Car with counterweight: Dynamics

[yolo123]

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.

 

[haruspex]

What total mass is affected by applying the brakes?

 

[yolo123]

2000 kg!

 

[yolo123]

This is what I did. Can you be more specific?

 

[yolo123]

Hmm. That's weird. I get the right answer if I plug 3800kg. But, why is just looking at the car (2000kg) wrong!?

 

[haruspex]

Hmm. That's weird. I get the right answer if I plug 3800kg. But, why is just looking at the car (2000kg) wrong!?

Does applying the brakes affect the acceleration of the counterweight?

 

[yolo123]

I think it does not.

 

[haruspex]

I think it does not.

So, even with the brakes applied, the counterweight will accelerate up the slope, making the cable go slack?

 

[yolo123]

Oh no. It will not. The rope stays taught. But the brakes still apply force only on the car.

 

[haruspex]

Oh no. It will not. The rope stays taught. But the brakes still apply force only on the car.

Sure, but that reduces the tension in the cable, so slowing the acceleration of the counterweight. The reduced tension means you have to apply the brakes harder than you thought because the counterweight isn't pulling back as hard.
As long as the cable is taut, the situation is no different from the two masses being connected by a rod, as a truck with a trailer. You need to add their masses when calculating acceleration forces.