Q6E - University Physics with Modern Physics

Question

A large wrecking ball is held in place by two light steel cables (Fig. E5.6). If the mass m of the wrecking ball is 3620 kg, what are 

(a) the tension T_B in the cable that makes an angle of 40° with the vertical and 

(b) the tension T_A in the horizontal cable?

Step-by-Step Solution

Answer

(a) The tension in cable B is 46310.6 N .

(b) The tension in cable A is 29968 N .

Step 1: Equilibrium of Wrecking ball

Given Data:

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

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.

Step 2: Determine the tension in the cable B by using vertical equilibrium (a)

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

$T_B\cos \theta =mg$ 

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, $\theta$ 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 .

Step 3: Determine the tension in cable A (b)

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

$T_A=T_B\sin \theta$ 

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 .