What is the force exerted by the cable? **Problem Statement:**A 5.0 m long beam that weighs 10,125 N is attached to a wall by a pin connection that allows the beam to rotate. The far end is supported by a cable that makes a 53-degree angle with the horizontal. A 545 N person is standing 1.50 m from the pin. The beam is not moving. Find the force exerted by the cable.**Explanation:**This problem involves a static equilibrium scenario where the forces and moments acting on the beam must balance to keep it stationary. To solve for the force exerted by the cable, we need to consider the forces acting on the beam, including its weight, the weight of the person, and the tension in the cable. By applying the principles of equilibrium, we can sum the moments about the pivot point to determine the necessary tension in the cable.

Given that : Length of beam ( l ) = 5 m Weight of the beam ( FB ) = 10125 N Angle = 53 degrees Weight of the person ( FP ) = 545 N Distance of person from the pin ( d ) = 1.50 m We know that : Torque = Force * distance The weight of beam acts in the center since the beam is uniform. This is 2.5 m meter from the pin. Torque due to the beam = FB * ( l / 2 ) Beam's Torque = 10125 * 2.5 = 25312.5 NTorque due to the person standing on beam = FP * d Substituting values : Torque due to the person standing on beam = 545 * 1.5 Torque due to the person standing on beam = 817.5 Torque due to the cable = T * l * sin 53 where T is the Tension force due to the cable. Only vertical component of the force is considered here, as the cable is exerting the force in vertical direction. Substituting the values : Torque due to the cable = T * 5 * sin 53 As per given question, the beam is not moving. Hence, the total torque due to cable = Torque due to weight of beam + torque due to the man Hence, we get : T * 5 * sin 53 = 817.5 + 25312.5 T = 817.5 + 25312.55 * sin 53 T = 7145.625 N is the force exerted by cable.