91影视

Given Data:

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

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.

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

The tension in the cable is calculated as:

$mg\sin 伪=T\cos 尾$

Here $伪$is the angle of the ramp, and $尾$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,

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

Therefore, the tension in the cable is 5460 N.

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

$F=mg\cos 伪-T\sin 尾$

Here $伪$is the angle of the ramp and $尾$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:

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

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