Chapter 9, Problem 64GP

To determine

To Find: The value of $F_{T1}\&F_{T2}$ in terms of $mg$ and the height of the tower $h$ is needed for equilibrium.

Answer to Problem 64GP

  $\begin{array}{l}{F}_{T1}=4.58mg\\ F_{T2}=5.0mg\\ h=158\text{m}\end{array}$

Explanation of Solution

Given:

The length of the right-hand section of the beam, $l_1=343.0\text{m}$

Formula used:

From the equilibrium condition in x-direction:

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_{T1}\cos 19°-F_{T2}\sin 60°=0\\ F_{T1}=F_{T2}\frac{\sin 60°}{\cos 19°}\mathrm{....}(1)\end{array}$

From the equilibrium condition in y-direction:

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ F_{T2}\cos 60°-F_{T1}\sin 19°-mg=0\\ F_{T1}=\frac{F_{T2}\cos 60°-mg}{\sin 19°}.\mathrm{....}(2)\end{array}$

Consider the torque counter clockwise as positive. The net torque at the meeting point of two roadways is zero.

  $mg\left(\frac{l_1}{2}\right)+F_{T2}\sin 60°(h)-F_{T2}\cos 60°(l_1)=0\mathrm{....}(3)$

Calculation:

From equation (1) & (2):

  $\begin{array}{l}{F}_{T2}\frac{\sin 60°}{\cos 19°}=\frac{F_{T2}\cos 60°-mg}{\sin 19°}\\ F_{T2}(0.9159)=\frac{F_{T2}(0.5)-mg}{0.32557}\\ 0.3F_{T2}=0.5F_{T2}-mg\\ 0.2F_{T_2}=mg\\ F_{T2}=5.0mg\end{array}$

Substitute the value of $F_{T2}$ in equation (1):

  $\begin{array}{l}{F}_{T1}=5mg\times \frac{\sin 60°}{\cos 19°}\\ =4.58mg\end{array}$

Substitute the value of forces in the equation (3):

  $\begin{array}{l}mg\left(\frac{l_1}{2}\right)+F_{T2}\sin 60°(h)-F_{T2}\cos 60°(l_1)=0\\ l_1=343\\ F_{T2}=5mg\\ mg\left(\frac{343}{2}\right)+(5mg)\sin 60°(h)-(5mg)\cos 60°(343)=0\\ 171.5+4.33h-857.5=0\\ 4.33h=686\\ h=\frac{686}{4.33}=158.4\text{m}\approx 158\text{m}\end{array}$

Conclusion:

The value $F_{T1}\&F_{T2}$ is $4.58mg\&5.0mg$ respectively.

The height of the tower $h$ is $158\text{m}$ .