Answer
(a) Please refer to the free-body diagram.
(b) $T = 5460~N$
(c) $F_N = 7220~N$
Work Step by Step
(a) Please refer to the free-body diagram.
(b) The component of the tension directed up the ramp is equal in magnitude to the component of the car's weight directed down the ramp.
$T~cos(31^{\circ}) = mg~sin(25^{\circ})$
$T = \frac{(1130~kg)(9.80~m/s^2)~sin(25^{\circ})}{cos(31^{\circ})}$
$T = 5460~N$
(c) The normal force from the ramp $F_N$ plus the component of the tension perpendicular to the ramp is equal in magnitude to the component of the car's weight directed into the ramp.
$F_N + T~sin(31^{\circ}) = mg~cos(25^{\circ})$
$F_N = (1130~kg)(9.80~m/s^2)~cos(25^{\circ}) - (5460~N)~sin(31^{\circ})$
$F_N = 7220~N$
