A1170-kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure (figure 1) . the cable makes an angle of 31.0 ∘ above the surface of the ramp, and the ramp itself rises at 25.0 ∘ above the horizontal. draw a free-body diagram for the car.

Refer to the diagram shown below.

The mass of the car is 1170 kg, therefore its weight is
W = (1170 kg)*(9.8 m/s²) = 11466 N

The component of the weight acting down the incline is
F = W sin(25°) = (11466 N)*sin(25°) = 4845.7 N

The normal reaction from the inclined plane is
N = W cos(25°) = (11466 N) cos(25°) = 1039.2 N

T =  tension in the cable, acting at 31° above the surface of the ramp.

The Free Body Diagram on the right shows all the forces (friction is ignored)
and they FDB is sufficient for determining the value of T which establishes equilibrium.