91Ó°ÊÓ

Given Data:

• The mass of the wrecking ball is $m=3620\mathrm{kg}$.
• The angle of cable B with vertical is $\mathrm{θ}=40°$.

The vertical component of the wrecking ball balances the weight of the wrecking ball, and the horizontal component of tension in cable B balances tension in cable A.

Write the equation for vertical equilibrium to find tension in cable B:

$T_B\cos \mathrm{θ}=mg$

Here $g$is the gravitational acceleration, and its value is $9.8\mathrm{m}/{\mathrm{s}}^2$, $m$is the mass of the wrecking ball,$T_B$is the tension in cable B,$\mathrm{θ}$is the angle of cable B with vertical.

Substitute all the values in the above equation.

$$\begin{gathered} T_B\left(\cos 40°\right)=\left(3620\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ {T}_B=46310.6\mathrm{N} \end{gathered}$$

Therefore, the tension in cable B is 46310.6 N .

Write the equation for horizontal equilibrium to find tension in cable A:

$T_A=T_B\sin \mathrm{θ}$

Here $T_A$is the tension in cable A.

Substitute all the values in the above equation.

$$\begin{gathered} T_A=\left(46310.3\mathrm{N}\right)\left(\sin 40°\right) \\ {T}_A=29968\mathrm{N} \end{gathered}$$

Therefore, the tension in cable A is 29668 N .