Chapter 2.3, Problem 2.45P

(a)

To determine

Find the tension $\left(T_{AC}\right)$ in the cable AC.

Answer to Problem 2.45P

The tension $\left(T_{AC}\right)$ in the cable AC is $\underset{\_}{2.5\text{kN}}$.

Explanation of Solution

Given information:

The force (F) in point C is 1.98 kN.

The horizontal (AC) length between point A and C is 4.8 m.

The horizontal (CB) length between point C and B is 3 m.

The height (A) at point A is 3.4 m.

The height (B) at point B is 3.6 m.

Calculation:

Calculate vertical distance $\left(h_1\right)$ between point A and C using the relation:

$h_1=A-d$

Substitute 3.4 m for A and 2 m for d.

$\begin{array}{c}{h}_1=3.4-2\\ =1.4\text{m}\end{array}$

Calculate vertical distance $\left(h_2\right)$ between point B and C using the relation:

$h_2=B-d$

Substitute 3.6 m for B and 2 m for d.

$\begin{array}{c}{h}_2=3.6-2\\ =1.6\text{m}\end{array}$

Show the free body diagram of two cable as in Figure 1.

Calculate the angle $\left(\alpha \right)$ using the relation:

$\tan \alpha =\frac{h_1}{AC}$

Substitute 1.4 m for $h_1$ and 4.8 m for AC.

$\begin{array}{c}\tan \alpha =\frac{1.4}{4.8}\\ \alpha ={\tan }^{-1}\left(0.29167\right)\\ \alpha =16.26°\end{array}$

Calculate the angle $\left(\beta \right)$ using the relation:

$\tan \beta =\frac{h_2}{BC}$

Substitute 1.6 m for $h_2$ and 3 m for BC.

$\begin{array}{c}\tan \beta =\frac{1.6}{3}\\ \beta ={\tan }^{-1}\left(0.533\right)\\ \beta =28.07°\end{array}$

Calculate the angle $\left(\theta \right)$ using the relation:

$\theta =90°-\alpha$

Substitute $16.26°$ for $\alpha$.

$\begin{array}{c}\theta =90°-16.26°\\ =73.74°\end{array}$

Calculate the angle $\left(\varphi \right)$ using the relation:

$\varphi =90°-\beta$

Substitute $28.07°$ for $\beta$.

$\begin{array}{c}\varphi =90°-28.07°\\ =61.927°\end{array}$

Calculate the angle $\left(\phi \right)$ using the relation:

$\phi =180°-\theta -\varphi$

Substitute $73.74°$ for $\theta$ and $61.927°$ for $\varphi$.

$\begin{array}{c}\phi =180°-73.74°-61.927°\\ =44.333°\end{array}$

Show the force triangle as in Figure 2.

Write the below expression by applying law of sines for Figure 2.

$\frac{T_{AC}}{\sin \varphi }=\frac{T_{BC}}{\sin \theta }=\frac{F}{\sin \phi }$ (1)

Calculate the tension $\left(T_{AC}\right)$ in cable AC:

Eliminate the $\left(T_{BC}\right)$ force in equation (1).

Substitute 1.98 kN for F, $44.333°$ for $\phi$ and $61.927°$ for $\varphi$ in equation (1).

$\begin{array}{c}\frac{T_{AC}}{\sin \left(61.927°\right)}=\frac{1.98}{\sin \left(44.333°\right)}\\ T_{AC}=\sin \left(61.927°\right)\times 2.833\\ =2.499\text{kN}\simeq 2.50\text{kN}\end{array}$

Therefore, the tension $\left(T_{AC}\right)$ in the cable AC is $\underset{\_}{2.5\text{kN}}$.

(b)

To determine

The tension $\left(T_{BC}\right)$ in the cable BC.

Answer to Problem 2.45P

The tension $\left(T_{BC}\right)$ in the cable BC is $\underset{\_}{2.72\text{kN}}$.

Explanation of Solution

Given information:

The force (F) in point C is 1.98 kN.

The horizontal (AC) length between point A and C is 4.8 m.

The horizontal (CB) length between point C and B is 3 m.

The height (A) at point A is 3.4 m.

The height (B) at point B is 3.6 m.

Calculation:

Calculate the tension $\left(T_{BC}\right)$ in cable BC:

Eliminate the $\left(T_{AC}\right)$ force in equation (1).

Substitute 1.98 kN for F, $44.333°$ for $\phi$ and $73.74°$ for $\theta$ in equation (1).

$\begin{array}{c}\frac{T_{BC}}{\sin \left(73.74\right)}=\frac{1.98}{\sin \left(44.333°\right)}\\ T_{BC}=\sin \left(73.74\right)\times 2.833\\ =2.7196\text{kN}\approx \text{2}\text{.72}\text{kN}\end{array}$

Therefore, the tension $\left(T_{BC}\right)$ in the cable BC is $\underset{\_}{2.72\text{kN}}$.