Mathematics 9709/43 — October/November 2015

Walkthrough

First, we look at particle $P$. Since it hangs vertically and the system is in equilibrium, the upward tension in the string must exactly balance the downward weight of $P$. This gives $T = 3g = 30 \text{ N}$.

Next, we focus on ball $B$ on the inclined plane. The problem explicitly states that the string parts are vertical. This is the key geometric detail: the tension force on $B$ is directed vertically upwards, not parallel to the slope. The weight of $B$ is also directed vertically downwards. To find the normal reaction $R$, we resolve all forces perpendicular to the inclined plane.

The angle between the vertical and the normal to the plane is $\alpha$. Therefore, both the weight ($4g$) and the tension ($T$) have a component perpendicular to the plane equal to their magnitude multiplied by $\cos \alpha$. The weight component pushes into the plane ($4g \cos \alpha$), while the tension component pulls away from the plane ($T \cos \alpha$). The normal reaction $R$ also pulls away from the plane. Setting the sum of forces away from the plane equal to the sum of forces into the plane gives $R + T \cos \alpha = 4g \cos \alpha$, which simplifies to $R = (4g - T) \cos \alpha$. Substituting the known values yields $R = 8 \text{ N}$.

Key Takeaways

• In equilibrium problems, always start with the simplest particle (here, the vertically hanging $P$) to find unknown forces like tension.
• When resolving forces on an inclined plane, carefully identify the direction of each force. A vertical force on an incline must be resolved using the angle $\alpha$ between the vertical and the normal to the plane.
• The component of a vertical force (weight or vertical tension) perpendicular to an incline of angle $\alpha$ is always $F \cos \alpha$.

Common Mistakes

• Assuming the string attached to $B$ is parallel to the inclined plane. If so, the tension would not have a perpendicular component, and the equation would incorrectly be $R = 4g \cos \alpha$.
• Forgetting to resolve the vertical tension $T$ perpendicular to the plane. Since the string is vertical, it contributes $T \cos \alpha$ to the forces perpendicular to the plane.
• Using $g = 9.8 \text{ m/s}^2$ instead of $g = 10 \text{ m/s}^2$. The mark scheme implies $g = 10 \text{ m/s}^2$ by giving the tension as exactly $30 \text{ N}$.

Things to Be Careful About

• Always read the problem statement for geometric details like "its straight parts vertical". This changes the force resolution significantly compared to a string parallel to the slope.
• Ensure that the angle used for resolving vertical forces on an inclined plane is $\alpha$ (the angle between the vertical and the normal to the plane), not $90^\circ - \alpha$.
• The question asks for the "normal component of the contact force", which is the normal reaction $R$. Do not confuse this with the total contact force, which would include friction.