Transcript S1: Chapter 1 Data: Location

M1: Chapter 5
Moments
Dr J Frost ([email protected])
Learning Objectives:
• Understand what is meant by the moment of a force.
• Use moments to solve problems involving unknown
lengths, forces and masses.
Last modified: 2nd April 2014
Intro
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.
?
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).
?
Moments
m = 70kg
10m
Moment = 𝟕𝟎𝒈 × 𝟏𝟎 =
? 𝟔𝟖𝟔𝟎 𝑵𝒎
!
The ‘moment’ of a force…
… measures the turning effect of the force on the body on which it is
?
acting.
𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒇𝒐𝒓𝒄𝒆 = 𝒇𝒐𝒓𝒄𝒆 × 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
about a point
perpendicular distance
Moments - Examples
1
2
𝐹 = 20𝑁
3
10𝑁
𝐹 = 20𝑁
10𝑚
60°
𝑃
5𝑠𝑖𝑛60
𝑃
Moment of force F
about point P
= 𝟐𝟎𝟎𝑵𝒎 clockwise
?
Moment of force F
about point P
= 𝟎𝑵𝒎
𝑃
We see what
force is acting
perpendicularly.
Moment of force F
about point P
= 𝟓𝟎𝒔𝒊𝒏𝟔𝟎 𝑵𝒎
?
clockwise
?
There is no ‘turning’ effect.
4
5𝑚
5
10𝑁
5𝑁
30°
2𝑚
3𝑚
𝑃
Moment:
𝟐𝟎𝒔𝒊𝒏𝟑𝟎 𝑵𝒎 ?
𝒂𝒏𝒕𝒊𝒄𝒍𝒐𝒄𝒌𝒘𝒊𝒔𝒆
65°
𝑃
Moment:
𝟏𝟓𝒄𝒐𝒔𝟔𝟓 𝑵𝒎 ?
𝒂𝒏𝒕𝒊𝒄𝒍𝒐𝒄𝒌𝒘𝒊𝒔𝒆
Summing moments
We can also find the overall moment by summing them – just treat one of the directions
(clockwise or anticlockwise) as negative.
3𝑁
5𝑁
2
1
20𝑁
𝑃
2𝑚
1𝑚
1𝑚
10𝑚
10𝑁
Sum of moments (about 𝑃):
15𝑚
𝟏𝟓 − 𝟒 − 𝟑
4𝑁
?
𝑃
= 𝟖𝑵𝒎 𝒄𝒍𝒐𝒄𝒌𝒘𝒊𝒔𝒆
Sum of moments (about 𝑃)
= 𝟐𝟎𝟎 − 𝟏𝟓𝟎
= 𝟓𝟎𝑵𝒎?clockwise
3
1𝑚
𝑚 = 50𝑘𝑔
𝑃
5𝑁
𝑥
4
4𝑚
𝑚 = 30𝑘𝑔
𝑃
3𝑚
𝑚 = 30𝑘𝑔
(Rod is
light)
Sum of moments (about 𝑃):
𝟑 × 𝟑𝟎𝒈 − 𝟏 × 𝟓𝟎𝒈
= 𝟒𝟎𝒈 𝑵𝒎?𝒄𝒍𝒐𝒄𝒌𝒘𝒊𝒔𝒆
Sum of moments (about 𝑃):
𝟑𝟎𝒈 𝒙 − 𝟒 𝑵𝒎
?
This whole chapter in a nutshell…
m = 70kg
10m
2m
! If a rigid body is in equilibrium then:
a The resultant force in any direction is 0.
b The sum of the moments about any point is 0.
i.e. Forces up = forces down
i.e. Clockwise moments =
anticlockwise moments
Example
Lewis and Tom are on a uniform seesaw of mass 20kg, where the pivot is not central. Lewis
weighs 70kg and is 10m from the pivot. Tom is 2m from the pivot. The seesaw is horizontal.
Determine the reaction force at the pivot of the seesaw.
Determine how heavy Tom is.
𝑅
m = 70kg
m=?
Lewis
Tom
2m
10m
20𝑔
70𝑔
Bro Tip: First draw on forces.
Click to Brosketch
Method 1:
Equating moments about pivot:
𝟐𝟎𝒈 × 𝟒 + 𝟕𝟎𝒈 × 𝟏𝟎 = 𝒎𝒈 × 𝟐
𝟕𝟖𝟎𝒈 = 𝟐𝒎𝒈 → 𝒎 = 𝟑𝟗𝟎𝒌𝒈
?
Equating forces in vertical direction, 𝑹 =
𝟒𝟖𝟎𝒈 = 𝟒𝟕𝟎𝟒𝑵
𝑚𝑔
Hint: Choose a suitable point to
take moments about.
Method 2:
Equating moments about Tom:
𝟐𝑹 = 𝟔 × 𝟐𝟎𝒈 + 𝟏𝟐 × 𝟕𝟎𝒈
𝑹 = 𝟒𝟖𝟎𝒈 = 𝟒𝟕𝟎𝟒𝑵
Equating forces in vertical direction:
𝟕𝟎𝒈 + 𝟐𝟎𝒈 + 𝒎𝒈 = 𝑹
𝒎 = 𝟑𝟗𝟎
?
Moments about which point?
Key Point: You can choose different points to take your moments about,
depending on which variable we wish to ‘ignore’.
𝑅
m = 70kg
m=?
Lewis
10m
70𝑔
Tom
2m
20𝑔
𝑚𝑔
If we take moments about the pivot:
We don’t have to worry about 𝑅 since it
? about the pivot.
doesn’t have any moment
If we take moments about Tom:
We don’t have to worry about Tom’s mass 𝑚 because Tom’s
? moment about Tom.
weight doesn’t contribute to the
Test Your Understanding
Q A uniform rod of mass 20kg is supported at each end so that it is horizontal.
The bar is 4m long and a mass of 10kg is placed 1m from one end.
Find the magnitude of the reactions at each support.
4𝑚
2𝑚
𝑅1
? Diagram
20𝑔
Moments about second support:
10𝑔 + 40𝑔 = 4𝑅1
50
𝑅1 =
𝑔 = 12.5𝑔
4
? Working
Similarly taking moments about first:
40𝑔 + 30𝑔 = 4𝑅2
𝑅2 = 17.5𝑔
10𝑔
1𝑚
𝑅2
Finding centre of mass
Q A non-uniform rod 𝐴𝐵 of length 4m and mass 5kg is in equilibrium in horizontal position resting on
two supports at C and D where 𝐴𝐶 = 1𝑚 and 𝐴𝐷 = 2𝑚. The magnitude of the reaction at 𝐶 is twice
the magnitude of the reaction at 𝐷. Find the distance of the centre of mass of the rod from 𝐴.
𝑥
2𝑚
2𝑚
1𝑚
𝑅
2𝑅
? Diagram
𝐴
𝐶 𝐶
𝐷
𝐵
5𝑔
Let centre of mass be 𝑥 metres from 𝐴.
Forces up = forces down:
3𝑅 = 5𝑔
Moments about 𝐴:
→
5
𝑅= 𝑔
3
? Working
2𝑅 + 2𝑅 = 5𝑔𝑥
4𝑅 4
𝑥=
=
5𝑔 3
Exam Questions
On provided hand-out.
(Solutions on following slides)
Answers
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Answers
Hint:
When the beam is on the verge of
tilting about P, there is no reaction
force at Q.
?
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Answers
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Answers
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Answers
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Answers
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Summary
Moment of a force = 𝒇𝒐𝒓𝒄𝒆 × ?
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
C LO: Understand what is meant by
the moment of a force.
What are the two crucial things to consider for systems in equilibrium?
1. Forces up = Forces down
2. Moments clockwise = ?
Moment anticlockwise
C LO: Solve problems
involving moments.
Where tends to be best to take moments about?
• When there is an unknown distance:
Sometimes about the point where the distance is measured from (but if
?
in doubt, take from the end).
• When there are two unknown forces:
About the point where one of the forces is acting, so that we don’t
consider this force in our moments equation
and thus have less
?
unknowns.
How do we deal with rods with unknown centres of mass?
Create a distance say 𝒙 for the centre of mass. Usually best to find
moments about the end of the rod. ?
If a rod is on two supports P and Q, and is on the verge of tilting about P?
There is no reaction force at Q.
?