Chapter 6, Problem 6.89P

The cable of length 15 m supports the forces $W_1=W_2=W_3$ at B and C. (a) Derive the simultaneous equations for ${\beta }_1,{\beta }_2,$ and ${\beta }_3.$ (b) Show that the solution to these equations is ${\beta }_1=41.0°,{\beta }_2=9.8°,$ and ${\beta }_3=50.5°.$ (c) Compute the force in each segment in terms of W.

To determine

(a)

Derive the simultaneous equations for ${\beta }_1,{\beta }_2\text{ and }{\beta }_3$.

Answer to Problem 6.89P

Three simultaneous equations denoted by (a), (b) and (c) are derived.

Explanation of Solution

Given information:

Length of the cable is 15 m.

Forces at B and C are $W_1=W_2=W$

Calculations:

From geometry:

$\begin{array}{l}6\sin {\beta }_1-5\sin {\beta }_2-4\sin {\beta }_3=\text{0}\mathrm{.........}\text{(a)}\\ 6\cos {\beta }_1+5\cos {\beta }_2+4\cos {\beta }_3=\mathrm{12.........}(b)\end{array}$

Apply Eq. (6.19) to joints B and C

$\begin{array}{l}{T}_0\left(\tan {\theta }_2-\tan {\theta }_1\right)=W\therefore T_0\left[\tan {\beta }_2-\tan \left(-{\beta }_1\right)\right]=W\\ T_0\left(\tan {\theta }_3-\tan {\theta }_2\right)=W\therefore T_0\left[\tan {\beta }_3-\tan {\beta }_2\right]=W\end{array}$

Equating two equations:

$\tan {\beta }_1+2\tan {\beta }_2-\tan {\beta }_3=\mathrm{0..........}(c)$

Conclusion:

Three simultaneous equations relating ${\beta }_1,{\beta }_2\text{ and }{\beta }_3$ are derived.

To determine

(b)

Show that the solution of the equations derived in part (a) are ${\beta }_1={41}^{\circ },{\beta }_2={9.8}^{\circ },\text{ and }{\beta }_3={50.5}^{\circ }.$

Answer to Problem 6.89P

Solution of the three equations defined in part (a) are ${\beta }_1={41}^{\circ },{\beta }_2={9.8}^{\circ },\text{ and }{\beta }_3={50.5}^{\circ }.$

Explanation of Solution

Given information:

Length of the cable is 15 m.

Forces at B and C are $W_1=W_2=W$.

Equations derived in part (a):

$\begin{array}{l}6\sin {\beta }_1-5\sin {\beta }_2-4\sin {\beta }_3=\text{0}\mathrm{.........}\text{(a)}\\ 6\cos {\beta }_1+5\cos {\beta }_2+4\cos {\beta }_3=\mathrm{12.........}(b)\\ \tan {\beta }_1+2\tan {\beta }_2-\tan {\beta }_3=\mathrm{0..........}(c)\end{array}$

Calculations:

Substituting the values ${\beta }_1={41}^{\circ },{\beta }_2={9.8}^{\circ },\text{ and }{\beta }_3={50.5}^{\circ }.$ in equation (a), (b) and (c).

$6\sin {41.0}^o-5\sin {9.8}^o-4\sin {50.5}^o=-\text{0}\text{.001}\approx \text{0 }\therefore \text{OK}$

$\begin{array}{l}6\cos {41.0}^o+5\cos {9.8}^o+4\cos {50.5}^o=12\therefore OK\\ \tan {41.0}^o+2\tan {9.8}^o-\tan {50.5}^o=\text{0}\text{.002}\approx \text{0 }\therefore OK\end{array}$

Conclusion:

Since all three equations are satisfied, thus ${\beta }_1={41}^{\circ },{\beta }_2={9.8}^{\circ },\text{ and }{\beta }_3={50.5}^{\circ }.$ are the solutions of the three equations.

To determine

(c)

Find the force in each segment in terms of W.

Answer to Problem 6.89P

Force in segment AB: $T_{AB}\text{=1}\text{.272W}$

Force in segment BC: $T_{BC}=0.974W$

Force in segment CD: $T_{CD}=1.509W$

Explanation of Solution

Given information:

Length of the cable is 15 m.

Forces at B and C are $W_1=W_2=W$.

Equations derived in part (a):

$\begin{array}{l}6\sin {\beta }_1-5\sin {\beta }_2-4\sin {\beta }_3=\text{0}\mathrm{.........}\text{(a)}\\ 6\cos {\beta }_1+5\cos {\beta }_2+4\cos {\beta }_3=\mathrm{12.........}(b)\\ \tan {\beta }_1+2\tan {\beta }_2-\tan {\beta }_3=\mathrm{0..........}(c)\end{array}$

Calculations:

Using Eq. 6.19 to joint B:

$\begin{array}{l}{T}_0\left(\tan {\theta }_2-\tan {\theta }_1\right)=W\\ \therefore T_0\left[\tan {\beta }_2-\tan \left(-{\beta }_1\right)\right]=W\end{array}$

$T_0=\frac{W}{\tan {\beta }_1+\tan {\beta }_2}=\frac{W}{\tan {41.0}^o+\tan {9.8}^o}=0.9597W$

Applying Eq. (6.18) to joints B, C and D (note that ${\theta }_1=-{\beta }_1$ )

$\begin{array}{l}{T}_{AB}\text{=}\frac{T_0}{\cos \left(-{\beta }_1\right)}\text{=}\frac{0.9597W}{\cos \left(-{41.0}^o\right)}\\ \Rightarrow T_{AB}\text{=1}\text{.272W}\\ T_{BC}\text{=}\frac{T_0}{\cos {\beta }_2}=\frac{0.9597W}{\cos {9.8}^o}\\ \Rightarrow T_{BC}=0.974W\\ T_{CD}\text{=}\frac{T_0}{\cos {\beta }_3}=\frac{0.9597W}{\cos {50.5}^o}\\ \Rightarrow T_{CD}=1.509W\end{array}$

Conclusion:

Force in segments AB, BC and CD are $T_{AB}\text{=1}\text{.272W}$, $T_{BC}=0.974W$ and $T_{CD}=1.509W$ respectively.