Chapter 8, Problem 99A

To determine

The tension in the cable supporting the pole.

Answer to Problem 99A

The tension in the cable supporting the pole is $420\text{ N}$ .

Explanation of Solution

Given info:

The length of the pole is $2.10\text{ m}$ .

The weight of the pole is $175\text{ N}$ .

The weight of the banner is $105\text{ N}$ .

Formula used:

The expression for the torque is,

  $T=Fr$

Here, $F$ is the applied force and $r$ is the radial distance (length).

Calculation:

From a horizontal, pivoted pole a banner is suspended as shown in figure below.

  

Consider the location of the axis of rotation in order to obtain zero torque.

Consider the vertical component of the tension force as $F_{\text{Ty}}$ .

Apply the total torque on the system as follows:

  $\begin{array}{l}{\tau }_{\text{Cable}}={\tau }_{\text{Pole}}+{\tau }_{\text{Banner}}\\ F_{\text{Ty}}{r}_{\text{Cable}}=F_{\text{Pole}}{r}_{\text{Pole}}+F_{\text{Banner}}{r}_{\text{Banner}}\\ F_T\sin 25°\times 2.10\text{ m}=\left(\text{175 N}\times 1.05\text{ m}\right)+\left(\text{105 N}\times 1.80\text{ m}\right)\\ F_{\text{T}}=420\text{ N}\end{array}$

Conclusion:

Therefore, the tension in the cable supporting the pole is $420\text{ N}$ .