# What is the magnitude of the force of the car on the truck?

1. Sep 22, 2007

### aligass2004

1. The problem statement, all variables and given/known data
A 1100 kg car pushes a 1700 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500N. a.) What is the magnitude of the force of the car on the truck? b.) What is the magnitude of the force of the truck on the car?

2. Relevant equations
F=ma

3. The attempt at a solution
I tried drawing a free body diagram for this, but it didn't work out well.

2. Sep 22, 2007

### learningphysics

First do the freebody diagram of the truck+car system. That gives you acceleration...

Then do the freebody diagram of the car alone. that'll give the force of the truck on the car.

3. Sep 23, 2007

### aligass2004

I got the acceleration to be 1.607 m/s^2. I tried using F=ma, but it didn't work.

4. Sep 23, 2007

### learningphysics

The acceleration is correct. What equation did you get when you used F=ma for the car?

5. Sep 23, 2007

### aligass2004

The number I got was 1767.7N

6. Sep 23, 2007

### learningphysics

That's not right. What are the forces acting on the car?

7. Sep 23, 2007

### aligass2004

The forces acting on the car are weight, the normal force, and the force of the truck.

8. Sep 23, 2007

### learningphysics

You're missing one force. The main one.

9. Sep 23, 2007

### aligass2004

I don't understand.

10. Sep 23, 2007

### learningphysics

The 4500N.

11. Sep 23, 2007

### aligass2004

Ok, so the normal force equal 4500N as well. So how do I go about finding the force of the truck on the car with all of that information?

12. Sep 23, 2007

### learningphysics

It's not the normal force... It is a forward force exerted at the back of the car... (it's the same force you used for the truck+car part).

Write out the F = ma equation for the car in the horizontal direction.

13. Sep 23, 2007

### aligass2004

I am still confused.

14. Sep 23, 2007

### learningphysics

There's a 4500N forward force at the back of the car... there's a backward force from the truck.

$$\Sigma{F} = ma$$

$$4500 - F_{truck} = m_{car}a$$

15. Sep 23, 2007

### aligass2004

I got the answer to be 2732.3, but I still understand the forward force at the back of the car thing.

16. Sep 23, 2007

### learningphysics

How did you do the first part?

17. Sep 23, 2007

### aligass2004

I found the acceleration.

18. Sep 23, 2007

### learningphysics

Yes, how did you find the acceleration? You used the 4500N force... how did you use it... how did you know how to use it?

19. Sep 23, 2007

### aligass2004

I used F=ma and manipulated the equation around to a=F/m. I don't know why I knew to use it though.

20. Sep 23, 2007

### learningphysics

Draw a picture... the 4500N force is pushing the car forward... and the car is pushing the truck forward...