A uniform, horizontal beam weighing 40 N is hinged to the wall at point A and su

Question

A uniform, horizontal beam weighing 40 N is hinged to the wall at point A and supported by a rope attached to the beam at point D. A box with a weight of 30 N hangs at point C, which is also the center of gravity of the beam. To prevent the beam from breaking off the wall a person applies a 20 N force horizontally at point B. The total length of the beam is 5 m, and the distance from D to B is 1 m. A Calculate the tension of the rope and the reaction force on the beam at point A. The angle that the rope makes with the horizontal beam is 52 degrees.

Explanation / Answer

Here ,

W = 40 N

W1 = 30 N

F = 20 N

let the tension in the rope is T

balancing the torque about wall hinge

T * sin(52 degree) * (5 -1) - (40 + 30) * 5/2 = 0

solving for T

T = 55.5 N

the tension in the string is 55.5 N

Noww, for the reaction force

vertical reaction = 70 - 55.5 * sin(52) = 26.6 N

horizontal reaction = 20 + 55.5 * cos(52) = 54.17 N

net reaction force = sqrt(26.6^2 + 54.17^2)

net reaction force = 60.3 N

the net reaction force is 60.3 N

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