# Need help with net force

1. Oct 21, 2008

### frosti

1. The problem statement, all variables and given/known data
A 930 kg car is pulling a 320 kg trailer. Together the car and trailer move forward with an acceleration of 2.16 m/s2. Ignore any frictional force of air drag on the car and all frictional forces on the trailer. determine the net force on the car.

2. Relevant equations
f = ma

3. The attempt at a solution
I thought the forces that are acting on the car is a force moving car forward and a force applied by the trailer slowing the car down so the net force on the car should be 930 x 2.16 - 320 x 2.16 = 1317.6. However, it is wrong. I don't really know how to think about this problem, can anyone please provide me with some insight?

2. Oct 21, 2008

### jenglish

I believe you're over thinking this one. The problem clearly states to ignore any frictional force of air drag or frictional forces of the trailer. Therefore, thinking about the system as a whole, the net force would simply equal the mass of the system as a whole times the acceleration of the system.

F=ma

3. Oct 21, 2008

### borgwal

jenglish is wrong, frosti is getting there, kind of: there is one friction force pushing the car forward, and a force exerted by the trailer on the car, backward. But there is also a force on the trailer forward. Draw a pucture with these 3 forces, and you 'll find the answer.

4. Oct 21, 2008

### jenglish

those two "extra" forces you mentioned are the tension on the connecting cable...they should ultimately cancel

5. Oct 21, 2008

### borgwal

No: the question is about the net force on the car, which includes the force the trailer (or the connecting cable if you wish) exerts. The net force on the car is not equal to "the mass of the system times the acceleration of the system": the latter quantity would be the net force on the system of car + trailer, not the net force on the car.

In any case, the question is simple, but not in that way.

Last edited: Oct 21, 2008
6. Oct 21, 2008

### frosti

borgwal, dont force exerted by the trailer on the car backward and the force on the trailer forward on the car cancel out? so all that left me is one force 930 x 2.16?

jenglish, you are saying the net force should be (930 + 320) x 2.16 right? but why would the net force on the car also include mass of the trailer?

I'm confused here on which one sounds more right. Could you guys elaborate a bit more?

7. Oct 21, 2008

### borgwal

Yes, the net force on the car is simply 930 x 2.16 9 (it's simple after all). There is no force the trailer exerts on the car in the forward direction, only one in the backward direction. Since the net force on the total system [a friction force on the car, in the forward direction] is (930+320) x 2.16 it means the trailer is pulling back the car with a force 320 x 2.16, and the car is pulling the trailer forward with that same force in the opposite direction [its an action-reaction pair]. If you draw all the forces it will be clearer, I suppose.

8. Oct 21, 2008

### jenglish

I suppose I didn't read this question thoroughly enough, assuming it was asking for the force on the system

9. Oct 21, 2008

### borgwal

By the way: the action-reaction pair of forces cancels out only when the car and trailer are seen as 1 system. When considering the car by itself there is no such cancellation.

10. Oct 21, 2008

### frosti

thank you guys so much. I got it now.