Problem 189
Easy Translational Dynamics

To measure the coefficient of friction between asphalt and…

To measure the coefficient of friction between asphalt and rubber, a rubber-covered body of mass $m = 100\ \mathrm{kg}$ is placed on the road. A horizontal pulling force $F = 196\ \mathrm{N}$ moves it uniformly along the road. What is the coefficient of friction?

Explanation of Method

Measuring the friction of rubber on asphalt.

Step 1: Uniform motion

Constant velocity means zero net force. Horizontally the pull balances friction; vertically the road balances the weight: $F = \mu N$.

\[ N = mg. \]

Step 2: Coefficient

\[ \mu = \frac{F}{mg} = \frac{196}{100 \cdot 9.81}. \]

\[\mu \approx 0.20\]

$F = 196\ \mathrm{N}$ $F_f = \mu mg$ $N$ $mg$ $v = \text{const}$ $100\ \mathrm{kg}$ $\text{rubber on asphalt}$
Uniform motion means zero net force: the pull exactly balances the sliding friction, and the road's push balances the weight.

Result

\[\boxed{\,\mu \approx 0.20\,}\]

Remark

Check. The original result is correct (with $g = 9.8\ \mathrm{m/s^2}$ it is exactly 0.2). This is the coefficient of sliding friction; starting the body from rest would take a somewhat larger force.