# Moments and Laminas

## Moments and Laminas Revision

**Moments and Laminas**

**Note: This topic is not part of the Edexcel spec.**

A **lamina** is simply a flat 2D object with **no thickness**. For** laminae**, we wish to find the **centre of mass** in order to resolve our **moment****s**.

Make sure you are happy with the following topics before continuing.

**Pivoting Around the Centre of Mass**

So, for example, say we have a coin balanced on its side. Its **centre of mass** is right in its centre, so we can conclude that the **moments are equal** on either side of the coin, and the **coin will not rotate**.

Now, let’s suggest we have a coin made of copper on one half, and much heavier gold on the other. Clearly, the **centre of mass** will be **skewed** towards the heavier gold half.

Since the** pivot** is at the point where the coin meets the surface, the only moment in the system is **clockwise**, given by the off-centred **centre of mass**.

The coin will roll until the **centre of mass** lies **directly above the pivot**, as this will guarantee that there is **no clockwise or anticlockwise moment**, and the system will be in **equilibrium**.

**Example: Resolving Moments of Laminae**

Say we have a uniform square with weight \textcolor{red}{700}\text{ N}. The square is held up by a piece of rope supplying **tension** \textcolor{blue}{T} at point A, and its **point of tilting** is at D. Given that the system is in **equilibrium**, what is the value of \textcolor{blue}{T}?

**[4 marks]**

This is a little trickier. When we looked at **Moments** previously, we assumed that our bodies of mass were on the line of action – that’s **no longer the case**.

First and foremost, we need to find the **centre of mass**. Since the square is uniform, the **centre of mass** is at the **centre of the lamina**.

Weight now provides** both a clockwise and anticlockwise moment**.

**Clockwise**, we now have

0.25 \times \textcolor{red}{700}\sin 15° + 0.5\textcolor{blue}{T}

And **anticlockwise**, we have

0.25 \times \textcolor{red}{700}\cos 15°

Since the system is in **equilibrium**, we have 45.29 + 0.5T = 169.04, meaning \textcolor{blue}{T} = 247.5\text{ N}.

## Moments and Laminas Example Questions

**Question 1: **A shop sign is suspended by two pieces of rope at its top corners, A and B. The sign can be modelled as a uniform rectangular lamina with mass 5\text{ kg}. If the rope attached at A snaps, what force must be applied horizontally at the corner beneath B, to keep the sign in equilibrium?

**[3 marks]**

The pivot will be at point B. The centre of mass of the sign is 40\text{ cm} along AB.

Therefore, we have:

Anticlockwise: 0.4 \times 5 \times 9.8 = 19.6\text{ N}

Clockwise: F

So we require

F = 19.6\text{ N}

**Question 2:** A 4\text{ cm} tall plank of wood is resting on a table. The plank is 40\text{ cm} long, with 15\text{ cm} off of the table, and uniform. If the plank weighs 150 \text{ N}, how many 1\text{ kg} weights can be placed on the edge of the plank before it falls over?

**[4 marks]**

The plank’s centre of mass is 5\text{ cm} from the edge of the table, and the edge of the plank is 15\text{ cm} from the edge of the table.

Since the system is explicitly perpendicular, we do not need to use trigonometric functions to correct the system.

Take the corner of the table to be our pivot, and let x be the number of 1\text{ kg} weights applied.

Clockwise: 0.15 \times 9.8x

Anticlockwise: 0.05 \times 150 = 7.5\text{ Nm} – this is supplied by the centre of mass of the plank.

Setting the two to be equal, we have 7.5 = 1.47x, which gives x = 5.102, meaning the system can support 5 weights. Adding a sixth will violate the equilibrium.

**Question 3:** A uniform rectangular lamina of mass 500\text{ g} is pivoted at B.

A string is used to pull the lamina up at A with tension T.

Given that the lamina is at an angle of 20°, what tension should be applied to keep the system in equilibrium?

**[3 marks]**

Taking moments about B:

Clockwise: 0.3 \times 0.5 \times 9.8\cos 20°

Anticlockwise: 0.3 \times 0.5 \times 9.8\sin 20° + 0.6T

Setting the two to be equal, we have

1.47\cos 20º = 1.47\sin 20° + 0.6T

This gives T = 1.464\text{ N (to }3\text{ dp)}.

## MME Premium Membership

£19.99

/monthLearn an entire GCSE course for maths, English and science on the most comprehensive online learning platform. With revision explainer videos & notes, practice questions, topic tests and full mock exams for each topic on every course, it’s easy to Learn and Revise with the MME Learning Portal.

Sign Up Now