Question

The kinetic coefficient friction of a rubber tire on wet concrete is 0.70. What is the mass of a tractor that with its brakes locked, can be dragged by a tow truck exerting a horizontal force of 1.0 x104 N?

Answer 1

The friction force with increasing horizontal 'drag' force will be increase too until it reaches the maximum value (Amonton-Coulomb law)

$$\left| {{{\vec F}_{{\text{fr}}{\text{.max}}}}} \right| = \mu N$$

where $N$ - the absolute value of normal force, by the third Newton's law equal to the gravitational force $N = mg$

Then the condition that the tractor can be dragged by a truck is

$$F > \left| {{{\vec F}_{{\text{fr}}{\text{.max}}}}} \right| = \mu mg$$

Thus

$$m < \frac{F}{{\mu g}}$$

Using $g \approx 9.8[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}]$ we've got

$$m < \frac{{1 \cdot {{10}^4}[{\text{N}}]}}{{0.7 \cdot 9.8[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}]}}$$

And

$$m < 1457.7[{\text{kg}}]$$