Please give me concept to solve this.

The difference in tension,  between adjacent sections of the pull cable at the given conditions is 17.701 kN

Explanation:

We take the cars as moving upwards such that the resultant pulling force on the car, F, along the cable is given by the relation

= Upward tension force,   - Downward tension force, - Component of the weight of the car along the taut cable

The parameters given are;

Mass of car, m = 2750 kg

Angle of inclination of taut cables, θ = 35°

The upward acceleration of the car, a = 0.81 m/s²

Given that the weight is acting vertically downwards, we have;

Component of the weight of the car along the taut cable = m × g × sin(θ)

∴ Component of the weight of the car along the taut cable = 2750 × 9.81 × sin (35°) = 15473.66 N

We therefore have;

=   - - 15473.66 N

 = m × a = 2750 × 0.81 =   - - 15473.66

∴   - = 2750 × 0.81 + 15473.66 = 17701.16 N

Hence the difference in tension,   - between adjacent sections of the pull cable if the cars are at the maximum permissible mass and are being accelerated up the incline = 17701.16 N or 17.701 kN.

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered