force question?

Question

A 920 kg car is pulling a 290 kg trailer. Together, the car and trailer have an acceleration of 2.24 m/s2 in the forward direction. Neglecting frictional forces on the trailer, determine the following (including sign).

(c) the force exerted by the trailer on the car
____N
(d) the resultant force exerted by the car on the road
magnitude
____N
direction
____° measured from the left of vertically downwards

I am having trouble with these problems; can someone please explain how to do them\? thank you

Answer 1

(c) f = -m(trailer)*a
(d) Assume the car's moving to the right, and that "measured from the left of vertically downwards" means measured toward the left from downwards (i.e., in the CW direction from -y).
First solve for vector in conventional x-y system and its magnitude. Then find vector angle CW from -y:
fx = -m(car)*g
fy = -m(car+trailer)*a
mag(f) = sqrt(fx^2+fy^2)
theta = -arctan(fx/fy)
The angle is not the more familiar arctan(fy/fx) because we want this output in the CW direction relative to -y. The minus sign changes CCW to CW, and the fx and fy interchange accomplishes the 90 deg rotation of the reference axis.
If you have trouble understanding the signs, draw a picture of, or imagine, the car and trailer going right (in +x direction), and the car's f vector along -y (weight) and -x (car pushing back on road to accelerate forward).
The problem itself is based on f=ma; the idea of a vector in any direction being the sum of its x and y components; the Pythagorean theorem (for magnitude); and trigonometry (for angle).