A 3.00 m-long, 190 N, uniform rod at the zoo is held in a horizontal position by two ropes at its ends (Fig. E 11.19). The left rope makes an angle of${\mathbf{150}}^{\mathbf{\circ }}$with the rod, and the right rope makes an angle $\mathit{\theta }$with the horizontal. A 90 N howler monkey (Alouatta seniculus) hangs motionless 0.50 m from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle . First make a free-body diagram of the rod.

The length of the uniform rod is: L = 3 m .

The weight of the uniform rod is: W = 190 N .

The angle made by the left rope with the rod is: ${\theta }_1=150°$.

The angle made by the right rope with the horizontal is: $\theta$.

The weight of the howler monkey is: localid="1668353109959" $W_1=90N$.

The distance of the howler monkey from the right end of the rod is: a = 0.50 m .

When a load is applied at the free end of a rope, its fibers get stretched, then the force experienced by the fibers of the rope is described as the ‘tensile force’.

For any system of forces, the value of unknown forces can be calculated by balancing all the forces acting on the system in the horizontal and vertical directions.

The free-body diagram of the rod is given by,

Here, $T_1$ is the tension in the left rope and $T_2$is the tension in the right rope.

Calculating torque about right end of the rope,

$W\times \frac{L}{2}+W_1\times a=T_1\cos 60{}^{\circ }\times L$

Putting the values,

$$\begin{gathered} 190\text{N}\times \frac{3\text{m}}{2}+90\text{N}\times 0.50\text{m}=T_1\times \frac{1}{2}\times 3\text{m} \\ 285\text{N}·\text{m}+45\text{N}·\text{m}=1.5T_1\text{m} \\ {T}_1=\frac{330\text{N}·\text{m}}{1.5\text{m}} \\ {T}_1=220\text{N} \end{gathered}$$

Balancing all the forces in the vertical direction,

$$\begin{gathered} T_1\cos 60{}^{\circ }+T_2\sin \theta =W+W_1 \\ 220\text{N}\times \frac{1}{2}+T_2\sin \theta =190\text{N}+90\text{N} \\ {T}_2\sin \theta =280\text{N}-110\text{N} \\ {T}_2\sin \theta =170N \end{gathered}$$

Balancing all the forces in the horizontal direction,

$$\begin{gathered} T_2\cos \theta =T_1\sin 60{}^{\circ } \\ {T}_2\cos \theta =220\text{N}\times \frac{\sqrt{3}}{2} \\ {T}_2\cos \theta =190.53N \end{gathered}$$

Squaring and adding equations (i) and (ii),

$$\begin{gathered} {T_2}^2{\sin }^2\theta +{T_2}^2{\cos }^2\theta ={\left(170\text{N}\right)}^2+{\left(190.53\text{N}\right)}^2 \\ {T_2}^2\left({\sin }^2\theta +{\cos }^2\theta \right)=\text{65201.68}{\text{N}}^2 \\ {T_2}^2={\text{65201.68N}}^2 \\ {T}_2=255.35\text{N} \end{gathered}$$

Hence, the tensions in the left and right ropes are 220 N and 255.35 N respectively.

Putting the value of $T_2$in the equation (i),

$$\begin{gathered} 255.35\text{N}\times \sin \theta =170\text{N} \\ \sin \theta =\frac{170}{255.35} \\ \theta ={\sin }^{-1}\left(0.67\right) \\ \theta =42{}^{\circ } \end{gathered}$$

Hence, the angle made by the right rope with the horizontal is$42°$