Answer
(a) $39N$
(b) $39N$
(c) $36N$
Work Step by Step
(a) We can find the required tension as follows:
$T=\frac{r_{weight}}{r_{wire}mg}$
$\implies T=\frac{\frac{1}{2}Lcos\theta}{h}mg$
We plug in the known values to obtain:
$T=\frac{\frac{1}{2}(1.2)cos25^{\circ}}{0.51}(3.7)(9.81)$
$T=39N$
(b) We can find the horizontal force as
$\Sigma F_x=H_x-T=0$
$implies H_x=T$
$\implies H_x=39N$
(c) We can find the required vertical component of the force as
$\Sigma F_y=H_y-mg=0$
$\implies H_y=mg$
We plug in the known values to obtain:
$H_y=(3.7)(9.81)=36N$
