Question

A mobile at the art museum has a $2.0kg$steel cat and a $4.0kg$steel dog suspended from a lightweight cable, as shown in FIGURE EX$7.23.$It is found that ${\theta }_1=20°$when the center rope is adjusted to be perfectly horizontal. What are the tension and the angle of rope $3$?

Short answer

The tension and the angle of the rope $3$are$67N$and$36°$respectively.

Step-by-step solution

Given information

Mass of the cat $m_1=2kg$

Mass of dog $m_2=4kg$

${\varphi }_1=20°$

Explanation

The horizontal component of tension in rope 1 will balance the horizontal tension in rope 2. The vertical component of tension in rope 1 will balance the weight of 2.0 kg cat.

$$\begin{gathered} T_1\cos {\varphi }_1=T_2\dots \dots \left(1\right) \\ {T}_1\sin {\varphi }_1=m_{cat}g\dots \dots \left(2\right) \end{gathered}$$

Divide equation $\left(1\right)$with equation $\left(2\right)$and brings $T_2$at one side as $T_2$

localid="1649068370720" $T_2=\frac{m_{cat}g}{\tan {\theta }_1}\dots \dots \left(3\right)$

Similarly, the second knot, which joins rope $2$and $3$, Horizontal component of rope 2 will balance T₂. The vertical component will balance the dog.

localid="1649069322916" $$\begin{gathered} T_3\cos \left({\theta }_3\right)=T_2\dots \dots \left(4\right) \\ {T}_3\sin \left({\theta }_3\right)=m_{dog}g \\ {T}_3=\frac{m_{dog}g}{\sin \left({\theta }_3\right)}\dots \dots \left(5\right) \\ \text{Substituteequaion(3)in(4),weget} \\ {T}_3\cos \left({\theta }_3\right)=\frac{m_{cat}g}{\tan \left({\theta }_1\right)}\dots \dots \left(6\right) \\ \text{Dividingequation(5)withequation(6)} \\ \tan \left({\theta }_3\right)=\frac{m_{dog}\tan \left({\theta }_1\right)}{m_{cat}} \\ {\theta }_3={\tan }^{-1}\left(\frac{m_{dog}\tan \left({\theta }_1\right)}{m_{cat}}\right) \\ {\theta }_3={\tan }^{-1}\left(2\times \tan \left(20\right)\right) \\ =36° \end{gathered}$$

Substituting the given data in equation (7), we get

localid="1649613546771" $$\begin{gathered} T_3=\frac{m_{dog}g}{\sin \left({\theta }_3\right)} \\ {T}_3=\frac{\left(4kg\right)\left(9.8m/s^2\right)}{\sin \left(36°\right)} \\ {T}_3=67N \end{gathered}$$

The tension and the angle of the rope $3$are $67N\text{and36°}$respectively