# Force of an object "skidding"?

Let's say I put the brakes on really suddenly in a car that was moving at 40mph. If I know how much force it takes to break the adhesion of the wheels, how do I find the force the car is exerting? Since I'm not giving it any gas, I won't be accelerating, but it has to be exerting force or I wouldn't be skidding.

In short, how do I find the force of an object that's coasting? I assume it has something to do with momentum, but I'm a newbie so I'm a little lost.

Edit: sorry, this was worded badly. I meant how do I find the force that the object will exert on you if you try to stop it. I knew the engine in the car isn't producing force, I just don't understand what force is causing the car not to adhere to the road. I guess what I'm trying to ask is "where is the force coming from that causes a coasting car to skid?" It makes sense that the car is only slowing down because of the brakes, but what causes the tires to lose adhesion if you apply the brakes too fast and lock up the wheels?

• It's more accurate to ask what force the friction between the locked wheels and the road is and what the influence of that force is on the speed of the car.
– Gert
Oct 30 '15 at 16:54
• There's not enough information. One would need to know how long the skid was (time or distance), or the coefficient of kinetic friction between the tires and the road. The latter would undoubtedly be a dynamic number, changing with, e.g. temperature. The force to break the adhesion relates to the coefficient of static friction, which not helpful in answering your question. Oct 30 '15 at 16:59

Just because an object is in motion, that doesn't mean that there is a force acting on it. Newton's 1st law states that an object in motion will stay in motion unless acted on by an outside force. So the car's momentum is what keep's it going as it coasts. The forces acting on the car would be gravity, the normal force (the ground pushing up on it), and friction slowing it down, but no force is pushing it forward. The skidding is coming from the friction of the wheels touching the road.To sum things up, no force is being applied to the car by you.

First you need to find the force of friction on the wheels of the car using the weight of the vehicle + the passenger. Then, calculate the acceleration. Then, calculate the force needed to accelerate the passenger by the same amount and you've found the force on your passenger.

If I understand the question you are partially asking what causes a skid.

You only get so much friction between two surfaces. After that they slide. Even after they slide there is still friction. Static and kinetic friction

The car is exerting a force. It has a negative acceleration. The car is loosing speed / momentum.

The force that causes the car to skid comes from the brakes.

• Yes, you understood my question correctly. Sorry, I couldn't word it well to save my life :P So the car isn't actually exerting a force just because it's moving, it's the braking force (which causes it to slow down) that produces a load that the wheels can't handle? Oct 31 '15 at 4:39

@K4KFH Your understanding is partially correct. The car's wheels are 'inadvertently' exerting kinetic friction force on the ground (as is the ground equally and in the opposite direction on the car's wheels). The quarter car model, that is the equations of motion modeling each wheel's dynamics in the simplest possible sense are given as $$\tau = I_w \dot{\omega}_w - F_w R_w,$$ where $$\tau$$ is the accelerating, braking or net torque on the wheel (acted through the driveshaft), $$I_w$$ is the moment of inertia of the wheel, $$\omega_w$$ is the wheel speed, $$F_w = \mu F^z_w$$ is the friction force that the ground acts on the wheel contact surface, with $$\mu$$ being the (kinetic when skidding) friction coefficient and $$F^z_w$$ being the normal force acted on the wheel by the car pressing down on it. If there is no skidding, $$v_g = \omega_w R_w$$, where $$v_g$$ is the ground speed of the car's center of mass. Therefore, the wheel slip is defined as $$\lambda = \frac{\omega_w R_w - v_g}{v_g}.$$ Imagine that for a given constant negative torque $$\tau$$ (as in braking), the wheel encounters ice, thus decreasing $$\mu$$ by a large value. Consequently, the value of $$\omega$$ will drop in the short term (over the next few seconds) and cause the wheel to slip ($$\lambda < 0$$). This is the technical description of a skid. ABS and traction control are systems which contain such undesired phenomenon.