Question

A 20-kg wagon is pulled along the level ground by a rope inclined at 30° above the horizontal. A friction force of 30 N opposes the motion. How large is the pulling force if the wagon is moving with an acceleration of 0,40 m/s2? Give the answer with one decimal and no units.

Answer 1

The horizontal component of the force is:

$$F_h = F\cos 30\degree = \dfrac{F\sqrt{3}}{2}$$

where $F$ is the required pulling force. According to the second Newton's law along the horizontal line, have:

$$F_h-F_{fr} = ma$$

where $F_{fr} = 30N$ is the friction force, $m = 20kg$ is the mass and $a = 0.4m/s^2$ is the accleration of the wagon. Substituting the expression for $F_h$ and expressing $F$, obtain:

$$\dfrac{F\sqrt{3}}{2} - F_{fr} = ma\\ \dfrac{F\sqrt{3}}{2} = ma+F_{fr} \\ F = \dfrac{2}{\sqrt3}(ma+F_{fr})\\ F = \dfrac{2}{\sqrt3}(20\cdot 0.4+30) \approx 43.9N$$

Answer. 43.9.