Mechanics 1 Chapter 8 :: Moments
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Motivating problem
This… is a door.
Why do you think the handle is put on the other side of the door from the hinge?
Increasing the distance of the force applied from the point of rotation increases the ‘turning effect’ of the force.
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If I double the distance of my finger from the hinge, what happens to the force required to keep the door open?
As the distance doubles, the force required halves (we’ll see why).
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Rigid Bodies and Overview
We previously dealt with particles, where each object was modelled just as a single point, and considered forces acting on each point separately.
In this chapter we consider rigid bodies (in this case rods), which takes into account the size of the object.
This means we can consider other properties, e.g. rotation of the body.
Clockwise moment = Anticlockwise moment
1:: Moments in equilibrium
For a ‘non-uniform’ rod we can’t model its weight as acting at the centre.
2:: Centre of Mass
3:: On the point of Tilting
Body rotating about pivot.
Moments
about a point
perpendicular distance
10m
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🖉
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8m
Tom has a mass of 75kg.
700 N
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The anticlockwise moment is greater, so the seesaw will tilt in an anticlockwise direction.
Notice that the measured distance is perpendicular to the force being considered.
(Disclaimer: ex-Tiffinians featured with their permission, and much to their delight)
Quickfire Examples
1
2
3
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We see what force is acting perpendicularly.
4
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5
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Test Your Understanding
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Terminology: A lamina is a 2D object whose thickness can be ignored.
Exercise 8A
Pearson Mechanics 1
Page 131
Resultant moments
If we have multiple coplanar forces, we can also find the overall moment by adding them – just treat one of the directions (clockwise or anticlockwise) as negative. This is similar in Year 1 to finding the resultant force.
1
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2
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3
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4
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(Rod is light)
Angled examples
Always start by drawing in perpendicular distances.
Click to Fro-sketch perpendicular distances
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Test Your Understanding
Click to Fro-sketch perpendicular distances
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Exercise 8B
Pearson Mechanics 1
Page 132-133
This whole chapter in a nutshell…
10m
70kg
🖉 If a rigid body is in equilibrium then:
The resultant force in any direction is 0.
The resultant moment about any point is 0.
a
b
8m
i.e. Forces up = forces down, as per Year 1
In other words, clockwise moments = anticlockwise moments
You will typically use both these properties to solve exam questions.
Example
10m
70kg
8m
Lewis and Tom are having fun on a uniform seesaw of mass 20kg. Lewis weighs 70kg and is 10m from the pivot. Tom is 8m from the pivot. The seesaw remains horizontal.
a) Determine the reaction force at the pivot of the seesaw.
b) Determine Tom’s mass.
Fro Tip: First draw on forces.
Click to Frosketch
Lewis
Tom
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Hint: Choose a suitable point to take moments about.
9m
If a rod is uniform its centre of mass is at its centre, so we model its weight as acting at that point.
Two Pivots
Note: A rod has no mass, so there is no reaction force of the rod on the man, only reaction forces from the pivots.
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Test Your Understanding
Edexcel M1(Old) May 2013(R) Q8
a
b
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Exercise 8C
Pearson Mechanics 1
Page 135=136
Centres of Mass
So far we have assumed that the rod is uniform, that is, its mass is equally distributed across the rod, such that the centre of mass is the centre.
But this may not be the case, and for non-uniform rods we may wish to find where the centre of mass lies, or we will be told where it lies.
Tom Tamsin
Lewis Sam
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Test Your Understanding
Edexcel M1(Old) May 2012 Q2
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? Diagram
Exercise 8D
Pearson Mechanics 1
Page 137-139
Tilting
Lewis
🖉 When a rigid body is on the point of tilting about a pivot, the reaction at any other support (or tension in any other wire/string) is zero.
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Example
Lewis
5m
9m
2m
3m
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Suspended System Example
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Test Your Understanding
Edexcel M1(Old) May 2013 Q6
a
b
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Exercise 8E
Pearson Mechanics 1
Page 140-141