Forces and Equilibrium: Question 4

Worked solution

Setting up components

Using the given axes, the $40\text{ N}$ force lies entirely along the $x$-direction, so its components are: $(40, 0)$

The $30\text{ N}$ force acts at $70^\circ$ to the horizontal, on the same side as the $40\text{ N}$ force, so its components are: $(30\cos70^\circ,\ 30\sin70^\circ)$

Part (a): Sum of components of the two known forces

$30\cos70^\circ = 30 \times 0.3420\ldots \approx 10.3\text{ N}$ $30\sin70^\circ = 30 \times 0.9397\ldots \approx 28.2\text{ N}$

Adding the $x$-components of the two known forces: $40 + 10.3\ldots \approx 50.3\text{ N}$

Adding the $y$-components (the $40\text{ N}$ force has no $y$-component): $0 + 28.2\ldots \approx 28.2\text{ N}$

So the resultant of the $40\text{ N}$ and $30\text{ N}$ forces has components approximately $(50.3,\ 28.2)$.

Part (b): Magnitude of $F$

For the point to be in equilibrium, all three forces must sum to zero. This means the three forces, drawn tip-to-tail, form a closed triangle (the “triangle of forces”), so $F$ must be equal in magnitude and exactly opposite in direction to the resultant of the other two: $(F_x, F_y) = -(50.26\ldots,\ 28.19\ldots) = (-50.26\ldots,\ -28.19\ldots)$

The magnitude of $F$ is found using Pythagoras: $F = \sqrt{50.26^2 + 28.19^2} = \sqrt{2526.1\ldots + 794.7\ldots} = \sqrt{3320.8\ldots} \approx 57.6\text{ N}$

Part (c): Angle $F$ makes with the horizontal

Both components of $F$ are negative, so $F$ points down and to the opposite side from where the $40\text{ N}$ and $30\text{ N}$ forces act. The angle $\alpha$ this makes below the horizontal satisfies: $\tan\alpha = \frac{|F_y|}{|F_x|} = \frac{28.19\ldots}{50.26\ldots}$ $\alpha = \arctan(0.5610\ldots) \approx 29.3^\circ$

So $F$ acts at approximately $29.3^\circ$ below the horizontal, on the opposite side to the $40\text{ N}$ and $30\text{ N}$ forces.

Final answers

• (a) Sum of $x$-components $\approx \boxed{50.3}\text{ N}$; sum of $y$-components $\approx \boxed{28.2}\text{ N}$
• (b) $F \approx \boxed{57.6}\text{ N}$
• (c) $F$ acts at $\approx \boxed{29.3^\circ}$ below the horizontal, on the opposite side to the other two forces