Chapter 9.1, Problem 74E

Equilibrium of Tensions The cranes in the figure are lifting an object that weighs 18,278 lb. Find the tensions T₁ and T₂.

To determine

To find: The tensions $T_1$ and $T_2$ in the strings of the cranes as givens in figure.

Answer to Problem 74E

The tension $T_1$ in the string is $-14,118.7i+5,790.5j$ and tension $T_2$ in the string is $14,114.89i+12,487.8j$.

Explanation of Solution

Consider a coordinate system with origin $O$. The components of tensions $T_1$ and tension $T_2$ are shown below in 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 w and sum of horizontal components of tensions $T_1$ and $T_2$ should be zero

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

$\begin{array}{c}\theta =180°-22.3°\\ =157.7°\end{array}$

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 $\left|T_1\right|\cos 157.7°$.

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 $\left|T_1\right|\sin 157.7°$.

Angle made by $T_2$ with positive $x$ -axis is $41.5°$.

So, horizontal component of $T_2$ is $\left|T_2\right|\cos 41.5°$.

Vertical component of $T_2$ is $\left|T_2\right|\sin 41.5°$.

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

$\left|T_1\right|\cos 157.7°+\left|T_2\right|\cos 41.5°=0$ (1)

Sum of vertical components of tensions $T_1$ and $T_2$ should be equal to the weight. Weight is $18,278\text{lb}$.

$\left|T_1\right|\sin 157.7°+\left|T_2\right|\sin 41.5°=18,278$ (2)

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

$-0.9252\left|T_1\right|+0.7489\left|T_2\right|=0$ (3)

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

$0.3794\left|T_1\right|+0.6626\left|T_2\right|=18,278$ (4)

Solve equation (3).

$\begin{array}{c}0.7489\left|T_2\right|=0.9252\left|T_1\right|\\ \left|T_2\right|=1.235\left|T_1\right|\end{array}$ (5)

Substitute $1.235\left|T_1\right|$ for $\left|T_2\right|$ in equation (4).

$0.3794\left|T_1\right|+0.6626\times 1.235\left|T_1\right|=18,278$

Solve further,

$\begin{array}{c}0.3794\left|T_1\right|+0.8183\left|T_1\right|=18,278\\ 1.1977\left|T_1\right|=18,278\\ \left|T_1\right|=15,260\end{array}$ (6)

Substitute $15,260$ for $\left|T_1\right|$ in equation (5),

$\begin{array}{c}\left|T_2\right|=1.235\times 15,260\\ =18,846.1\end{array}$ (7)

A vector v is represented as,

$v=\left(\text{Horizontal}\text{component}\right)i+\left(\text{Vertical}\text{component}\right)j$ (8)

Therefore, Tension $T_1$ is,

$T_1=\left|T_1\right|\cos 157.7°i+\left|T_1\right|\sin 157.7°j$

Substitute $15260$ for $\left|T_1\right|$ from equation (6) in above equation.

$\begin{array}{c}{T}_1=\left|T_1\right|\cos 157.7°i+\left|T_1\right|\sin 157.7°j\\ =15,260\times \left(-0.9252\right)i+15,260\times 0.3794j\\ =-14,118.7i+5,790.5j\end{array}$

Tension $T_2$ is,

$T_2=\left|T_2\right|\cos 41.5°i+\left|T_2\right|\sin 41.5°j$

Substitute $18846.1$ for $\left|T_2\right|$ from equation (7) in above equation.

$\begin{array}{c}{T}_2=\left|T_2\right|\cos 41.5°i+\left|T_2\right|\sin 41.5°j\\ =18,846.1\times \left(0.7489\right)i+18,846.1\times 0.6626j\\ =14,114.89i+12,487.8j\end{array}$

Thus, The tension $T_1$ in the string is $-14,118.7i+5,790.5j$ and tension $T_2$ in the string is $14,114.89i+12,487.8j$.