Answer
(a) Please refer to the free-body diagram.
$F_T = 470~N$
(b) $F_N = 163~N$
Work Step by Step
(a) Please refer to the free-body diagram.
We can find the angle $\theta$ that the wire makes with the wall.
$sin(\theta) = \frac{16.0~cm}{46.0~cm}$
$\theta = arcsin(\frac{16.0}{46.0})$
$\theta = 20.35^{\circ}$
$F_T~cos(20.35^{\circ}) = mg$
$F_T = \frac{(45.0~kg)(9.80~m/s^2)}{cos(20.35^{\circ})}$
$F_T = 470~N$
(b) $F_N = F_T~sin(20.35^{\circ})$
$F_N = (470~N)~sin(20.35^{\circ})$
$F_N = 163~N$
