Transcript Moments and Couples and Reactions

Last Lecture Review 1
Two horizontal forces act on a block that is sliding on ice.
Assume that a third horizontal force F also acts on the
block.
What is the magnitude and direction of F when the block is:
a) Stationary
b) Moving to the left at constant speed?
3N
5N
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Last Lecture Review 2
In 3 minutes, describe to the person sitting next to you:
a) What is a force?
b) What change does a force (or system of unbalanced
forces) make to a body?
c) What is equilibrium?
Then discuss with the class.
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Last Lecture Review 3
Select the correct FBD at particle A.
30
A 40
100 N
F1
A)
A
F2
B)
30
40°
100 N
A
F
C)
30°
D)
F2
40°
F1
30°
A
A
100 N
100 N
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Today:
• Mechanical Units, Scalars and Vectors;
• Statics: Forces and Equilibrium, Moments
and Couples and Reactions, Centre of
Gravity and Centroids, Distributed Force;
• Dynamics: Constant velocity and constant
acceleration in linear systems, circular
systems;
• Work, Power and Energy
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Moments and Couples
Objectives :
Students will be able to:
a) Calculate a moment and it’s direction;
b) Solve problems involving moments of a force and couples in 2D
c) Calculate reaction forces
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Moment in 2-D
F
M
d
O
The moment of a force about a point is a measure
of the tendency for rotation (sometimes called a
torque).
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The magnitude of the moment about point O is:
Mo = F.d
where d is the perpendicular distance from point O to the line of action
of the force F.
In 2-D, the direction of MO is either clockwise or anti-clockwise
depending on the direction of the tendency for rotation.
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a
For example, MO
= F.d
F
b
and the direction is anti-clockwise.
O
d
However, it is sometimes easier to determine MO by using the
components of F as shown.
F
Fy
a
b
O
Fx
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Moment at Knee with Distance
Compare the applied
moment at these leg
positions.
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F
Fy
Here,
a
b
Fx
MO = (FY a) – (FX b)
O
Note the different signs on the terms.
The sign convention for a moment in 2-D is anti-clockwise is positive.
We can determine the direction of rotation by imagining the body pinned
at O and deciding which way the body would rotate because of the
force.
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Check-in Quiz
F = 10 N
A
d=3m
What is the moment of the 10 N force about point A?
a) 10 N·m anti-clockwise
b) 30 N·m anti-clockwise
c) 13 N·m clockwise
d) (10/3) N·m clockwise
e) 7 N·m clockwise
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Check-in Quiz
Which of these exercises cause a moment at the
shoulder?
a)
d)
b)
c)
e)
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Check-in Quiz - Answer
Which of these exercises cause a moment at the
shoulder?
a)
d)
b)
c)
e)
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Practice Problem
What is the
moment of the
30N force on the
wheel nut?
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Equilibrium of Moments
When a system is in equilibrium, the moments, as well as
the forces acting on the system must be balanced.
i.e.
Ʃ Fx =
Ʃ Fy =
ƩM =
0
0
0
For equilibrium, the sum of the moments must be equal to
zero.
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Worked Example
0.6 m
0.52 m
Fe
W
R
The wheelbarrow shown carries a load of 58 kg. Calculate the force Fe
required to hold the handles in a stationary position.
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Solution
First, calculate the weight of the load:
W
=
=
=
m.g
58 x (9.8)
568 N
Now, equate the moments of the two forces about the pivot point, which
is the axis of the wheel, and solve for the unknown force Fe
568 x 0.52
=
Fe x 1.12
Fe
=
(568 x 0.52)/1.12
=
264 N
Hence the total effort required is 264 N, or 132 N per handle.
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Check-in-Quiz
• The drawing shows two different wheelbarrow designs.
• Which design requires more human force to hold steady?
• If the wheels, axle, and handle of a wheelbarrow weigh W = 57.0N
and the load weighs WL = 510N, what is the difference in force? 18
Reactions
When a force F is applied at a point away from the
centre of rotation on a rigid body, the body
experiences a force as well as a moment.
300
mm
60
N
Force applied to
spanner to turn a nut
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The reaction at the nut is therefore equal and opposite
for the moment and the force, so that the system is
equilibrium.
300
mm
18 N.m
60 N
Reaction force
and moment on
the nut if the
system is in
equilibrium
60 N
Force applied to
spanner to turn a nut
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Example
A bolt is being tightened by a force of 60 N applied to a spanner at a
distance of 300 mm. What is the reaction force and moment from the
bolt at point B if the system is in equilibrium?
The force-moment system at B comprises a force of 60 N and a turning
moment equal to
M =
60 x 0.3
=
18 N.m
The reaction force is equal and opposite in direction to the tightening
force, and the turning moment is in the opposite direction.
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Reactions of Supports
Smooth
surface
Roller
support
Cable or
link
Sliding
pin
Rough
surface
Angle depends
on friction
Pinned
support
Fixed
support
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Exercise
A force F is applied at the end of the
rib in an accident.
Determine the reaction force and
moment at the spine end of the rib
if the rib does not move or rotate.
M
Fw
0.3 m
F = 1.3 kN
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Check-in Quiz
y
1m
P
1
m
30 N
x
40 N
30 N
For this force system, the reaction at the wall P is ______ .
A) FRP = 40 N (along - x dir.) and MRP = 60 N.m anticlockwise
B) FRP = 0 N and MRP = 30 N.m anti-clockwise
C) FRP = 30 N (along - y dir.) and MRP = 30 N.m clockwise
D) FRP = 40 N (along - x dir.) and MRP = 30 N.m anticlockwise
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Problem Question
y
100mm
x
Given:
A 80 N force is applied to
the leg. The length of the
leg is 900mm and the
length of the foot is
100mm. The leg is in
equilibrium.
Find:
The moment of the applied
force at the hip and the
reaction at the hip.
900mm
F
20º
Plan: Draw a FBD, resolve the force along x and y axes,
and determine MO using scalar analysis.
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Problem Solution
y
100mm
x
900mm
F
Fy = 80 cos 20° N
Fx = 80 sin 20° N
20º
R is equal and
opposite to F. as it is
the only force acting.
M
Hip joint
R
+ MH due to applied F = [(80 cos 20°)(100) + (80 sin 20°)(900)] N·mm
= 24700 N·mm
= 24.7 N·m.
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To keep the leg in equilibrium, reaction moment is equal and opposite: -24.7Nm
Moment of a Couple
A couple consists of two forces which have:
1. The same magnitude
2. Parallel lines of action
3. Opposite direction
When your hands are on two opposite points of the steering
wheel of a car, one hand pushing up and the other pulling
down with equal but opposing forces, the result is a couple.
It causes rotation but no translation.
D
F
F
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Moment of a Couple
There is no net force acting to translate the wheel. However,
there is a moment generated.
Calculating the moments at the centre of the wheel, (the axis
of rotation).
M
= F x D/2
+ F x D/2
D
F
= F (D/2 + D/2)
=FxD
F
The moment of the couple does not depend on the
choice of the reference point.
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Worked Example
Determine the moment of
the couple acting on the
scapula (shoulder blade) by
muscles Upper trapezius
and Serratus anterior.
F1 = - F2 = 100N and d =
5cm.
Then, determine the forces
required to produce the
same moment if the
muscles were located at
2.5cm from the centre of
rotation (COR) of the
scapula.
d
d
COR
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Solution
Moment of the couple:
M = FxD
= 100 x 0.05m
= 5 N.m anti-clockwise
To apply the same moment when D = 0.025 m, the forces required
can be calculated from M = F x D
5 = F x 0.025
F = 5/0.025
= 200 N each
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Check-in-Quiz
This karate throw uses a force
couple, one at the hand, and
one at the foot to rotate the
opponent.
Explain in which direction the
forces act, and what these
forces are called.
How could the throw be made
more effective?
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Summary
A couple is a pair of equal and opposite forces acting
along parallel lines.
Moment of a couple is a turning effect produced by the
couple. A couple has no resultant force in any direction and
the magnitude of the moment of a couple is equal to the
forces multiplied by the distance between the forces.
Reactions for a body in equilibrium are equal and opposite
to the forces and moments generated at that point.
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Next Lecture:
• Mechanical Units, Scalars and Vectors;
• Statics: Force and Equilibrium, Moments
and Couples and Reactions, Centre of
Gravity and Centroids, Distributed Force;
• Dynamics: Constant velocity and constant
acceleration in linear systems, circular
systems;
• Work, Power and Energy
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