Question

A large wrecking ball is held in place by 2 light steel cables. If the tension Ta in the

horizontal cable is 580N,

a. What is the tension Tb in the cable that makes an angle of 40° with the vertical?

b. What is the mass of the wrecking ball?

Answer 1

Let’s write the conditions of the equilibrium for the wrecking ball:

$$\sum F_x=0, \sum F_y=0.$$

Let’s consider the forces that act on the wrecking ball in the horizontal $x$- and vertical $y$-direction:

$$T_a-T_bsin\theta=0, (1)$$ $$T_bcos\theta-mg=0. (2)$$

a) We can find the tension $T_b$ in the cable that makes an angle of 40° with the vertical from the first equation:

$$T_b=\dfrac{T_a}{sin\theta}=\dfrac{580\ N}{sin40^{\circ}}=902\ N.$$

b) We can find the mass of the wrecking ball from the second equation:

$$T_b\cos\theta=mg,$$ $$m=\dfrac{T_bcos\theta}{g}=\dfrac{902\ N\cdot cos40^{\circ}}{9.8\ \dfrac{m}{s^2}}=70.51\ kg.$$

Answer:

a) $T_b=902\ N.$

b) $m=70.51\ kg.$