Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 147: 60

$a_t=16.1ft/s^2$, $T=46.9lb$

We know that $\Sigma F_t=ma_t$ $\implies Wcos60=ma_t~~~~~$eq(1) The mass of the boy is given as $m=\frac{W}{g}=\frac{60}{32.2}=1.863$ Now from eq(1), we have $60cos60=1.863a_t$ $\implies a_t=16.1ft/s^2$ We also know that $\Sigma F_n=ma_n$ $\implies 2T-Wsin60=m\frac{v^2}{\rho}$ $\implies 2T-60sin60=1.863\frac{(15)^2}{10}$ This simplifies to: $T=46.9lb$