Chapter 9.1, Problem 73E

Equilibrium of Tensions A 100-lb weight hangs from a string as shown in the figure. Find the tensions T₁ and T₂ in the string.

To determine

To find: To find the tensions $T_1$ and $T_2$ in the string of the given figure.

Answer to Problem 73E

The tension $T_1$ in the string is $-56.5i+67.4j$ and tension $T_2$ in the string is $56.5i+32.6j$

Explanation of Solution

Given:

Tension $T_1$ and tension $T_2$ support a $100\text{lb}$ weight as shown in figure.

Calculation:

Consider a coordinate system with origin $O$ as shown in Figure (1).

Figure (1)

As the system is in equilibrium, the resultant of forces acting in horizontal and vertical direction should be zero.

Sum of vertical components of tensions $T_1$ and $T_2$ should be equal to the weight and sum of horizontal components of tensions $T_1$ and $T_2$ should be zero. Components of tension forces are shown in figure (2).

Figure (2)

Angle made by $T_1$ with positive direction of $x$-axis is,

$\begin{array}{c}\theta =180°-50°\\ =130°\end{array}$

The horizontal component of the vector v, which makes angle $\theta$ with positive direction

of $x$-axis is $\left|v\right|\cos \theta$.

Therefore, horizontal component of $T_1$ is $\left|T_1\right|\cos 130°$.

Vertical component of vector v, which makes angle $\theta$ with positive direction of $x$-axis is $\left|v\right|\sin \theta$

Therefore, vertical component of $T_1$ is $\left|T_1\right|\sin 130°$.

Angle made by $T_2$ with positive $x$ -axis is $30°$.

So, horizontal component of $T_2$ is $\left|T_2\right|\cos 30°$.

Vertical component of $T_2$ is $\left|T_2\right|\sin 30°$.

Sum of horizontal components of tensions $T_1$ and $T_2$ should be zero as system is in equilibrium.

$\left|T_1\right|\cos 130°+\left|T_2\right|\cos 30°=0$ (1)

Sum of vertical components of tensions $T_1$ and $T_2$ should be equal to the weight. Weight is $100\text{lb}$.

$\left|T_1\right|\sin 130°+\left|T_2\right|\sin 30°=100$ (2)

Put values of cosine and sine terms in equation (1)

$-0.642\left|T_1\right|+0.866\left|T_2\right|=0$ (3)

Use values of cosine and sine terms in equation (2).

$0.766\left|T_1\right|+0.5\left|T_2\right|=100$ (4)

Solve equation (3).

$\begin{array}{c}0.866\left|T_2\right|=0.642\left|T_1\right|\\ \left|T_2\right|=0.741\left|T_1\right|\end{array}$ (5)

Substitute $0.741\left|T_1\right|$ for $\left|T_2\right|$ in equation (4).

$0.766\left|T_1\right|+0.5\times 0.741\left|T_1\right|=100$

Solve further,

$\begin{array}{c}1.136\left|T_1\right|=100\\ \left|T_1\right|=87.99\end{array}$ (6)

Substitute $87.99$ for $\left|T_1\right|$ in equation (5),

$\begin{array}{c}\left|T_2\right|=0.741\times 87.99\\ =65.2\end{array}$ (7)

A vector v is represented as,

$v=\left(\text{Horizontal}\text{component}\right)i+\left(\text{Vertical}\text{component}\right)j$

Therefore, Tension $T_1$ is,

$T_1=\left|T_1\right|\cos 130°i+\left|T_1\right|\sin 130°j$

Substitute $87.99$ for $\left|T_1\right|$ from equation (6) in above equation.

$\begin{array}{c}{T}_1=\left|T_1\right|\cos 130°i+\left|T_1\right|\sin 130°j\\ =87.99\times \left(-0.6427\right)i+87.99\times 0.766j\\ =-56.5i+67.4j\end{array}$

Tension $T_2$ is,

$T_2=\left|T_2\right|\cos 30°i+\left|T_2\right|\sin 30°j$

Substitute $65.2$ for $\left|T_2\right|$ from equation (7) in above equation.

$\begin{array}{c}{T}_2=\left|T_2\right|\cos 30°i+\left|T_2\right|\sin 30°j\\ =65.2\times \left(0.866\right)i+65.2\times 0.5j\\ =56.5i+32.6j\end{array}$

Thus, the tension $T_1$ in the string is $-56.5i+67.4j$ and tension $T_2$ in the string is $56.5i+32.6j$.