Post a New Question


posted by .

Two factors:
1) the weight of the car is now partially down the cable, it adds to tension: mgsinTheta is the component of weight pulling back on the cable.
2) going up you have mass*acceleration.

Total tension: ma +mgsinTheta
check my thinking, draw a diagram.

A section of an alpne cable-car system has a maximum permissible mass of each car with occupants in 2800kg. The cars riding on a support calbe ar pulled by a second cable attached to each pylon(support tower); assume the cables are straight. What is the difference in tension b/w adjacent sections of pull cable if the cars are at the maxiumum permissible mass and are being accelerated up the 35 degree incline at .81 m/s^2?

What is the equation that would be used to solve this problem?

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question