Answer
$N_p=2.65KN$, $\rho=68.3m$
Work Step by Step
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:
$\rho=68.3m$
