Moments

The Principles of Moments

Moments are quite simple to calculate. There are two things we need to know:

• Each body’s perpendicular distance from the point of rotation
• The force applied by each body

Imagine a seesaw with one child at either side.

If both children are sat as far from the centre as each other, but one child weighs more, the seesaw will lower on their side.

If both children weigh the same, but one child comes closer to the centre as the other stays in place, the seesaw will lower on the side of the furthest of the two.

Hence, the larger the force, and the greater the distance, the larger the moment.

Moments can be calculated by:

Moment $=$ Force $times$ Perpendicular Distance from the force’s Line of Action

Weightless Beams

So, let’s say we’ve got a $color{limegreen}2text{ m}$ long beam, balanced about its centre. We have a coin on one end of the beam, with a mass of $textcolor{purple}{10text{ g}}$. Assuming $g = 10text{ ms}^{-2}$, how much downward force must we apply $textcolor{orange}{40text{ cm}}$ away from the centre of the beam to keep the system in equilibrium?

Let the coin be acting in a clockwise direction.

Clockwise moment: $1 times textcolor{purple}{0.01} times 10 = 0.1text{ Nm}$

Anticlockwise moment: $textcolor{orange}{0.4} times F = 0.1text{ Nm}$

This means that we require $F = dfrac{textcolor{purple}{0.1}}{textcolor{orange}{0.4}} = 0.25text{ N}$.

Uniform Rods

The previous example we assumed that the beam was weightless. While that does make our calculation easier, it isn’t quite right to use in a practical scenario.

Let’s now say that the beam is balanced $color{limegreen}1.5text{ m}$ away from the coin, as below.

The beam now weighs $textcolor{blue}{20text{ g}}$. Assume, again that $g = 10text{ ms}^{-2}$ and that the system is in equilibrium.

We have to consider the centre of mass of the beam, as a new source of clockwise moments.

The beam’s centre of mass is $textcolor{orange}{0.5text{ m}}$ away from the pivot.

Clockwise moment: $(textcolor{orange}{0.5} times textcolor{blue}{0.02} times 10) + (textcolor{limegreen}{1.5} times textcolor{purple}{0.01} times 10) = 0.25text{ Nm}$

Anticlockwise moment: $0.5 times F = 0.25text{ Nm}$

So we require $F = dfrac{0.25}{0.5} = 0.5text{ N}$.

Non-Uniform Rods

Let’s say the rod still weighs $textcolor{blue}{20text{ g}}$, but its centre of mass is no longer at the centre of the rod. Now, we exert a force of $F = 0.58text{ N}$.

Given that the rod is still balanced $textcolor{limegreen}{1.5text{ m}}$ away from the coin and the system is still in equilibrium, find out where the centre of mass of the rod lies.

Let $x$ be the distance from the line of action.

Clockwise moment: $(x times textcolor{blue}{0.02} times 10) + (textcolor{limegreen}{1.5} times textcolor{purple}{0.01} times 10)$

Anticlockwise moment: $0.5 times 0.58 = 0.29text{ Nm}$

$x = dfrac{0.29 – (textcolor{limegreen}{1.5} times textcolor{purple}{0.01} times 10)}{(textcolor{blue}{0.02} times 10)} = 0.7text{ m}$.

Point of Tilting

Sometimes, a rod will rest on multiple supports. A rod is said to be at the point of tilting about a support if the reaction force at all other supports is $0$. Even at the point of tilting, the rod is about to tilt rather than actually tilting, so moments are still balanced around the pivot.

Example: The rod below is at the point of tilting around support $A$. There are two weights attached to it, which are shown. One weight has a mass of $color{limegreen}2color{grey}kg$. Find the mass of the other weight, and the reaction force at support $A$.

Since we are at the point of tilting about $A$, we know that $R_{B}=0$.

Take moments about $A$.

$color{orange}0.1color{grey}Mg=color{blue}0.15color{grey}timescolor{limegreen}2color{grey}g$

$0.98M=2.94$

$M=dfrac{2.94}{0.98}$

$M=3$

Resolve vertically:

$R_{A}+R_{B}=Mg+color{limegreen}2color{grey}g$

$R_{B}=0$

$M=3$

$R_{A}=3g+color{limegreen}2color{grey}g$

$R_{A}=5g$

$R_{A}=49$