## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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.