A 1170-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.

Question

Physics

Orion

22 October, 04:28

A 1170-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.

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Answers (1)

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Rebekah Arellano · Answer 1

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.

A 1170-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.

A 1170-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.