Question

Figure shows a section of a cable-car system. The maximum permissible mass of each car with occupants is 2800 kg. The cars, riding on a support cable, are pulled by a second cable attached to the support tower on each car. Assume that the cables are taut and inclined at angle $\mathit{\theta }\mathbf{=}\mathbf{35}\mathbf{°}$. What is the difference in tension between adjacent sections of pull cable if the cars are at the maximum permissible mass and are being accelerated up the incline at 0.81 m/s²?

Short answer

The difference in the tension between adjacent sections of pull cable is$1.8\times {10}^4\mathrm{N}$

Step-by-step solution

Given information

It is given that,

The mass of each car with occupants is$M=2800\mathrm{kg}$

The acceleration of each car is$a=0.81\mathrm{m}/{\mathrm{s}}^2$

The angle made by cable with horizontal is$\theta =35°$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagram the difference in the tension between adjacent sections of the pull cable can be found.

Formula:

The Newton’s second law is,

$F_{net}=\underset{}{\sum Ma}$

where,$F_{net}$is the net force, Mis mass and a is an acceleration.

Determining the difference in the tension between adjacent sections of pull cable

FBD for the system is,

According to the Newton’s second law,

$F_{net}=\sum _{}Ma$

From FBD, write for the net force acting in x-direction as,

$$\begin{gathered} F_{net,x}=T_2-T_1-Mg\sin \theta \\ =Ma \\ {T}_2-T_1=Ma+Mg\sin \theta \\ {T}_2-T_1=2800\left(0.81+9.8\sin \left(35°\right)\right) \\ {T}_2-T_1=1.8\times {10}^4 \end{gathered}$$

Hence, the difference in the tension between adjacent sections of pull cable is$1.8\times {10}^4\mathrm{N}$.