Mathematics 9709/41 — May/June 2019

Walkthrough

The problem asks us to prove that a system of three coplanar forces is in equilibrium. A system is in equilibrium if and only if the vector sum of all forces acting on it is zero. The most direct way to verify this is to resolve each force into its horizontal and vertical components and check that they sum to zero in both directions.

Step 1: Trigonometric ratios. The problem gives $\tan \alpha = \frac{12}{5}$ and $\tan \theta = \frac{4}{3}$. To resolve the forces, we need $\sin$ and $\cos$ for these angles. Using Pythagoras' theorem on the implied right-angled triangles, we find the hypotenuses are $13$ and $5$ respectively. This gives us the exact fractions: $\sin \alpha = \frac{12}{13}$, $\cos \alpha = \frac{5}{13}$, $\sin \theta = \frac{4}{5}$, and $\cos \theta = \frac{3}{5}$. Using exact fractions rather than decimal approximations is crucial to avoid rounding errors and to show the components cancel perfectly.

Step 2: Horizontal resolution. We set up a horizontal axis. Taking right as positive, the $50\text{ N}$ force pushes right ($+50 \cos \theta$), the $78\text{ N}$ force pulls left ($-78 \cos \alpha$), and the $112\text{ N}$ force is vertical so it contributes nothing. Substituting the values gives $30 - 30 = 0$.

Step 3: Vertical resolution. Taking upwards as positive, both the $78\text{ N}$ and $50\text{ N}$ forces have upward components ($+78 \sin \alpha$ and $+50 \sin \theta$), while the $112\text{ N}$ force pulls down ($-112$). Substituting the values gives $72 + 40 - 112 = 0$.

Step 4: Conclusion. Because both the net horizontal and net vertical forces are exactly zero, the resultant force is the zero vector, which is the definition of equilibrium for a particle under coplanar forces.

Key Takeaways

• To prove equilibrium, resolve all forces into two perpendicular directions (usually horizontal and vertical) and show that the resultant in each direction is zero.
• When given $\tan \theta$, always construct the right-angled triangle to find exact $\sin \theta$ and $\cos \theta$ values to ensure precise cancellation in equilibrium problems.
• Sign conventions (e.g., right is positive, up is positive) must be applied consistently when resolving forces.

Common Mistakes

• Using decimal approximations for the angles (e.g., $\alpha \approx 67.4^\circ$) and then multiplying, which can lead to small rounding errors that make the components look like $0.01$ instead of $0$. Always use exact fractions.
• Forgetting the negative sign for a force component that acts in the negative direction of the chosen axis (e.g., taking the leftward $78\text{ N}$ component as positive in the horizontal resolution).
• Confusing $\sin$ and $\cos$ when resolving: the horizontal component uses $\cos$ (adjacent to the angle with the horizontal axis) and the vertical component uses $\sin$ (opposite to the angle with the horizontal axis).

Things to Be Careful About

• Ensure the angle given in the diagram is correctly matched to the trigonometric ratio. Here, $\alpha$ is measured from the negative horizontal axis, so its horizontal component is $78 \cos \alpha$ and vertical is $78 \sin \alpha$. Similarly, $\theta$ is measured from the positive horizontal axis.
• The mark scheme also accepts Lami's theorem as an alternative method. If using Lami's theorem, one must correctly identify the angles between the forces: $180^\circ - (\alpha + \theta)$, $90^\circ + \theta$, and $90^\circ + \alpha$, and then show that $\frac{F_1}{\sin A_1} = \frac{F_2}{\sin A_2} = \frac{F_3}{\sin A_3}$.
• Explicitly state the conclusion $X = 0$ and $Y = 0$ (or equivalent) to earn the final mark for showing equilibrium; simply calculating the components is not enough.