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 ?.
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".
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