Find Tension in Cable for 4.0m Beam w/ 1200N Weight

[DH214]

A uniform, 4.0-meter long beam weighing 1200 N is supported by a cable. The beam pivots at the bottom, and a 2100 N dead weight hangs from its top. Since the beam is not accelerating, you know that the net torque on it is zero. Find the tension in the cable, which is oriented at 25° from the horizontal i.e., perpendicular to the beam.

 

[tiny-tim]

A uniform, 4.0-meter long beam weighing 1200 N is supported by a cable. The beam pivots at the bottom, and a 2100 N dead weight hangs from its top. Since the beam is not accelerating, you know that the net torque on it is zero. Find the tension in the cable, which is oriented at 25° from the horizontal i.e., perpendicular to the beam.

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[thomate1]

To find the tension in the cable, we can use the equation for torque, which is equal to the force applied multiplied by the perpendicular distance from the pivot point. In this case, the force applied is the weight of the beam (1200 N) and the distance is the length of the beam (4.0 meters).

Since the beam is not accelerating, the net torque must be equal to zero. This means that the torque due to the weight of the beam must be equal and opposite to the torque due to the hanging weight. We can set up the following equation:

(1200 N)(4.0 m) = (2100 N)(4.0 m)sin(25°) + Tension(4.0 m)cos(25°)

Solving for Tension, we get Tension = 2559 N. This is the tension in the cable that is perpendicular to the beam. We can also calculate the vertical component of the tension using the following equation:

Tension vertical = Tension cos(25°) = 2559 N cos(25°) = 2309 N

Therefore, the tension in the cable is 2559 N perpendicular to the beam and 2309 N in the vertical direction. This tension is necessary to balance the weight of the beam and the hanging weight, ensuring that the beam remains in equilibrium and does not accelerate.