Mathematics 9709/43 — October/November 2018

Walkthrough

First, we identify all the forces acting on the ring $R$. The ring has a mass of $0.2 \text{ kg}$, so its weight is $0.2g$ acting vertically downwards. Using $g = 10 \text{ m/s}^2$, the weight is $2 \text{ N}$. A horizontal force $P$ acts to the left. The string $ARB$ passes through the smooth ring, meaning the tension $T$ is the same in both sections $AR$ and $BR$.

Next, we determine the directions of the tension forces. Section $AR$ is given as inclined at $70^\circ$ to the horizontal. From the diagram, $A$ is above and to the left of $R$, so the tension along $AR$ pulls the ring up and to the left at $70^\circ$ to the horizontal. Section $BR$ is perpendicular to the roof. The roof is inclined at $45^\circ$ to the horizontal. A line perpendicular to a $45^\circ$ slope is at $45^\circ$ to the horizontal. From the diagram, $B$ is above and to the right of $R$, so the tension along $BR$ pulls the ring up and to the right at $45^\circ$ to the horizontal.

To find the tension $T$, we resolve forces vertically. The upward vertical components of the two tension forces must balance the downward weight. This gives the equation $T \sin 70^\circ + T \sin 45^\circ = 0.2g$. Substituting $g = 10$ and solving for $T$ yields $T \approx 1.21 \text{ N}$.

To find $P$, we resolve forces horizontally. The rightward horizontal component of the tension along $BR$ must balance the leftward force $P$ and the leftward horizontal component of the tension along $AR$. This gives the equation $T \cos 45^\circ = P + T \cos 70^\circ$. Rearranging for $P$ and substituting the value of $T$ yields $P \approx 0.443 \text{ N}$.

Key Takeaways

• In equilibrium problems involving a smooth ring on a string, the tension is uniform throughout the string.
• Forces can be resolved into perpendicular components (usually horizontal and vertical) to apply the equilibrium condition $\sum F = 0$.
• Geometrical relationships (like a line being perpendicular to a slope) must be used to correctly determine the angles of the forces.

Common Mistakes

• Forgetting that the tension is the same in both sections of the string because the ring is smooth.
• Incorrectly determining the angle of $BR$ with the horizontal; it is $45^\circ$, not $90^\circ$ or $45^\circ$ to the vertical (though $45^\circ$ to the vertical is the same as $45^\circ$ to the horizontal, care must be taken with the components).
• Sign errors when resolving horizontally; ensuring that leftward and rightward forces are correctly assigned.
• Using $g = 9.8$ or $g = 9.81$ instead of $g = 10$, which is standard for this syllabus unless specified otherwise (the mark scheme values confirm $g=10$ is used here).

Things to Be Careful About

• Always check the direction of the forces to ensure correct signs in the resolution equations.
• The angle $70^\circ$ is given to the horizontal, so the vertical component uses $\sin 70^\circ$ and the horizontal component uses $\cos 70^\circ$.
• Ensure the final answers are given to an appropriate number of significant figures (3 s.f. is standard).
• Verify that the geometry makes sense: the sum of the upward vertical components must equal the weight, and the sum of the leftward horizontal components must equal the sum of the rightward horizontal components.