# Resultant force on a car

1. Feb 8, 2008

### jelly1500

[SOLVED] Resultant force on a car

1. The problem statement, all variables and given/known data
A 1070 kg car is pulling a 320 kg trailer. Together the car and trailer move forward with an acceleration of 2.01 m/s2. Ignore any frictional force of air drag on the car and all frictional forces on the trailer. Determine the following.
(a) the net force on the car
(b) the net force on the trailer
(c) the force exerted by the trailer on the car
(d) the resultant force exerted by the car on the road
(e) direction (measured from the left of vertically downwards)

2. Relevant equations
F=ma

3. The attempt at a solution
I figured out A, B & C but can't seem to get D & E. I thought the resultant force exerted by the car on the road is Fg=mg=1070(9.8)=10486 but it is wrong.

And for the direction, I used inverse tan (10486/2150.7) but that was wrong too. (2150.7 being the net force on the car)

2. Feb 8, 2008

### andrevdh

The car is experiencing a forwards directing force, lets say F, from the ground and a backwards directed force coming from the trailer, lets say T. These two forces combined gives the resultant force that the car is experiencing.

3. Feb 8, 2008

### jelly1500

So would the forwards directing force be F=2150.7 (from answer a) and the backwards directed force coming from the trailer be T=643.2 (answer b), giving me a total resultant force of 2793.9.

Sorry I'm still a confused.

4. Feb 8, 2008

### andrevdh

Correct. Best make a drawing. That may crystalize the situation for you. What it means is that the force coming from the ground need to supply the resultant force accelerating the car and the trailer.

5. Feb 8, 2008

### jelly1500

Ok, thank you so much.

And for part e, I would then get the inverse tan of 10486/2793.9?

10486 being the force of gravity in the y direction and 2793.9 being the resultant force in the x direction.

6. Feb 8, 2008

### andrevdh

The direction of which force is required - it is not clear for me from the question as you put it?

7. Feb 8, 2008

### jelly1500

Oh sorry for not making that clear: D and E are correlated, so the direction of the resultant force exerted by the car on the road.

8. Feb 8, 2008

### andrevdh

I assume that it is the force from the road on the car. According to my drawing it looks like the tangent is the forward force divided by the normal force.

Last edited: Feb 8, 2008
9. Feb 8, 2008

### jelly1500

oh ok, thank you for all your help!

Last edited: Feb 8, 2008
10. Feb 8, 2008

### andrevdh

I assume you mean the resultant force from the road on the car?

In that case my drawing shows that the tangent is given by the ratio F/N , N being the normal force acting on the car.