Chapter 1, Problem 1.3.27P

A 150-lb rigid bar AB. with friction less rollers al each end. is held in the position shown in the figure by a continuous cable CAD. The cable is pinned at C and D and runs over a pulley at A.

(a) Find reactions at supports A and B.

(b) Find the force in the cable.

  

To determine

Reactions at supports A, and B.

Answer to Problem 1.3.27P

The correct answers are:

$A_x=-28.9\text{ lb, }{A}_y=50\text{ lb, }{B}_x=-65.0\text{ lb}\text{.}$

Explanation of Solution

Given Information:

You have a 150 lb rigid bar AB, with frictionless rollers at each end held with cables as shown in the figure below:

Calculation:

Consider the following diagram,

Calculate the position of D as,

$\begin{array}{c}\text{Position of }D=(4\cos {30}^{\circ }\text{ ft},4\sin {30}^{\circ }\text{ ft})\\ =(2\sqrt{3}\text{ ft},2\text{ ft})\end{array}$

Calculate the position of B as,

$\begin{array}{c}\text{Position of }B=(4\cos {30}^{\circ }\text{ ft},2+4\sin {30}^{\circ }\text{ ft})\\ =(2\sqrt{3}\text{ ft},4\text{ ft})\end{array}$

Calculate the position of C as,

$\begin{array}{c}\text{Position of }C=(4\cos {30}^{\circ }\text{ ft},2+3+4\sin {30}^{\circ }\text{ ft})\\ =(2\sqrt{3}\text{ ft},7\text{ ft})\end{array}$

Calculate the position of center of gravity as,

  $\begin{array}{c}\text{Position of center of gravity}=(\frac{4\cos {30}^{\circ }}{2}\text{ft},\frac{2+4\sin {30}^{\circ }}{2}\text{ ft})\\ =(\sqrt{3}\text{ ft},2\text{ ft})\end{array}$

Calculate angle $\angle BAD$ as,

$\begin{array}{c}\angle BAD=\left[{\tan }^{-1}\left(\frac{4}{2\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2}{2\sqrt{3}}\right)\right]\\ \text{=}\left[{49.1}^{\circ }-{30}^{\circ }\right]\\ \text{=19}{\text{.1}}^{\circ }\end{array}$

Calculate the angle $\angle CAD$ as,

$\begin{array}{c}\angle CAD=\left[{\tan }^{-1}\left(\frac{7}{2\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2}{2\sqrt{3}}\right)\right]\\ \text{=}\left[{63.67}^{\circ }-{30}^{\circ }\right]\\ \text{=33}{\text{.67}}^{\circ }\end{array}$

Consider the following free body diagram,

Member AB :

Take equilibrium of moments about point A,

  $\begin{array}{l}{\displaystyle \sum M_A=0}\\ \Rightarrow -4B_x-\sqrt{3}\times 150=0\\ \Rightarrow B_x=\frac{-\sqrt{3}\times 150}{4}=64.95\text{ lb}\text{. }\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of forces in $x-$ direction,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow A_x+B_x+F_{AC}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+7^2}}+F_{AD}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+2^2}}=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of forces in $y-$ direction in,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+F_{AC}\times \frac{7}{\sqrt{{(2\sqrt{3})}^2+7^2}}+F_{AD}\times \frac{2}{\sqrt{{(2\sqrt{3})}^2+2^2}}-150=0\mathrm{....}\text{(3)}\end{array}$

Analyze Joint C :

Consider free body diagram at C.

Take equilibrium of forces in $x-$ direction,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow C_x-F_{AC}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+7^2}}=0\mathrm{....}\text{(4) }\end{array}$

Take equilibrium of forces in $y-$ direction in,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow C_y-F_{AC}\times \frac{7}{\sqrt{{(2\sqrt{3})}^2+7^2}}=0\mathrm{....}\text{(5)}\end{array}$

Analyze Joint D :

Consider free body diagram at D.

Take equilibrium of forces in $x-$ direction,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow D_x-F_{AD}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+2^2}}=0\mathrm{....}\text{(6) }\end{array}$

Take equilibrium of forces in $y-$ direction in,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow D_y-F_{AD}\times \frac{2}{\sqrt{{(2\sqrt{3})}^2+2^2}}=0\mathrm{....}\text{(7)}\end{array}$

Also since the same cable rolls, hence

$F_{AC}=F_{AD}\mathrm{....}\text{(8)}$

Also take equilibrium of the forces along AD for member AB as, $$

$(A_x+B_x)\cos {30}^{\circ }+(A_y-150)\sin {30}^{\circ }+F_{AD}+F_{AC}\times \cos (33.67)=0\mathrm{....}\text{(9)}$

Now, solve equations (1-9) to get

$A_x=-28.9\text{ lb, }{A}_y=50\text{ lb, }{B}_x=-65.0\text{ lb}\text{.}$

Conclusion:

Thus the reaction forces are: $A_x=-28.9\text{ lb, }{A}_y=50\text{ lb, }{B}_x=-65.0\text{ lb}\text{.}$

To determine

Reaction force in cable.

Answer to Problem 1.3.27P

The tension force in cable is 71.6 lb.

Explanation of Solution

Given Information:

You have a 150 lb rigid bar AB, with frictionless rollers at each end held with cables as shown in figure below,

Calculation:

Consider the following diagram,

Calculate position of D as,

$\begin{array}{c}\text{Position of }D=(4\cos {30}^{\circ }\text{ ft},4\sin {30}^{\circ }\text{ ft})\\ =(2\sqrt{3}\text{ ft},2\text{ ft})\end{array}$

Calculate position of B as,

$\begin{array}{c}\text{Position of }B=(4\cos {30}^{\circ }\text{ ft},2+4\sin {30}^{\circ }\text{ ft})\\ =(2\sqrt{3}\text{ ft},4\text{ ft})\end{array}$

Calculate position of C as,

$\begin{array}{c}\text{Position of }C=(4\cos {30}^{\circ }\text{ ft},2+3+4\sin {30}^{\circ }\text{ ft})\\ =(2\sqrt{3}\text{ ft},7\text{ ft})\end{array}$

Calculate position of center of gravity as,

  $\begin{array}{c}\text{Position of center of gravity}=(\frac{4\cos {30}^{\circ }}{2}\text{ft},\frac{2+4\sin {30}^{\circ }}{2}\text{ ft})\\ =(\sqrt{3}\text{ ft},2\text{ ft})\end{array}$

Calculate angle $\angle BAD$ as,

$\begin{array}{c}\angle BAD=\left[{\tan }^{-1}\left(\frac{4}{2\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2}{2\sqrt{3}}\right)\right]\\ \text{=}\left[{49.1}^{\circ }-{30}^{\circ }\right]\\ \text{=19}{\text{.1}}^{\circ }\end{array}$

Calculate angle $\angle CAD$ as,

$\begin{array}{c}\angle CAD=\left[{\tan }^{-1}\left(\frac{7}{2\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2}{2\sqrt{3}}\right)\right]\\ \text{=}\left[{63.67}^{\circ }-{30}^{\circ }\right]\\ \text{=33}{\text{.67}}^{\circ }\end{array}$

Consider the following free body diagram,

Member AB :

Take equilibrium of moments about point A,

  $\begin{array}{l}{\displaystyle \sum M_A=0}\\ \Rightarrow -4B_x-\sqrt{3}\times 150=0\\ \Rightarrow B_x=\frac{-\sqrt{3}\times 150}{4}=64.95\text{ lb}\text{. }\mathrm{....}\text{(1)}\end{array}$

Take equilibrium of forces in $x-$ direction,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow A_x+B_x+F_{AC}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+7^2}}+F_{AD}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+2^2}}=0\mathrm{....}\text{(2)}\end{array}$

Take equilibrium of forces in $y-$ direction in,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+F_{AC}\times \frac{7}{\sqrt{{(2\sqrt{3})}^2+7^2}}+F_{AD}\times \frac{2}{\sqrt{{(2\sqrt{3})}^2+2^2}}-150=0\mathrm{....}\text{(3)}\end{array}$

Analyze Joint C :

Consider free body diagram at C.

Take equilibrium of forces in $x-$ direction,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow C_x-F_{AC}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+7^2}}=0\mathrm{....}\text{(4) }\end{array}$

Take equilibrium of forces in $y-$ direction in,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow C_y-F_{AC}\times \frac{7}{\sqrt{{(2\sqrt{3})}^2+7^2}}=0\mathrm{....}\text{(5)}\end{array}$

Analyze Joint D :

Consider free body diagram at D.

Take equilibrium of forces in $x-$ direction,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ \Rightarrow D_x-F_{AD}\times \frac{2\sqrt{3}}{\sqrt{{(2\sqrt{3})}^2+2^2}}=0\mathrm{....}\text{(6) }\end{array}$

Take equilibrium of forces in $y-$ direction in,

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow D_y-F_{AD}\times \frac{2}{\sqrt{{(2\sqrt{3})}^2+2^2}}=0\mathrm{....}\text{(7)}\end{array}$

Also since the same cable rolls, hence

$F_{AC}=F_{AD}\mathrm{....}\text{(8)}$

Also take equilibrium of the forces along AD for member AB as, $$

$(A_x+B_x)\cos {30}^{\circ }+(A_y-150)\sin {30}^{\circ }+F_{AD}+F_{AC}\times \cos (33.67)=0\mathrm{....}\text{(9)}$

Now, solve equations (1-9) to get

$A_x=-28.9\text{ lb, }{A}_y=50\text{ lb, }{B}_x=-65.0\text{ lb, }{F}_{AC}=71.6\text{ lb}\text{.}$

Conclusion:

The tension force in cable is 71.6 lb.