Chapter 9, Problem 19P

(a)

To determine

The tension in the horizontal massless cable CD.

Answer to Problem 19P

Solution:

410 N

Explanation of Solution

Given:

The uniform aluminum pole AB is 7.20 m long and has a mass 12 kg. The mass of the traffic light is 21.5 kg.

Formula used: For equilibrium:

  ${\displaystyle \sum \tau }=r_i\times F_i=0$

Here, $F_i$ is the applied force and $r_i$ represents its corresponding distance from the rotational axis.

Calculations:

  

The lengthof the cable connects to the hanging pole is the hypotenuse of a triangle which has a side 3.8 m and opposite angle of 37 degree, so the length of the hypotenuse is

  $\begin{array}{l}l\sin ({37}^{°})=3.80\text{m}\\ l\text{=6}\text{.3m}\end{array}$

Consider rotation through the base A, and by considering Newtons second law for angular rotation:

  $\begin{array}{l}\sum \tau =T\times 3.8\text{m}-\left(12\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(\cos 37°\right)\left(\frac{7.2\text{m}}{2}\right)\text{-}\left(21.5\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(7.2\text{m}\right)\left(\cos 37°\right)\text{=0}\\ T\approx \text{410 N}\end{array}$

Conclusion:

Hence, the tension in the horizontal massless cable, $T=410\text{N}$ .

(b)

To determine

The vertical and horizontal components of the force exerted by the pivot A on the aluminum pole.

Answer to Problem 19P

Solution:

The horizontal component of force is 410 N and vertical components of force is 328 N.

Explanation of Solution

The forces in the x and y directions, and since the system is static, these sums are zero,

In which case,

  $\begin{array}{l}\sum F_x=0\\ F_x-T=0\\ F_x=T=410\text{N}\end{array}$

And

  $\begin{array}{l}\sum F_y=0\\ F_y-Mg-mg\\ F_y=(M+m)g\end{array}$

  $\begin{array}{l}{F}_y=\left(21.5+12\right)\left(9.8\right)\\ F_y=328\text{N}\end{array}$

Conclusion:

The horizontal component of force is 410 N and vertical components of force is 328 N.