A tractor of mass m_2 drags a rock of mass m_1 = 7 t over…

A tractor of mass $m_2$ drags a rock of mass $m_1 = 7\ \mathrm{t}$ over level ground. The coefficient of friction between the rock and the ground is $\mu_1 = 0.4$; between the tractor's wheels and the ground it is $\mu_2 = 0.2$. What is the smallest tractor mass for which it can drag the rock?

Explanation of Method

Smallest tractor that can drag a rock.

Step 1: What pushes the tractor forward

The engine turns the wheels, but the force that actually moves the tractor is the ground's static friction on the wheels. It can be at most $\mu_2 m_2 g$ (all the weight on the driving wheels).

Step 2: Condition

To drag the rock at constant speed, this must at least equal the rock's sliding friction: \[\mu_2 m_2 g \ge \mu_1 m_1 g \;\Longrightarrow\; m_2 \ge \frac{\mu_1}{\mu_2}\,m_1 = \frac{0.4}{0.2} \cdot 7\ \mathrm{t}.\]

\[m_{2,\min} = 14\ \mathrm{t}\]

$7\ \mathrm{t}$ $\mu_1 m_1 g$ $\le \mu_2 m_2 g$ $m_2$ $\mu_2 m_2 g \ge \mu_1 m_1 g \ \Rightarrow\ m_2 \ge 14\ \mathrm{t}$

The tractor is pushed forward only by the ground's static friction on its wheels, which can be at most $\mu_2 m_2 g$. That must exceed the sliding friction holding the rock back.

Result

\[\boxed{\,m_{2,\min} = 14\ \mathrm{t}\,}\]

Remark

Check. The original result is correct. A lighter tractor would just spin its wheels, however strong its engine. Getting the rock moving from rest needs a little more, since static friction is larger than sliding friction.