# Force exerted from car on trailer question

1. Oct 12, 2013

### Joshuava

1. The problem statement, all variables and given/known data

A 1480(k)g car pulls a 300(k)g trailer. The car exerts a horizontal force of 3900(N) against the ground in order to accelerate.

What force does the car exert on the trailer? Assume an effective friction coefficient of 0.15 for the trailer.

2. Relevant equations
My question really is how to solve this problem correctly, but more so, correcting the mistakes I have made in my attempt in gathering a correct solution. I will just put all my work that way anyone who looks at this can easily follow my path and be able to correct my work.

3. The attempt at a solution

Normal force on car = 1480(kg) * 9.81(m/s^2) = 14518.8(N)
Friction on car = 14518.8(N) * 0.15 = 2177.82(N)

Normal force on trailer = 300(kg) * 9.81(m/s^2) = 2943(N)
Friction on trailer = 2943(N) * 0.15 = 441.45(N)

3900(N) in my problem will be working in the positive direction as the other forces are resisting movement due to friction:

F = 3900(N) - 2177.82(N) - 441.45(N) => 1280.73(N)

Now, using Newton's 2nd law of F = m * a, I have rearranged to solve for acceleration:
a = F / m

a = 1280.73(N) / [1480(kg) + 300(kg)] =>0.7195(m/s^2)

F(trailer)=m(trailer) * a(trailer) + any other additional forces acting on the trailer?

F(trailer) = 300(kg) * 0.7195(m/s^2) + 441.45(N) => 657.3(N)

660(N) is my solution.

Would anyone be kind enough to help correct my mistakes? I appreciate you taking the time to take a look!

Last edited: Oct 12, 2013
2. Oct 12, 2013

### lep11

It is said in the problem statement that 0.15 is for the trailer only so does that mean 3900N is a net force so kinetic friction on the car can be neglected?

Last edited: Oct 12, 2013
3. Oct 12, 2013

### Joshuava

This was copied directly from masteringphysics so I really cant answer that question...I just attempted to solve it as best I could, but my answer is not what it is looking for.

That would change the answer significantly.

The new F would be 3458.55(N)

acceleration would change to 1.94(m/s^2)

The new F(trailer) would be 300(kg) * 1.94(m/s^2) + 441.45 => 1024.35(N)?

Last edited: Oct 12, 2013
4. Oct 12, 2013

### lep11

What is it looking for, then?

5. Oct 12, 2013

### Joshuava

I believe the question is looking for the amount of force that is exerted onto the trailer by the car. It wants the answer in 2 significant digits, and with my experience thus far, I don't imagine an answer would be 1.0E3(N)

6. Oct 12, 2013

### lep11

Yes, but do you know what the answer should be? I got T≈1024N

7. Oct 12, 2013

### lep11

Draw free-body diagrams for the car and trailer

Newton second
for Car F-T=mcar*a
for Trailer T-Fμ=mtrailer*a
they have equal acceleration a and T is the force the car exerts on the trailer or vice versa(Newton III) You don't even have to calculate a, just solve for T and substitute it into car equation.

8. Oct 12, 2013

### Joshuava

Awesome, now I understand it all. Thanks so much for your help! By the way, I put in 1000 for the solution and it accepted it, even though it was asking for 2 sig figs.