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.
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.
