A uniform, 17.1 kg beam of length 2.80 m is hanging horizontally as shown in the

Question

A uniform, 17.1 kg beam of length 2.80 m is hanging horizontally as shown in the diagram below. It is pinned to a vertical wall at point B and supported by a thin, massless wire connected at point C on the beam and at point A on the wall. [Use g = 9.80 m/s^2] There is a load W, with a weight of 289 N, hanging at the end of the beam. The distance AB is 1.57 m, and the distance BC is 2.04 m. (The distance BD is 2.80 m.) What is the tension in the wire? Find the horizontal and vertical components of the force on the horizontal beam at the joint B. Use "+" for forces up or to the right, "-" for forces down or to the left. B_x = B_y =

Explanation / Answer

direction is parallel to the i1 = 25 A i.e in upward direction

Since the beam is in equilibrium,then

Net torque acting on the beam is zero ,axis of rotation is about point B

Torque due to weight of the beam + Torque due to tension in the wire =0

(-27.6*9.8*(3.8/2)) -( T*3.12*sin(theta)) = 0

from the figure

tan(theta) = AB/BC = 2.9/3.12

theta = 43 deg

then

(-27.6*9.8*(3.8/2)) + ( T*3.12*sin(43)) = 0

T = 241.5 N

Bx = T*cos(theta) = 241.5*cos(43) = 176.62 N

By = T*sin(theta) - mg = (241.5*sin(43)) - (27.6*9.81)

By = -106.05 N

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