1. 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:
Stationary
Moving to the left at constant speed?
Answers:
A – if the block is stationary then it is in equilibrium. Therefore the sum of forces = 0. Therefore the force F is to the left & 2N to balance the other forces.
B – if the block is moving at a constant speed then there is no acceleration and therefore the block is also in equilibrium. Therefore the force F is again to the left & 2N to balance the other forces. 3N 5N

2. Last Lecture Review 2
In 3 minutes, describe to the person sitting next to you:
What is a force?
What change does a force (or system of unbalanced forces) make to a body?
What is equilibrium?
Then discuss with the class.
Suggest 1 minute on each. Then get students to share their answers. Things you are looking for:
A push or pull on an object that causes it to change velocity (magnitude or direction)
A force changes a body’s acceleration (f=ma)
Equilibrium is when the net force on an object is zero, or it’s acceleration (vector) is zero (i.e. stationary or moving at a constant speed).

3. Last Lecture Review 3 Select the correct FBD at particle A. A) B) C)
30
40
100 N
Select the correct FBD at particle A. A) B)
40°
F F2 C)
30° F F1 F2 D)
Answers:
1. D

4. 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

5. 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

6. Moment in 2-D F M O d
Ask students who’s heard of moments or torques. How have they heard those terms been used?
See if they can identify what is common to both these examples.
How have they applied moments in ordinary life (doors – where is easiest to open?, sports?)
The moment of a force about a point is a measure of the tendency for rotation (sometimes called a torque).

7. Mo = F.d The magnitude of the moment about point O is:
where d is the perpendicular distance from point O to the line of action of the force F.
Get a volunteer to lift another’s leg under the knee. Ask the class to observe which direction they pull and which direction the leg rotates.
Ask them to lift from the foot then from half way up and say which way is easier.
Get another volunteer from the class to be spun by another. Ask the class to observe which direction they push and which direction the person rotates.
Ask them to spin from the forearm then from shoulder and say which way is easier.
In 2-D, the direction of MO is either clockwise or anti-clockwise depending on the direction of the tendency for rotation.

8. and the direction is anti-clockwise.F a b d O
For example, MO = F.d
and the direction is anti-clockwise.
However, it is sometimes easier to determine MO by using the components of F as shown. a b O F
F x
F y

9. Moment at Knee with Distance
Compare the applied moment at these leg positions.
As the leg is lifted further off the ground, the weight force through the leg is applied at a larger perpendicular distance from the knee. Hence the applied moment at the knee increases.

10. MO = (FY a) – (FX b) F F y Here, F x
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.

11. Check-in Quiz Answers: 1. B 2. A F = 10 N A d = 3 m
What is the moment of the 10 N force about point A?
10 N·m anti-clockwise
30 N·m anti-clockwise
13 N·m clockwise
(10/3) N·m clockwise
7 N·m clockwise
Answers:
1. B
2. A

12. Check-in Quiz Which of these exercises cause a moment at the shoulder?
a) b) c)
d) e)
Answers:
1. A, B and E
2. A

13. Check-in Quiz - Answer
Which of these exercises cause a moment at the shoulder?
a) b) c)
d) e)
Answers:
1. A, B and E
2. A

14. Practice Problem What is the moment of the 30N force on the wheel nut?
Ask students, how they plan on solving the problem. – Recall problem solving approach.
Get students to have a go first, then get them to do on whiteboard.
Can find perp distance first using trig. Or some may know how to break into components.

15. 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 = 0
Ʃ Fy = 0
Ʃ M = 0
For equilibrium, the sum of the moments must be equal to zero.
Explain that this is an extension on what we already know (shown in red).
Get two volunteers to push against each other with small force. Ask students, are the forces in balance? (yes, since there’s no translation) Are the moments at the shoulder zero? (yes, since there’s no rotation).
In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces).
For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be equal to zero.
 F = 0 and  MO = 0

16. Worked Example
0.6 m
0.52 m Fe W
Do this example on the whiteboard. First convert the mass to Weight force (test students – what force in Newtons is W?)
Be sure to ask students: which force causes a moment around the wheel axis? Which direction is the moment (clockwise or anti-clockwise?) Note the reaction force does not cause a moment when we select the wheel axis since the force is directed at zero distance from the axis. R
The wheelbarrow shown carries a load of 58 kg. Calculate the force Fe required to hold the handles in a stationary position.

17. 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.
Ask students to reflect on the process. What would have happened had we tried to sum moments about another position on the wheelbarrow?

18. Check-in-Quiz The drawing shows two different wheelbarrow designs.
Ask students to think about (if they need help), when we sum moments about the wheel axis, what moments contribute that are acting against the moment created by the human force (in A, Wload creates an additional moment that must be counteracted by the human).
Answer
The positioning of the load over the wheel means that an additional moment is not generated that the person must make up for with their force. Therefore the second design does not require as much force to hold the wheel barrow stationary.
Design 1 :
510 x x 0.6 = F x 1.3
F =
Design 2
57 x 0.6 = F x 1.3
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?

19. 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
Before explaining this slide:
Before we have usually taken moments about the same place as the reaction forces that keep a body in equilibrium. So we haven’t had to look in detail at how big and in what direction the reaction force is.
However, when we come to the human body, reactions are very important because bones and ligaments act to keep our body stable when loads and moments are applied, and we need to know how big these can be before the load will cause us damage.
So now we’re going to talk about reactions for a system in equilbrium.
Force applied to spanner to turn a nut

20. 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
60 N
Note if the nut is actually being turned, then the moment generated by the applied force is greater than the reaction moment at the nut resisting rotation (friction/threads).
Point out the applied force. Then sum F = 0 to keep it in equilibrium, so F(nut) – 60N. Then since the 60N force applied at point away from the nut causes the nut to want to rotate, if the system is in equilibrium, the resisting moment at the nut is equal and opposite to the applied moment.
So altogether there is a reaction force and reaction moment at the nut resisting the applied force (and induced moment).
Reaction force and moment on the nut if the system is in equilibrium
Force applied to spanner to turn a nut 20

21. 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.

22. Reactions of Supports Smooth surface Roller support Cable or link
Sliding pin
Rough surface
Reinforce that here we are finding the reaction forces. And these are important for free body diagrams when we will use them to solve for unknowns.
Reactions of supports are easy when we think about –
What kind of resistance can we have at the support end?
E.g. Pin – can rotate. Hence no moment reaction is possible. Cannot move horizontally or vertically – Hence force reaction horizontal and vertical is possible.
E.g. Fixed support – cannot rotate or move. Hence forces and moment reactions are possible.
Ask students, what kind of supports are in the human body (e.g. knee, pin – no moment reaction AT knee joint itself – all moments must be generated at a distance from knee joint; e.g. rib cage – fixed support – cannot rotate around the spine or translate easily so hence must supply moment and force reaction at the spine in response to applied force on the rib cage – hence why can snap easily!!)
Pinned support
Fixed support
Angle depends on friction

23. 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
Sum Fy =0
Fw=0
Fw = kN
Sum M = 0
1300 x M = 0
M = N.m
F = 1.3 kN

24. Check-in Quiz P 1m
30 N
40 N x y
For this force system, the reaction at the wall P is ______ .
A) FRP = 40 N (along - x dir.) and MRP = 60 N.m anti-clockwise
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 anti-clockwise
First ask students:
What type of reactions are possible at the wall (fixed support)?
Answer:
1. D

25. Problem Question y
100mm
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. x
900mm F
20º
Get students to try this on their own. Then get a member of the class to come up and show their answer on the board. Make sure you get class to acknowledge his/her participation – this can be tricky stuff!
Plan: Draw a FBD, resolve the force along x and y axes, and determine MO using scalar analysis.

26. R is equal and opposite to F. as it is the only force acting.
Problem Solution y
100mm x
Fy = 80 cos 20° N
Fx = 80 sin 20° N
900mm F
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
= N·mm
= N·m.
To keep the leg in equilibrium, reaction moment is equal and opposite: -24.7Nm

27. Moment of a Couple A couple consists of two forces which have:
The same magnitude
Parallel lines of action
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.
So far we have discussed turning effects of single forces. We now turn our attention to a very special combination of forces called a couple.
Get students to kinaesthetically demonstrate.
Ask students, what would happen if one force was much larger than the other and applied at the same distance? – Try this with 3 volunteers.
Ask students, what would happen if one force was smaller but applied at a greater distance? – Try this with 3 volunteers.
Either way, the force wants to move the whole wheel one way or another. A true couple does not do this since the forces cancel out – it only causes rotation. F D

28. 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/ F x D/2
= F (D/2 + D/2)
= F x D
The moment of the couple does not depend on the choice of the reference point. F D
Reiterate that if a net force (unbalanced force) acts on something then it changes velocity. Here, the algebraic sum of the two forces is equal to zero.

29. 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.
The wheel diameter is 350 mm and the forces applied are each 5 N. d
COR d
Note to students that we will approximate F1 to be equal and opposite to F2 even though in real life it is not quite opposite direction.
Demonstrate the first part on the whiteboard.
For second part, ask students to try on their own. Then compare answers. Get a student to shout the answer to both, and explain the answer to the second part (doubles since the distance halved.)

30. Solution Moment of the couple: M = F x D = 100 x 0.05m
= 5 N.m anti-clockwise
To apply the same moment when D = m, the forces required can be calculated from M = F x D
5 = F x 0.025
F = 5/0.025
= 200 N each

31. 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?
Force at the hand is down, and at the foot is up. This creates a moment about the opponent’s centre of gravity to rotate his body.
Here the man throwing could get closer to his opponent and get his hips into the motion, this increases the F in the moment Fd.
He could also get under his opponent more fully to get his leg further from his hand and this increases the d in the moment Fd.

32. 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.
May not include this but An equivalent couple is any other couple with the same magnitude and rotational sense.

33. 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