Today
as I was driving, I noticed that I drive slower with snow on the ground than
when it is not on the ground. Through previous experiences, I knew the the road
feels “slippery” when there is snow on the ground. I believe that this is because
there is less friction with a snow-covered road. This would be caused by a
difference in the coefficient of friction.

In
fact, according “engineering toolbox”, the coefficient of friction of a dry
road is 1, while the coefficient of friction of a wet road is 0.2. This means
the force due to friction for a given object is 5 times greater on a dry road.

F

_{friction dry }/ F_{friction wet }= (μ_{kdry}F_{N})_{ }/ (μ_{kwet}F_{N})
F

_{friction dry }/ F_{friction wet }= (1 F_{N})_{ }/ (0.2 F_{N})
F

_{friction dry }/ F_{friction wet }= (1)_{ }/ (0.2)
F

_{friction dry }/ F_{friction wet }= 5
I
thought it was thus useful to see how the breaking force is different between a
wet and dry road. In order to do this, I modeled the car as a box with four
forces acting on it: friction, applied (brake), normal, and gravity. For the
purpose of this model, there will be no y-acceleration or velocity (F

_{normal}= F_{gravity}). As such, I will only be examining the frictional and applied forces (which act opposite the direction of motion)
For
my first example, I wanted to see how much more force would be needed to stop a
50 kg car moving 15 m/s in 10 m (v

_{i }= 15 m/s and x = 10 m). Again, the only forces I will be exploring are the applied force of the brake and the frictional force. It is important to note the the normal force is equal and opposite to the force of gravity. As such, F_{N}is equal to m*g. Moreover, F_{B}will be used to denote the brake force; it is assumed that the brake force is uniformly applied over the stopping distance.*Dry Road*

ΔKE=−ΔPE+W

_{NC}
½ m (v

_{f}^{2}– v_{i}^{2}) = -mg(h_{f}-h_{i}) + μ_{k}F_{N *}x_{ }+ F_{B}* x_{ }½ (50 kg) [(0 m/s)

^{2}– (15 m/s)

^{2}] = 0 - 1 (50 kg) (9.8 m/s) 10 m + F

_{B}10 m

_{ }-5625 Nm = -4900 Nm + F

_{B }10 m

-725 N m /
10 m = F

_{B}_{ }F

_{B}= -72.5 N

*Wet Road*

ΔKE=−ΔPE+W

_{NC}
½ m (v

_{f}^{2}– v_{i}^{2}) = -mg(h_{f}-h_{i}) + μ_{k}F_{N *}x_{ }+ F_{B}* x_{ }½ (50 kg) [(0 m/s)

^{2}– (15 m/s)

^{2}] = 0 - .2 (50 kg) (9.8 m/s) 10 m + F

_{B}10 m

_{ }-5625 Nm = -980 Nm + F

_{B }10 m

-4645 N m /
10 m = F

_{B}_{ }F

_{B}= -464.5 N

With
a wet road, the break force must be 392 N greater (still opposing the direction
of motion). In other words, because the friction is much less on a wet road,
the brake force much be greater to stop a given car with the the same initial
speed in the same distance.

However,
if the applied force from the brake is at most 100 N, the car would not be
apple to stop in 10 m! As such, it is worth examining how much further the car
would take to stop on a wet road (as compared to a dry road) if the same brake
force was applied. For this example, the 50 kg car still initially had a velocity
of 15 m/s. The applied brake force is assumed to be -100 N.

*Dry Road*

ΔKE=−ΔPE+W

_{NC}
½ m (v

_{f}^{2}– v_{i}^{2}) = -mg(h_{f}-h_{i}) + μ_{k}F_{N *}x_{ }+ F_{B}* x_{ }½ (50 kg) [(0 m/s)

^{2}– (15 m/s)

^{2}] = 0 - 1 (50 kg) (9.8 m/s) x - 100 N x

_{ }-5625 Nm = -490 N x - 100 N x

-5625 Nm = -590 N x

x = 9.5 m

_{}*Wet Road*

ΔKE=−ΔPE+W

_{NC}
½ m (v

_{f}^{2}– v_{i}^{2}) = -mg(h_{f}-h_{i}) + μ_{k}F_{N *}x_{ }+ F_{B}* x_{ }½ (50 kg) [(0 m/s)

^{2}– (15 m/s)

^{2}] = 0 – 0.2 (50 kg) (9.8 m/s) x - 100 N x

_{ }-5625 Nm = -98 N x - 100 N x

-5625 Nm = -198 N x

x = 28.4 m

x

_{wet}/ x_{dry}= 28.4 m / 9.5 m = 2.98
For
a given braking force, this particular car would take approximately 3 times the
distance to slow down on a wet road as compared to a drive road. This explains
why it is helpful to drive slow in snow, as it will take a much greater
distance to slow down for a given velocity. A lower initially velocity would
require a shorter distance as the change in kinetic energy would be lower—thus,
the decreased friction can be offset by driving slower!

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