Applying Newtons Laws in a 'cable car' situation?

Question

http://img.photobucket.com/albums/v638/ravenatic20/cablecars.jpg
Like my MS Paint skills? haha
The image shows a section of an alpine 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 each pylon (support tower); assume the cables are straight. 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 35 degree incline at 0.81 m/s^2 ? Explain.

Answer 1

From lowest to highest, the pull cable sections carry 1, 2 and 3 cars. Each car requires F=m*g*sin(35deg) + m*a. You can do the math:
F is in N, m=2800 kg, g=9.8 m/s^2, a=0.81 m/s^2.
Elegant picture!