Question

A system of two cables supports a 150 N ball. the left hand cable is parallel to the x-axis and the right hand cable is 30 degrees above the x-axis. What is the tension in the right-hand cable?

A) 87 N
B) 150 N
C)170 N
D) 260 N
E) 300 N

Answer 1

The tension in the right-hand cable is 112.5 N.

Step-by-step explanation:

To find the tension in the right-hand cable, we can break the 150 N ball into its x and y components. The x component of the ball's weight is 150 N * cos(30°) = 129.90 N. As the left-hand cable is parallel to the x-axis, its tension will be equal to the x component of the ball's weight: 129.90 N.

The y component of the ball's weight is 150 N * sin(30°) = 75 N. The tension in the right-hand cable will be equal to the sum of the y component of the ball's weight and the additional tension needed to support the ball: 75 N + additional tension = tension in right-hand cable.

Since the tension in the left-hand cable is twice that in the right-hand cable, we can set up the equation: 2 * additional tension = 75 N.

Solving for additional tension, we find that it is equal to 37.5 N. Therefore, the total tension in the right-hand cable is 75 N + 37.5 N = 112.5 N.