# How to compute forces acting on car's wheels, when handbrake is on?

I'm trying to make a 2d top down driving game, and I want to be able to drift with handbrake. To make calculations easier and faster, the car has only two wheels. If I can solve the following problem, I know how to compute the rest:

The car is standing still, and there is some force F and torque T acting on it. What forces (A and B) would the two wheels apply to keep the car in place? Friction is supposed to be limitless.

The problem is, that when I write the equations down, I get four unknown, but only three equations:

$$A_x + B_x = -F_x$$ $$A_y + B_y = -F_y$$ $$-r_1 \vec A_x + r_2 \vec B_x = -T$$ Unknowns are $$A_x, A_y, B_x$$ and $$B_y$$.

Here's a picture. X is right and Y up.

• Show your three equations Sep 27 '20 at 13:58
• You also need the height of the center of mass above the roll axis of the car. Sep 27 '20 at 17:27
• Oh, I forgot to tell that the car has only two wheels. Sorry. I now added that to the question. Sep 27 '20 at 18:06
• Where is the steering angle shown on the sketch? Sep 28 '20 at 22:08
• Nowhere. I don't think it matters, because the handbrake doesn't allow the wheels to rotate. It does matter, when grip is lost, but in the problem I'm supposing friction to be limitless (limits can be applied afterwards). It was unclear, so I edited the question. Sep 29 '20 at 4:23

Hand brake actually only locks rear wheels (or one rear wheel in my case), which means that there is one less unknown (exactly what was needed):

$$A \vec N + \vec B = - \vec F$$ $$-r_a A N_x + r_b B_x = -T$$ where $$\vec N$$ is unit vector perpendicular to wheel A, and $$A$$ force in that direction.

Basically the equations say that forces $$A \vec N$$ and $$\vec B$$ must result in force and torque exactly opposite to F and T.

Some algebra: $$A \vec N + \vec B = - \vec F \; | -A \vec N \Rightarrow \\ \vec B = - \vec F -A \vec N$$ $$-r_a A N_x + r_b B_x = -T \Rightarrow \\ -r_a A N_x - r_b(A N_x + F_x) = -T \; | \cdot (-1) \Rightarrow \\ r_a A N_x + r_b(A N_x + F_x) = T \; | -r_b F_x \Rightarrow \\ r_a A N_x + r_b A N_x = T - r_b F_x \Rightarrow \\ (r_a + r_b) N_x A = T - r_b F_x \; | \div (r_a + r_b) N_x \Rightarrow \\ A = \frac {T - r_b F_x} {N_x (r_a + r_b)}$$

Final equations are these: $$A = \frac {T - r_b F_x} {N_x (r_a + r_b)}$$ $$\vec B = - \vec F -A \vec N$$

Force $$A$$ as a vector is $$A \vec N$$.

Here's a picture again: