Edexcel 2024 Paper 3 Question 6 Solution

Edexcel 2024 Paper 3 Question 6 – Moments and Equilibrium Explained

❓ The Question

💡 Teacher Explanation

This is a typical rod question, but with a few small details that matter.

You’ve got forces at different points, a reaction at the peg, and friction at the ground. The key is to keep the diagram clear and not mix up directions.

Part (a) is really just moments about one point. After that, part (b) becomes about resolving forces and linking everything together.

Where people struggle is trying to do too much at once. It’s better to build it step by step.

🎯 How To Recognise This Question Type

Rod + forces + angle → draw diagram → moments → resolve → use $F = \mu R$

🧠 Step By Step Solution

Step 1: Take moments about A (part a)

Take moments about point A to remove unknown reactions there.

You get:
$S \times 1.5a = Mg \times a \cos\theta$

Given $\tan\theta = \frac{4}{3}$ , so:
$\cos\theta = \frac{3}{5}$

Substitute:

$S \times 1.5a = Mg \times a \times \frac{3}{5}$

Cancel $a$ :

$S \times 1.5 = \frac{3}{5}Mg$

So:

$S = \frac{2}{5}Mg$

Step 2: Resolve horizontally (part b)

The only horizontal forces are friction and part of S:

$F = S\sin\theta$

Step 3: Resolve vertically

$R = Mg – S\cos\theta$

At this point, don’t rush — keep everything in terms of S first.

Step 4: Use trig values

From earlier:

$\sin\theta = \frac{4}{5}$
$\cos\theta = \frac{3}{5}$

Substitute into both equations.

Step 5: Use limiting equilibrium

$F = \mu R$

Substitute your expressions for F and R.

Solve carefully — this is where algebra slips happen.

You should reach:

$\mu = \frac{8}{19}$

✅ Final Answer

(a) $S = \frac{2}{5}Mg$
(b) $\mu = \frac{8}{19}$

✔ This would score full marks

🎯 Mark Scheme Breakdown

(a) Moments (3 marks)

M1:
Takes moments about A — must be clearly shown
A1:
Correct moment equation with $\cos\theta$
A1*:
$S = \frac{2}{5}Mg$ seen
Must show use of $\cos\theta = \frac{3}{5}$

✔ Missing the trig value loses the final mark

(b) Resolving and solving (6 marks)

M1:
Resolves horizontally
(e.g. $F = S\sin\theta$ )
A1:
Correct horizontal equation
M1:
Resolves vertically
(e.g. $R = Mg – S\cos\theta$ )
A1:
Correct vertical equation
DM1:
Uses $F = \mu R$ and combines equations
A1:
Final answer
$\mu = \frac{8}{19}$ (≈ 0.42)

What matters here

You don’t have to substitute S immediately — structure matters more
Equations must be clear before solving
If you write multiple equations, only the ones used are marked

Where marks are lost

Wrong direction for forces
Mixing up sin and cos
Trying to take moments badly instead of resolving
Skipping steps when solving for μ

Total: 9 marks

⚠️ Examiner Insight

This question split students a bit more than earlier ones. Some set it up cleanly and picked up most of the marks. Others got stuck early and never really recovered.

Part (a) was usually fine, as long as moments were taken about A. The main issue there was missing the trig value or not showing it clearly.

Part (b) is where things became less consistent. The stronger answers kept things simple — resolve horizontally, resolve vertically, then use $F = \mu R$ . The weaker ones tried more complicated approaches, like taking extra moments, which often led to errors.

A common theme was confusion with directions and trig. Without a clear diagram, it’s easy to mix things up.

This is one of those questions where a calm setup makes a big difference.

💬 Need help with questions like this?

If questions involving forces and moments feel unclear, it’s often because the setup isn’t fully secure.
Working with an A Level maths tutor online support can help you build clearer diagrams and more reliable methods.

🔗 Next Steps

← Question 5 Solution
→ Full Paper Summary

👨‍🏫Author Bio

S Mahandru is an A Level Maths specialist who focuses on helping students improve exam technique and handle multi-step problems with more confidence.

❓ Frequently Asked Questions

📌Why take moments about A in part (a)?

Because it removes unknown forces at A straight away. That makes the equation simpler and avoids extra variables.

📌Do I have to resolve horizontally and vertically?

Not always, but it’s usually the safest method. Other approaches exist, but they’re more likely to lead to mistakes.

📌Why is limiting equilibrium important here?

It tells you that $F = \mu R$ . Without that, you can’t link friction to the normal reaction.

📌What is the most common mistake in this question?

Mixing up directions or trig values. That usually comes from not having a clear diagram at the start.