

weight of boom = 200 lb

length of boom = 24 ft

since the system is in equlibrium so net torque = 0

torque about hinge = 500 * 24 + 200 * 12 - T * sin(30) * 24

0 = 500 * 24 + 200 * 12 - T * sin(30) * 24

tension in the cable T = 1200 lb

horizontal component of the force on hinge = T * cos(30)

horizontal component of the force on hinge = 1200 * cos(30)

horizontal component of the force on hinge = 1039.23 lb

vertical component of the force on hinge = 500 + 200 - T * sin(30)

vertical component of the force on hinge = 500 + 200 - 1200 * sin(30)

vertical component of the force on hinge = 100 lb

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