Chapter 5, Problem 84A

a.

To determine

Tension in each wire.

Answer to Problem 84A

  $T=150\text{N}$

Explanation of Solution

Given:

Weight of lamp, $F_g=mg=150\text{N}$

Angle between wire, $\theta ={120}^o$

Formula used:

The formula used is

  $F_g=2T\cos (\theta /2)$

Here, $F_g$ is weight, $T$ is tension and $\theta$ is angle between wires.

Calculation:

The free body diagram of lamp is shown below:

  

From free body diagram,

Since, the component $T\sin (\theta /2)$ cancels out, so

  $\begin{array}{c}mg=2T\cos (\theta /2)\\ T=\frac{mg}{\cos (\theta /2)}\\ T=\frac{150}{2\cos \left(120°/2\right)}\\ T=150\text{N}\end{array}$

Conclusion:

Therefore, the tension in the wire is $T=150\text{N}$ .

b.

To determine

Tension in the wire.

Answer to Problem 84A

  $T=106\text{N}$

Explanation of Solution

Given:

Weight of lamp, $F_g=mg=150\text{N}$

Angle between wire, $\theta =90°$

Formula used:

The formula used is

  $F_g=2T\cos (\theta /2)$

Here, $F_g$ is weight, $T$ is tension and $\theta$ is angle between wires.

Calculation:

The free body diagram of lamp is shown below,

  

From free body diagram,

The component $T\sin (\theta /2)$ cancels out, so

  $\begin{array}{c}mg=2T\cos (\theta /2)\\ T=\frac{mg}{\cos (\theta /2)}\\ T=\frac{150}{2\cos \left(90°/2\right)}\\ T=106\text{N}\end{array}$

Conclusion:

Therefore, the tension in the wire is $T=106\text{N}$ .