Unit 5:
Turning Effect
Of Forces
Unit 5.1
Moments
Unit 5.2
Principle of
Moments
Unit 5.3
Centre of
Gravity
Unit 5.4
Stability
Learning Outcomes:
Candidates should be able to:
(a) Describe the moment of a force in terms of its turning effect and relate
this to everyday examples
(b) Recall and apply the relationship moment of a force (or torque) = force
x perpendicular distance from the pivot to new situations or to solve
related problems
(c) State the principle of moments for a body in equilibrium
(d) Apply the principle of moments to new situations or to solve related
problems
(e) Show understanding that the weight of a body may be taken as acting at
a single point known as its centre of gravity
(f) Describe qualitatively the effect of the position of the centre of gravity
on the stability of objects
Important Concepts
1)
The moment of a force (or torque) is the product of the force and the
perpendicular distance from the pivot to the line of action of the force.
** 2) The Principle of Moments states that when a body is in equilibrium, the sum
of clockwise moments about a pivot is equal to the sum of anticlockwise
moments about the same pivot.
3)
The centre of gravity of an object is defined as the point through which its
whole weight appears to act for any orientation of the object.
4)
Stability refers to the ability of an object to return to its original position after
it has been tilted slightly.
5)
An object is in stable equilibrium if it returns to its original position when it
is slightly displaced.
6)
An object is in unstable equilibrium if it topples when slightly displaced.
7)
An object is in neutral equilibrium if its centre of gravity neither rises nor
falls, and stays at the same height when the object is slightly displaced.
**: Definitions you must know
1) Force at 0cm – 4.0 N – Downward 2) Force R – Upward 3) Force W – Downward 4) Force at 80cm – 2.0 N Downward 4) For each force, identify the direction of the force. (Upward force / Downward force) b) Since the metre rule is in equilibrium, we can conclude, Sum of upward forces = sum of downward forces R = (4.0 N + w + 2.0 N) = (4.0 N + 1.0 N + 2.0 N) = 7.0 N 5) To solve for the other remaining unknown, equate the sum of forces in one direction (upwards) with sum of forces in the opposite direction (downwards). a) Taking the moments about the pivot, Sum of clockwise moments = sum of anticlockwise moments W x (20 cm) + (2.0 N)(50 cm) = (4.0 N)(30 cm) W = 1.0 N 3) Applying the principle of moments, generate an equation involving clockwise moments and anticlockwise moments and solve for the unknown force. Tip: Moments are taken about the pivot in order to obtain an equation that does not involve both unknowns. Hence the value of the other unknown can be found. 1) Force at Ocm – 4.0 N –Anticlockwise 2) Force R – no moments as R is acting directly on the pivot itself 3) Force W – Clockwise 4) Force at point 80cm – 2.0 N – Clockwise 2) For each force, identify the type of moment ( Clockwise / Anticlockwise), and classify all the clockwise moments and anticlockwise moments respectively. In this case, the pivot has already been identified in the question. 1) Identify the pivot / hinge / fixed point. Application to Above Example Steps
Worked Example Problem Solving Technique To solve for unknown forces, application of two principles is required: • Principle of Moments. Moments are taken about the pivot in order to obtain an equation that does not involve both unknowns. Hence the value of the other unknown can be found • Equating sum of forces in one direction (upwards) with sum of forces in the opposite direction (downwards). Conditions for Equilibrium 1) The resultant moment about the pivot is zero, i.e. the Principle of Moments must apply 2) All Forces acting on it are balanced; i.e. the resultant force is zero
Examples
CG of irregularshaped objects (2dimensional)
CG of regularshaped objects (2dimensional)
1) Uniform metre rule: Its CG is at the 50 cm mark and a uniform half metre rule will have its CG at the 25 cm mark 2) Circular disc: Its CG is at the centre of the circle 3) A rectangular laminar Its CG is at the centre of the laminar 4) A triangular laminar Its CG is at the centre of the laminar For an object with regular shape and uniform density: Its centre of gravity will be at the center of the object. Point of CG can lie outside the regularshaped object.Centre of Gravity (CG) – How to Obtain the CG of Different Objects
Steps taken to obtain the CG of a irregularshaped (planar) object: 1) Set up a retort stand at a certain height with a clamp. 2) Clamp the wooden cock with a pin onto the retort stand. 3) Puncture 3 small holes at three corners of the planar object. The holes should be as far apart as possible from each other. 4) Mark “A”, “B”, “C” beside the3 holes. 5) Hang the planar object at the pin (hole A) 6) Tie the plump line next to the planar object. Allow the plump line to come to a rest and put a mark “ A’ ” on the planar object where the plump line passes through at the bottom. 7) Repeat step 5 to 6 for holes B and C. 8) Remove the planar object from the pin and draw straight lines connecting A to A’, B to B’ and C to C’. 9) The intersecting points of the 3 lines will be the CG of the irregularshaped object.A
C
B
A’
B’
C’
Factors Affecting the Stability of an Object 1) The base area of the object 2) The height of CG above its base To Improve the Stability of an Object 1) Broaden / Widen the base area of the object 2) Add weights to the base of the object such that the CG becomes lower When the cone is tilted slightly, 1) Its centre of gravity neither rises nor falls; it remains at the same level above the surface supporting it, 2) The lines of action of the two forces W and R always coincide, 3) There is no moment provided by its weight W about the point of contact C to turn the paper cone. When the cone is titled slightly, 1) Its centre of gravity falls and continues to fall further, 2) The line of action of its weight W lies outside the base area of the cone 3) The clockwise moment of its weight W about the point of contact C causes toppling. When the cone is tilted slightly, 1) Its centre of gravity rises and then falls back again, 2) The line of action of its weight W lies inside the base area of the cone 3) The anticlockwise moment of its weight W about the point of contact C causes the cone to return to its original position. How to describe the stability of the object in terms of the position of its centre of gravity? Position remains the same when displaced CG out of the base when displaced CG within the base when displaced Neutral Equilibrium Unstable Equilibrium Stable Equilibrium Stability – 3 Cases of Equilibrium Clockwise moment Anticlockwise moment No moment
