Question

A 45 kg wagon is at rest on a rough surface. The coefficient of sliding friction between the wagon and the floor is 0.80. How much force is needed to start the wagon ?.

Answer 1

The force, needed to start the wagon should be larger than the static force of friction, which is $F_f = \mu N$, where $N$ is the normal force, and $\mu$ is the coefficient of friction. Here, the normal force is equal to force of gravity, i.e. $N = mg$. Hence, $F = 0.8 \cdot 45 kg \cdot 9.81 \frac{m}{s^2} \approx 353.2 N$.