The figure shows a 2000 kg cable car descending a high hill. A counterweight of mass 1800 kg on the other side of the hill aids the brakes in controlling the cable car's speed. The rolling friction of both the cable car and the counterweight are negligible. How much braking force does the cable car need to descend at constant speed?

Question

The figure shows a  2000 kg cable car descending a high hill.  A counterweight of mass  1800 kg on the other side of the hill aids the brakes in controlling the cable car's speed.  The rolling friction of both the cable car and the counterweight are negligible. How much braking force does the cable car need to descend at constant speed?

anonymous · Answer

attachment

anonymous · Answer

Hi,
There is something weird here:
Please look at problem and solution.

When I set my equations:
2000a=2000gsin(30°)-T
1800a=-1800gsin(20°)+T

I add them up:
3800a=2000gsin(30°)-1800gsin(20°)
a=1m/s^2

Now I want to equilibriate the car:
The force of brakes=-2000a=-2000a=-2000N.

But the book has a different answer. I understand their method. But why is my method wrong? Please try to explain to me in detail clearly. I have difficulty understanding.