Answer
$N_A=400lb$
$F_A=248lb$
$h=3.16ft$
Work Step by Step
We can determine the required force, reaction and height as follows:
We apply the equation of motion in x-direction
$\Sigma F_x=ma_x$
$\implies F_A=(ma)_{snow\space mobile}+(ma)_{rider}$
We plug in the known values to obtain:
$F_A=\frac{250}{32.2}(20)+(\frac{150}{32.2})(20)=248lb$
Similarly, the equation of motion in y-direction is given as
$\Sigma F_y=ma_y$
$\implies N_A-W_1-W_2=0$
We plug in the known values to obtain:
$N_A-250-150=0$
$\implies N_A=400lb$
We apply the sum of moments about A
$\Sigma M_A=\Sigma M_{kA}$
$\implies W_2(0.5)+W_1(1.5)=m_2a(h)+m_1a(1)$
$\implies 150(0.5)+250(1.5)=\frac{150}{32.2}\times 20h\times \frac{250}{32.2}\times 20(1)$
This simplifies to:
$h=3.16ft$
