Answer
(a) Please refer to the free-body diagram.
(b) $F_N = 12.0~m$
(c) $F_T = 6.88~m$
Work Step by Step
(a) Please refer to the free-body diagram.
(b) $F_N~cos(35.0^{\circ}) = mg$
$F_N = \frac{mg}{cos(35.0^{\circ})} = \frac{(m)(9.80~m/s^2)}{cos(35.0^{\circ})}$
$F_N = 12.0~m$
(c) $F_T = F_N~sin(35.0^{\circ})$
$F_T = (12.0~m)~sin(35.0^{\circ})$
$F_T = 6.88~m$
