Simple forces diagram of a car maintaining speed up a slope I'm currently in training to be a high school science teacher in Australia and I'm looking at my year 7 class's test for this term. I think the answer to one of the questions on the answer sheet is wrong and I wanted to check to see what you all thought before I send my supervising teacher an email and look like an ass.
The question asks students to draw a 2D force diagram (free-body diagram) for a car maintaining speed while going up a slope. The answer given in the original answer sheet is in blue.
I think this answer is wrong for a couple of reasons:

*

*Intuition- when coming up to a slope in your car you need to push down on the accelerator harder to maintain your speed. The need for that much added force is doubtful to be from increased friction with the road or drag from air resistance. This is why I think the forward vector needs to be longer than the backwards vector.

*The vectors don't cancel out and therefore could not be maintaining the same velocity, i.e. balanced force. The answer in blue doesn't take into account that the normal force actually pushes the car backwards perpendicularly to the slope of the road, and therefore there needs to be additional force acting in the forwards direction to compensate.

I put my answer in red and proved the vectors cancel out next to it (I just focused on the wheel so I had a good reference point). Physics isn't my strong suit, I'm a bio and chem nerd, so I just wanted to check if I had some misunderstanding before shouting foul at my supervisor.
If anyone can confirm my thoughts or shoot me down, that would be much appreciated.
Thanks a lot
Mr. Weisbar
 A: I like the second diagram.  The downward force is the weight of  the car (from gravity). (That can be broken into components; one backwards and one perpendicular to the road. The normal force (up and to the left is from the road reacting to the perpendicular component of gravity. The backward force (from friction with the air and deformation of the tires) would work the backward component of gravity. Finally, the forward force is a static friction force from the road acting on the bottom of the tires (resulting from wheels being turned by the engine).  At a constant speed, the vector sum of these forces will be zero.
A: At the level you are asking about the system that you are considering (the car) should be treated as a point mass at the centre of mass of the car.
After all in a real car there are two (four) normal forces, all not necessarily equal, and the same number of frictional forces which are not shown separately.
So the first or second diagram might be acceptable but certainly not the third one.
Although the length of the vectors are often an indication of the relative magnitudes of the forces this is not always the case.
However, given that you have stated that your scheme of requires free body diagrams to show the relative magnitudes  then the second diagram does follow the convention used at your school although for year 7 students I would be pleased if they had drawn the first diagram (with perhaps a bonus for the second diagram).
Even then you would have to check that the lines of action of all the forces went though the same point (which is not true for diagrams one and two) and check the relative lengths of the normal and gravitational force vectors!
MIT have produced a booklet about Free Body Diagrams which you may find of use?
