Answer:
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
[tex]F_{car}[/tex] = Upward tension force, [tex]Tension_{(upwards)}[/tex] - Downward tension force, [tex]Tension_{(downwards)}[/tex] - 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;
[tex]F_{car}[/tex] = [tex]Tension_{(upwards)}[/tex] - [tex]Tension_{(downwards)}[/tex] - 15473.66 N
[tex]F_{car}[/tex] = m × a = 2750 × 0.81 = [tex]T_{upwards}[/tex] - [tex]T_{downwards}[/tex] - 15473.66
∴ [tex]Tension_{(upwards)}[/tex] - [tex]Tension_{(downwards)}[/tex] = 2750 × 0.81 + 15473.66 = 17701.16 N
Hence the difference in tension, [tex]Tension_{(upwards)}[/tex] - [tex]Tension_{(downwards)}[/tex] 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.
