forces of static friction question

Then from what you stated wouldn't the force of friction be causing it to move forward but their would have to be some other friction opposing that motion or is that taken into account for Ffs? And isn't static friction a amount ranging to some maximum unroll the object starts to move, so since their is an acceleration their isn't static but only kinetic friction?
 The friction opposing motion is actually the air resistance. As for your other question, it is static friction. There is no kinetic friction involved unless the wheels start slipping (you're peeling out.) If your wheels are not slipping, the wheels are not sliding across the road. In fact, if there were no acceleration, there is no friction at all, static or kinetic.
 K yess I think I just clicked the air was opposing it to move and the static friction was actually opposing this in the other direction making it the "applied" force. K I have one last question I think. Even though the Ffs is acting in the opposite direction wouldn't their be a little bit of friction acting in the same direction as air because the tires are touching the ground or is it because of the Ffs the car is able to move forward and their would only be friction in the same direction as Fair if the brakes were on and the tires were not rotating?
 Tire friction opposes the tire's direction of motion. There is no tire friction that goes in the same direction of motion of the tires. Friction never does that.
 Kk I'm good now thank you so much for the help I really appreciate it I wasn't thinking about the tires direction I was thinking about the car body itself.

Recognitions:
Homework Help
 Quote by chiuda 1. The problem statement, all variables and given/known data a 1450 kg car is towing a trailer of mass 454 kg. The force of air resistance is 7471N[backward]. If the acceleration of both vehicles is 0.225m/s^2[forward], what is the force of static friction on the wheels from the ground? 2. Relevant equations I am able to find Force normal(Fn) by multiplying its total mass by force of gravity because force normal is equal to force gravity (Fg) because the car is neither hovering or sinking in the surface of which it moves across. I can also find Fnet horizontal (Fnet,h) because i have mass and i have acceleration. but even with these values i am still missing the coefficient for static friction and force applied (Fapp) and force of friction static (Ffs) 3. The attempt at a solution Fnet,vert=Fn+(-Fg) 0=Fn-Fg Fn=Mxg Fn=1904kgx9.81m/s^2 Fn=18678N Fnet, horiz= MxA = 1904kgx9.81m/s^2 = 4.4N I then do not really no what to do other than: U(coefficient of static friction) Ffs=UxFn =Ux18678N Fnet, horiz=(-F,air)+(-Ffs)+Fapp 4.4N=-7471-Ffs+Fapp 7475.4=-Ffs+Fapp I am now stuck as you can see and have made many othe attempts in different ways but they all ended up fairly similar, ANY HELP AT ALL WILL BE VERY MUCH APPRECIATED!!!!!:)
back to square 1.

I think you have forgotten the original question.

The two bits I have highlited red show you the Net force.

Fnet = ma

m = 1904 kg
a = 0.225 ms-2

From that you can calculate the nett force.

Lets suppose the answer to that is 400N (you can calculate the real value)

What makes up the net force?

It is usually the sum of a Forward Force and a backwards force.

The backwards force is the air resistance - given as 7471N

If the net Force is indeed 400N (forward), there must be a forward force of 7841N ( you can again calculate what the real value has to be) acting on the car.

What could be supplying that Force?

Hint: Unless it is due to a gravitational Field [that is down; not forward], magnetic Field (no) or electric Field (no) it must be a CONTACT force.

What is touching the car - or some part of the car?

Hint 2: It is not the air touching the car - that has already been taken care of in the air resistance figure of 7471N [backward].

EDIT: Whoops, while I was composing I think you were finding the solution.

 Tags force, friction, frictional forces, kinetic friction, static fiction