# Wheel slips and weight

With rainy season starting, I have been thinking about traction and have a question:

Question: If two identical vehicles, one of mass $$m$$ and the other of mass $$2m$$ are starting from rest with equal acceleration, which vehicle's wheels are more likely to slip assuming no deformation of the tire?

I have been wondering about this and I'm thinking that the heavier vehicle's tires would have more friction because of increased normal force and because the angular acceleration of both is equal, the heavier vehicle is less likely to slip. Can someone please help me put this into equations?

The following may be useful. You basically stated the entire relationship which is:

$$F_{friction}=\mu N$$

where $$\mu$$ is the coefficient of static friction between tire and road and $$N$$ is the normal force. If we assume each tire gets $$1/4$$ the vehicle weight, then:

$$F_{friction}=\mu N = \mu \frac{mg}{4}$$

where $$m$$ is the mass of the vehicle, and $$g$$ is the acceleration of gravity.

Thus, as the mass increases, the maximum possible static friction force increases. If this force is exceeded then we get into slippage.

I hope this helps.