# Chapter 7 - Newton's Third Law - Exercises and Problems - Page 189: 48

The tension in the rope is 3590 N

#### Work Step by Step

The vertical component of the tension in each side of the rope provides the force to accelerate the person up in the air. Let $T$ be the tension in each side of the rope. Note that $T_y = T~sin(\theta)$. We can set up a force equation to find $T$: $\sum F = ma$ $2T_y - mg = ma$ $2T~sin(\theta) = m(g+a)$ $T = \frac{m(g+a)}{2~sin(\theta)}$ $T = \frac{(70~kg)(9.80~m/s^2+8.0~m/s^2)}{2~sin(10^{\circ})}$ $T = 3590~N$ The tension in the rope is 3590 N.

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