A uniform, horizontal beam of length 6.0 m and weight 140 N is attached at one end to a wall by a pin connection (so that it may rotate). A cable attached to the wall above the pin supports the opposite end. The cable makes an angle of 60° with the horizontal. What is the tension in the cable needed to maintain the beam in equilibrium?Select one:a.81 Nb.140 Nc.70 Nd.69 N

To solve this problem, we need to understand that the beam is in equilibrium, which means the sum of the forces and the sum of the moments (torques) are both zero.

Step 1: Identify the forces acting on the beam. There are three forces acting on the beam: the weight of the beam acting downwards at its center, the tension in the cable acting upwards and to the right, and the reaction force at the pin acting upwards and to the left.

Step 2: Set up the equilibrium equations. The sum of the forces in the vertical direction must be zero, and the sum of the moments about any point must also be zero.

Let's choose the pin as the point about which to calculate moments, since we know the moment of the tension force about this point (it's zero, because the line of action of the force passes through the point).

Step 3: Calculate the moment of the weight of the beam. The weight acts at the center of the beam, which is 3.0 m from the pin. So, the moment of the weight is (140 N)(3.0 m) = 420 N.m.

Step 4: Set up the moment equilibrium equation. The sum of the moments about the pin is zero, so the moment of the weight must be balanced by the moment of the tension in the cable. The tension acts at a distance of 6.0 m from the pin, and its vertical component creates a moment about the pin. So, we have T*sin(60) * 6.0 m = 420 N.m.

Step 5: Solve for the tension T. From the above equation, we get T = 420 N.m / (6.0 m * sin(60)) = 140 N.

So, the tension in the cable needed to maintain the beam in equilibrium is 140 N. The correct answer is (b) 140 N.