## Engineering Mechanics: Statics & Dynamics (14th Edition)

$N_p=2.65KN$, $\rho=68.3m$
We apply Newton's second law to determine the unknown values $\Sigma F_b=ma_b=0$ $\implies N_psin\theta=0$ $\implies N_p sin15=70\times 9.81$ This simplifies to: $N_p=2653.2=2.65KN$ and $\Sigma F_n=ma_n$ $\implies N_p cos\theta=m\frac{v^2}{\rho}$ We plug in the known values to obtain: $2653.2cos15=70\times \frac{(50)^2}{\rho}$ This simplifies to: $\rho=68.3m$