Answer
a) $N=0.983N$
b) $f_x=0.492N$
c) $f_y=1.25N$
Work Step by Step
(a) We can find the required force as follows:
$\Sigma \tau=0=-(\frac{1}{2}lcos\theta)mg+N(\frac{H}{sin\theta})$
This simplifies to:
$N=\frac{sin\theta (\frac{1}{2}lcos\theta)mg}{H}$
$\implies N=\frac{mglsin\theta cos\theta}{2R(1+cos\theta)}$
We plug in the known values to obtain:
$N=\frac{(0.214)(9.81)(0.436)sin30.0^{\circ}cos30.0^{\circ}}{2(\frac{1}{2}\times 0.216)(1+cos 30.0^{\circ})}$
$N=0.983N$
(b) We can find the required horizontal component of the force as
$F_x=f_x-Nsin\theta=0$
$\implies f_x=Nsin\theta$
We plug in the known values to obtain:
$f_x=(0.983N)sin30.0^{\circ}$
$f_x=0.492N$
(c) We can find the vertical component of the force as
$\Sigma F_y=f_y-mg+Ncos\theta=0$
$f_y=mg-Ncos\theta$
We plug in the known values to obtain:
$f_y=(0.214)(9.81)-(0.983)cos 30^{\circ}$
$f_y=1.25N$
