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A 1100-N uniform boom at phi = 58.0 degree to the horizontal is supported by a c

ID: 2142236 • Letter: A

Question

A 1100-N uniform boom at phi = 58.0 degree to the horizontal is supported by a cable at an angle ? = 32.0 degree to the horizontal as shown in the figure below. The boom is pivoted at the bottom, and an object of weight w = 2350 N hangs from its top. Find the tension in the support cable. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kN Find the components of the reaction force exerted by the pivot on the boom.

Explanation / Answer

Vertical component of tension = T sin 32
Horizontal component = - T cos 32.
Balancing forces in Y
Ry + T sin 32 = 1100 + 2350
Balancing forces in X
T cos 22 + Rx = 0

? R x = - T cos 22

Balancing torque about pivot

T * l - 1100 (l/2) cos 58 - 250 * l * cos 58 = 0

T * l = 2900 * l * cos 58

T = 2900 * cos 58 = 1536.76 N

? Rx = - ( - 1536.76 cos 32 ) = 1303 N

? Ry = 3450 - 1303 sin32 = 2760 N


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