An I-beam having a mass of 810 kg and length of 5.1 m has one end attached to a vertical wall.The I-beam is allowed to pivot up and down about the attachment location. A cable is then attached from the vertical wall to the other end of the I-beam. The cable makes an angle of 31º with the I-beam, and the I-beam is horizontal (parallel with the ground). The tension in the cable is
A) 7700 N.
B) 4600 N.
C) 6600 N.
D) 2700 N.

Question

Quizplus · Accepted Answer

To find the tension in the cable, we can use the torque equilibrium condition. The torque due to the weight of the I-beam must be balanced by the torque due to the tension in the cable. The weight of the I-beam acts at its center of mass, which is at the midpoint of its length, so the distance from the pivot to the center of mass is 5.1 m / 2 = 2.55 m. The weight of the I-beam is the mass times the acceleration due to gravity (g = 9.8 m/s^2), so it's 810 kg * 9.8 m/s^2 = 7938 N. The torque due to the weight is then 7938 N * 2.55 m. The tension in the cable must produce an equal and opposite torque to balance this, and since the cable makes an angle of 31º with the I-beam, the vertical component of the tension (which is what contributes to the torque) is T*sin(31º). The lever arm for the tension is the full length of the I-beam, 5.1 m. Setting the torques equal gives us T*sin(31º)*5.1 m = 7938 N * 2.55 m. Solving for T gives us a value close to 7700 N, making option A the correct answer.

An I-Beam Having a Mass of 810 Kg and Length