-A 500 kg nonuniform boom,6.0 m long,is loosely pinned at the pivot at P.A 800 kg block is suspended from the end of the boom at A.The boom forms a 30° angle with the horizontal,and is supported by a cable,4.0 m long,between points D and B.Point B is 4.0 m from P,and point D is 4.0 m above P.The center of mass of the boom is at point C,which is 2.0 m from P.In the figure,the X-component of the pivot force on the boom at P is closest to:

Question

Quizplus · Accepted Answer

The correct answer is A because the X-component of the pivot force on the boom at P is closest to 12,000 N. This is because the boom is in equilibrium, so the sum of the forces in the x-direction must be zero. The only force in the x-direction is the pivot force at P, so it must be equal to the sum of the forces in the x-direction due to the weight of the boom and the weight of the block. The weight of the boom is 500 kg * 9.8 m/s^2 = 4900 N, and the weight of the block is 800 kg * 9.8 m/s^2 = 7840 N. The sum of these forces is 4900 N + 7840 N = 12,740 N. The X-component of the pivot force on the boom at P is closest to 12,000 N because it must be equal to the sum of the forces in the x-direction due to the weight of the boom and the weight of the block.

-A 500 Kg Nonuniform Boom,6