Question

A driver is negotiating a turn on a mountain road that has a radius of 40.0m when the 1600 kg car hits a patch of wet road The coefficient of friction between the wet road and the wheels is 0.500 if the car is moving at 30.0km/h will the car skid off the road

Answer 1

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Answer on Question #72583, Physics / Mechanics | Relativity

A driver is negotiating a turn on a mountain road that has a radius of $R = 40.0 \, \text{m}$ when the $m = 1600 \, \text{kg}$ car hits a patch of wet road. The coefficient of friction between the wet road and the wheels is $\mu = 0.500$ if the car is moving at $v = 30.0 \, \text{km/h}$ will the car skid off the road?

Solution:

Car will skid off the road when the centripetal force would be more than friction force

$$F_{\mathrm{c}} > F_{\mathrm{frict}}$$

The centripetal force

$$F_{\mathrm{c}} = \frac{m v^{2}}{R} = \frac{1600 \times (30 \div 3.6)^{2}}{40.0} = 2777.8 \, \mathrm{N}$$

Friction force

$$F_{\mathrm{frict}} = \mu N = \mu m g = 0.5 \times 1600 \times 9.8 = 7840 \, \mathrm{N}$$

Thus

$$F_{\mathrm{c}} < F_{\mathrm{frict}}$$

So car will not skid off the road.

**Answer:** Car will not skid off the road.