# Car with counterweight: Dynamics

1. Apr 13, 2014

### 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

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.

#### Attached Files:

File size:
11.8 KB
Views:
76
• ###### Screen Shot 2014-04-13 at 4.49.03 PM.png
File size:
37.1 KB
Views:
78
2. Apr 13, 2014

### haruspex

What total mass is affected by applying the brakes?

3. Apr 13, 2014

### yolo123

2000 kg!

4. Apr 13, 2014

### yolo123

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

5. Apr 13, 2014

### yolo123

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

6. Apr 13, 2014

### haruspex

Does applying the brakes affect the acceleration of the counterweight?

7. Apr 13, 2014

### yolo123

I think it does not.

8. Apr 13, 2014

### haruspex

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

9. Apr 13, 2014

### yolo123

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

10. Apr 13, 2014

### haruspex

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.