Q8E - University Physics with Modern Physics

Question

A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.8). 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. 

(a) Draw a free-body diagram for the car. 

(b) Find the tension in the cable. 

(c) How hard does the surface of the ramp push on the car?

Step-by-Step Solution

Answer

(a) The free-body diagram

 

(b) The tension in the cable is 5460 N   

(c) The surface of the ramp is pushing the car with 7224 N .

Step 1: Tension in rope and push of surface on the car

Given Data:

• The mass of the car,$m=1130\mathrm{kg}.$
• The angle of the ramp with horizontal,$\alpha =25°$ .
• The angle of cable from the surface of the ramp, $\beta =31°$ .

 

Tension in rope and push of surface on the car:

The tension in the cable can be found by considering the vertical equilibrium of the car on the ramp. The push of the ramp surface on the car can be found by considering the equilibrium of forces on the car along the horizontal direction.

Step 2: Draw the free body diagram of the car (a)

The free-body diagram of the car is given below:

Step 3: Determine the tension in the cable by considering the horizontal equilibrium of the car along the ramp (b)

The tension in the cable is calculated as:

 

 $mg \sin \alpha =T \cos \beta$ 

 

Here $\alpha$ is the angle of the ramp, and $\beta$ is the angle of the cable from the ramp, $T$ is the tension in the cable.

 

Substitute all the values in the above equation, and we get,

$$\begin{gathered} \left(1130\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\sin {25}^{°}\right)=T\left(\cos {31}^{°}\right) \\ T=5460 \mathrm{N} \end{gathered}$$  

 

Therefore, the tension in the cable is 5460 N.

Step 4: Determine the force by which the ramp surface pushes the car (c)

The push force of the ramp surface on the car is given as:

 

$F=mg \cos \alpha -T\sin \beta$  

 

Here $\alpha$ is the angle of the ramp and $\beta$ is the angle of the cable from the ramp, $F$ is the push force of the ramp surface.

 

Substitute all the values in the above equation, and we get:

$$\begin{gathered} F=\left(1130\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\cos {25}^{°}\right)-\left(5460 \mathrm{N}\right)\left(\sin {31}^{°}\right) \\ F=7224 \mathrm{N} \end{gathered}$$  

 

Therefore, the ramp surface is pushing the car with 7224 N.