1. Sean Dalton www.itsligo.ie/staff/sdalton
Mechanics
Sean Dalton
Course Format
Lectures/Tutorials
Laboratory Work
Course Assessment
Laboratory Work (20%)
Assessments (20%)
Final Exam (60%)
Requirements
Mechanical Engineering Science (Hannah+Hillier)
Engineering Science (Hughes and Hughes)
Scientific Calculator
Science Notebook
Your goal as the instructor is to give a quick overview of the interface.
One important factor to consider is time. Module 1 should only take about 25 minutes max. to lecture. This leaves 5 minutes for a quick demo, then about 20 minutes for the lab, and 10 minutes for a break. Total time 1 hour.
Don’t try to go into too much detail on every slide, or you’ll run out of time. Focus on the ones they’ll need or over-look, like:
Prompt field,
Ribbon bar changes,
Window area, fit, zoom, pan,
opening, saving, new documents.
Don’t waist time on details of help or tutorials, they’ll figure it out.

2. Course Content Forces systems: Triangle of Forces, Bowes notation
Moments: Principle of moments, levers etc
Centroid of Area, Centre of gravity
Friction:Horizontal and Inclided surfaces
Stress and strain: Elastic/Plastic, Factor of Safey
Linear Motion: Force and Acceleration
Rotary motion and Torque
Centrifugal Force
Simple Machines
Mechanics is divided into two areas:
Statics is the area of mechanics which allows us to work out forces in stationary structures.
Dynamics allows us to determine to effect of a force on an object which is free to move.

3. Steps when answering Questions
The following steps should be followed when solving mechanics problems
State what is being calculated
Write down the formula
Rearrange the formula ……………..(if necessary)
Insert Values
Calculate the answer
Add units
Underline the answer
Calculate the diameter of a circle of area 1m2
Calculating diameter
A = /4 D2
D =  4A 
=  4x1
= m

4. Units All Quantities in Mechanics must conform to some system of units
The Imperial system
The Metric system
International system of Units SI units
Base Units Vs. Derived Units
There are 7 base units (5 reqd. in Mechanic)

5. Units All remaining units are called derived units
Achieved by combining the base units

6. Multiples of Base Units

7. Mass Vs. Weight
Mass is defined as the amount of matter (material) in a body (and is expressed in kilograms)
Weight is the force exerted by gravity acting on a body (and is expressed in Newtons)
Whether on the earth or the moon the amount of sugar in a bag of sugar is constant This is a measure of its mass. in Kilograms
However it will feel heavier on the earth than on the moon This is a measure of its weight. In Newtons

8. Pressure/stress
Pressure is a measure of the force per unit area (units N/m2)
P = F Force
A Area
A man and woman of equal mass (80kg) stand on a floor. If the man is wearing flat shoes and woman is wearing stiletto heals as shown calculate the maximum pressure exerted on the floor in each case.
Area (man) =  302 / x = = 1170mm2
Area (woman) =  82 /2 + 3 x = = 148mm2
Pressure (man) = 784.8/1170 = 0.67 N/mm2 (MN/m2)
Pressure (woman) = 784.8/148 = 5.3 N/mm2 (MN/m2)
Weight = 80x9.81= 784.8N

9. Triangle/Polygon of Forces
A Force is that which changes or tends to change a bodies state of rest or of uniform motion in a straight line.
The unit of Force is called the Newton and is the force required to give a mass of 1kg an acceleration of 1m/s2.
A quantity which has both magnitude and direction is referred to as a Vector quantity (Force)
A quantity which has magnitude only is a Scalar quantity (e.g. Mass, Time)

10. Vector representation of forces
Since a force has both magnitude and direction it can be represented by a vector
The arrow indicates the direction of the force
The length of the arrow indicates the magnitude of the force (drawn to a suitable scale)
Force
Scale
2 N = 1 cm
Vector
10 N
5 cm
20 N
10 cm

11. Resultant/Equilibrant
When a number of forces act on a body the resultant force is that single force which would have the same effect
This can be found by representing the forces as vectors and adding them head to tail
The equilibrant is that force which must be applied to a system to produce equilibrium. (equal and opposite to resultant)
5 N
10 N
15 N
Scale: 1N = 1cm
10 cm
5 cm
Resultant
Equilibrant
15 cm

12. Resultant Find the resultant forces shown opposite
Find the resultant of the forces shown below

13. Triangle of Forces
If 3 forces are acting on an object but it remains stationary then it is said to be in equilibrium. (this implies ‘no resultant’)
In order for a body to be in equilibrium under the action of 3 coplanar forces:
the lines of action of the 3 forces must pass through a common point of concurrency.
the forces when represented by vectors added end to end will form a closed triangle.
The principle of concurrency
The triangle of forces
Point of concurrency
Closed Triangle

14. Example
A mass of 10 kg is suspended by two cords from points D and E. Calculate the tension in each chord.

15. Example
Determine the force in the rods AB and BC when carrying a load of 8kN.  = 60o
W = 8000 N
Fbc = N
Fab = N

16. Example
The 80kg is supported by 2 rods AB and BC. Determine the force in each rod.
W = 784.8N
FAB = 632.4N
FBC = 395.2N

17. Example
A jib crane has a jib 5 m long and a tie rod 3.5 m long attached to a post 2 m vertically above the foot of the jib. Determine the force in the jib and the tie rod when a mass of 3 tonnes is suspended from it. 2m 5m
3.5m
38.7o
56.9o
29.43kN
73.6kN
51.5kN
3 T

18. Example
Shown below is a jib crane. Determine the force in the tie and the magnitude and direction of the force at the lower wall joint

19. Resolution of Forces Fx = F Cos  Fy = F Sin 
Instead of solving forces graphically they can be solved mathematically (using trigonometry). Any single force acting at a point can be replaced by a pair of forces which have the same effect.
Where these two 'component forces' are at right angles to each other they can be found easily using trigonometry. When a force is broken down into its component forces the force is said to have been 'resolved into its components'.
Fx = F Cos 
Fy = F Sin 

20. Resolution of Forces
Four co-planer forces act at a point 0, the values and directions of the forces being as shown opposite. Calculate the magnitude and direction of the resultant.
Force Net horizontal force Net vertical force
50 50 Cos (0) = Sin (0) = 0
30 30 Cos (120) = Sin (120) = 26
20 20 Cos (210) = Sin (210) = -10
10 10 Cos (-90) = Sin (-90) = -10
17.68 N N

21. Resolution of Forces (Problems)
For the 5 forces shown determine the magnitude and nature of the resultant.

22. Resolution of Forces Question 2 (Huges P38 Q22)
The following forces act at a point.
20 N due north
30 N 20 degrees south of east
10 N south west
16 N 5 degrees south of west
Calculate the magnitude and direction of the resultant.
[5.36N at 13.8o north of east]
Question 3 (Hughes P38 Q21)
The following horizontal forces act at a point:
50N in a direction due east
80N in a direction due south
and 30 N in a direction 20o north.
Calculate the direction of the fourth force necessary to maintain equilibrium. [73N at 72.6 north of west]

23. Moments
The moment of a force about a point is a measure of its turning effect and is defined as the product of: The force 2 The shortest distance (perpendicular distance) to the line of action of the force
The point about which a body is free to rotate is called the fulcrum.
The leverarm is the shortest distance (perpendicular distance) from the fulcrum to the line of action of the force. F x F x

24. Principle of Moments
When a body is in equilibrium under the action of any number of forces, the sum of the clockwise moments about any point in the body is equal to the sum of the anticlockwise moments about the same point.
Clockwise Moments = Anticlockwise Moments
A lever is a simple machine, which applies the principle of moments to give a mechanical advantage.

25. Moments
A lever is pivoted at its mid point C. 5kg is suspended at E 180mm from C. Calculate the mass required at D to maintain balance.

26. Moments
Determine the position of the suspension point in order for the lever to be balanced.

27. Moments
For the Beam shown determine the reactions at the supports

28. Moments

29. Centroid of Area
Determine the location of the centroid of area of the shape shown below.
[300mm]
[137.2mm]

30. Friction
When an object is placed on a surface the object will exert a normal force(W) on the surface.
The surface prevents the object sinking into it by exerting a normal reaction force(Rn) upwards.
In order to move the object by applying a force P, a tangential force (called friction F acting in the opposite direction) must be overcome.

31. Laws of Friction
If a force P is applied to the book an equal but opposite force called friction is exerted by the table to maintain equilibrium.
If the value of P is increased the friction force must increase accordingly.
There is a limit beyond which the friction force cannot increase. If P exceeds this value sliding starts. The maximum force is called the limiting friction force.
The value of the limiting friction force is proportional to the normal force Rn and is independent of the area of contact.
 = F = limiting friction where:  = coeff. of friction
Rn Normal reaction
F =  Rn
It will also be noticed that it takes a larger force to initiate movement than it does to maintain it.
st > sl st = coefficient of static friction
sl = coefficient of sliding friction

32. Angle of Repose
The angle of inclination of the surface just prior to slippage is called the angle of repose.
Force up slope = Force down slope
F = W Sin 
 Rn = W Sin 
 W Cos  = W Sin 
 = W Sin  W Cos 
= Tan 

33. Angle of repose
Box A weighs 100 N and box B weighs and the coefficient of friction between the box and the ramp is For what range of weights B will the block neither
(a) slide up the slope or
(b) slide down the slope.