# Chapter 7 - Newton's Third Law - Exercises and Problems: 18

The tension in rope 3 is 66.6 N and the angle of rope 3 is $36.1^{\circ}$

#### Work Step by Step

Let $T_3$ be the tension in rope 3. Let $T_1$ be the tension in rope 1. The vertical component of $T_1$ is equal to the cat's weight. $T_1~sin(\theta_1) = m_{cat}~g$ $T_1 = \frac{m_{cat}~g}{sin(\theta_1)}$ $T_1 = \frac{(2.0~kg)(9.80~m/s^2)}{sin(20^{\circ})}$ $T_1 = 57.3~N$ The horizontal component of the tension in rope 1 is equal to the tension in rope 2 which is equal to the horizontal component of the tension in rope 3. $T_{3x} = T_1~cos(\theta_1)$ $T_{3x} = (57.3~N)~cos(20^{\circ})$ $T_{3x} = 53.8~N$ The vertical component of $T_3$ is equal to the dog's weight. $T_{3y} = m_{dog}~g$ $T_{3y} = (4.0~kg)(9.80~m/s^2)$ $T_{3y} = 39.2~N$ We can find the tension $T_3$. $T_3 = \sqrt{(T_{3x})^2+(T_{3y})^2}$ $T_3 = \sqrt{(53.8~N)^2+(39.2~N)^2}$ $T_3 = 66.6~N$ We can find the angle $\theta$ of rope 3. $tan(\theta) = \frac{T_{3y}}{T_{3x}}$ $\theta = arctan(\frac{39.2~N}{53.8~N})$ $\theta = 36.1^{\circ}$ The tension in rope 3 is 66.6 N and the angle of rope 3 is $36.1^{\circ}$.

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