© john parkinson 1 what was that? © john parkinson 2 moments
TRANSCRIPT
©John Parkinson
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WHAT WAS THAT?WHAT WAS THAT?WHAT WAS THAT?WHAT WAS THAT?WHAT WAS THAT?WHAT WAS THAT?WHAT WAS THAT?
©John Parkinson
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MOMENTS
MOMENTS
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F
FF
F
r
MOMENT of FORCE = F x r
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THE MOMENT OF A FORCE ABOUT A POINT DEPENDS UPON:
MOMENT
UNITS ? NEWTON METRES (Nm)
THE SIZE OF THE FORCE
THE PERPENDICULAR DISTANCE OF THE FORCE’S LINE OF ACTION FROM THE POINT
about a point
= FORCE X PERPENDICULAR DISTANCE of the force from the point
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r
F
O
What is the moment of F about O?
d
Moment = F x d But d = r sin
Hence MOMENT = Fr sin
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FOR A SYSTEM TO BE IN EQUILIBRIUM, THE SUM OF THE CLOCKWISE MOMENTS
ABOUT ANY POINT MUST EQUAL THE SUM OF THE ANTICLOCKWISE MOMENTS
ABOUT THAT POINT
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Clockwise Moment = 6.0N x 4.0 cm =24.0 N cm
Anticlockwise Moment = 8.0N x 3.0 cm =24.0 N cm
THE SYSTEM IS IN EQUILIBRIUM
Clockwise moment
Anticlockwise moment
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Why are these systems balanced?
1
2
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Equilibrium of a Rigid Body Under Coplanar Forces
Equilibrium means that…..
• …there is no rotation. • …there is no acceleration.• …there is no net force acting on the object.
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CONDITIONS FOR THE EQUILIBRIUM OF A BODY
* The vector sum of all the forces acting
on the body is ZERO
* The algebraic sum of all the moments acting
about any point is ZERO
[Otherwise there would be translational motion]
[Otherwise there would be rotational motion]
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This uniform bridge is 20 m long with a mass of 10 tonnes. The lorry has a mass of 20 tonnes and its mass centre is situated 6 m from A.
Using g = 10 N kg-1, Find the reaction force at each support A and B.
A B200 000 N
R1 R2
100 000 N
10 m6 m
20 m
Vertically R1 + R2
= 100 000 + 200 000
= 300 000Taking moments about A eliminates R1
200 000 x 6 + 100 000 x 10 = 20 R2
R1 = 110 000 N
Taking moments about B eliminates R2
200 000 x 14 + 100 000 x 10 = 20 R2
R2 = 190 000 N
Check 190 000 + 110 000 = 300 000 !
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PAIRS OF EQUAL AND OPPOSITE FORCES
THAT PRODUCE ROTATION
F
F
PAIRS OF EQUAL AND OPPOSITE FORCES
THAT PRODUCE ROTATION
PAIRS OF EQUAL AND OPPOSITE FORCES
THAT PRODUCE ROTATION
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THE MOMENT OF A COUPLE
F
F
dd
C
G = MOMENT OF THE COUPLE
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dF
dFG
THE MOMENT OF a COUPLE is called a
TORQUE
HENCE G = F d
TORQUETORQUE
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B
N SI I
F
F
APPLYING FLEMING’S LEFT HAND RULE
THE TWO FORCES PRODUCE A COUPLE ON THE COIL CAUSING ROTATION
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• …of an object is the point at which the entire weight of an object may be considered concentrated.
• Objects can be balanced with a fulcrum under their centre of gravity.
• Objects in free fall rotate about their center of gravity.
CENTRE OF GRAVITY - CENTRE OF MASS
A regular object has its Centre of Mass at its geometrical centre.
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WV03 VXP
G
mg
d
mg x d is a clockwise moment
That’s OK!
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WV03 V
XP
mg x d is now an anticlockwise moment!
mg
G
d
WV03
VXP
WV
03 V
XP
WV
03 V
XP
WV
03 V
XP
Oh
dear!
But at least WE
know about
Moments
d
mgmg x d kicked it over !!