Chapter 6: Turning effect/Moment:

Turning effect of an object about a point is called moment. It can be calculated by the
product of the force and perpendicular distance from pivot to the applied force.

Moment = Force x perpendicular distance from point of action.

Unit: Nm
According to the equation above, to get same amount of moment: If perpendicular
distance increases, force decreases and vice versa.

Less

More

Moment of force in different situation.

In 2, there is no perpendicular distance between the center of gravity and the pivot, hence
no net moment.
About P/ Taking wooden peg P as pivot, there is no distance between R1∧P , therefore
Moment of R1 = R1 x 0 = 0
About P, the distance between R2and P is X so
Moment of R2 = R2 x X
Similarly, about Q,
Moment of R2 = R2 x 0 = 0
Moment of R1 = R1 x X

Principle of Moment: When any system is in equilibrium, the sum of clockwise moment
about a point is equal to the sum of anticlockwise moment about that point. Net moment
of the system is zero.

Condition of equilibria:
1) Net force acting on the object is zero (this is called translational equilibrium).
2) Net moment about a point is zero (this is called rotational equilibrium).
 A ladder is balanced on a point. Two forces are acting on the ladder and this ladder is in
rotational and translational equilibrium.

Clockwise Moment = F1 x X1
Anticlockwise moment = F2 x X2
Therefore: F 1 x X 1=F 2 x X 2

Also, F 1+ F 2=R

Centre of gravity: It is a point in an object through which the total weight of the object
appears to act.

Centre of Mass: It is a point in an object through which total mass of the object appears to
act.

Question:
A uniform ladder of 3m and mass 1.25kg is balanced on two supports as shown below.
There is a mass of 5kg hanging at a distance of 0.8m from support A. Support A is 0.2m
from the left edge of the ladder. Support B is 0.3m from the right edge of the ladder.

0.2m R1 0.8m R2 0.3m

A B
5kg

W
a) Using the principle of moment, calculate the reaction forces R1 and R2 acting on the
ladder.
b) State and explain what happen to the values of R1 and R2 if the 5kg mass is moved
towards B.

Answers:
a) In equilibrium, total clockwise moment = total anticlockwise moment
Taking A as pivot,
(5×9.81×0.8) + (1.25×9.81×1.3) = R2×2.5
⇒ 55.18 = 2.5R2
⇒ R2 = 22.07N

9.81×(1.25+5) = R2+R1
⇒ R1 = 9.81(1.25+5) – 22.07
⇒ R1 = 39.24N

b) If the 5kg mass is moved towards B, the distance between the mass and the pivot (A)
will increase and hence the clockwise moment will increase. Since the system still remains
balanced, the total anticlockwise moment also increases.
Anticlockwise moment about A = R2×2.5
Distance between A and B remains same, so the force R2 increases. The object still
remains in translational equilibrium so the total upward force remains equal to the total
downward force. So R1 decreases.

Determination of Centre of Gravity:

1. For regular shaped objects:
For a rectangular lamina or square: At first, we have to draw two diagonals on the
object. The intersection point of these two diagonals gives the centre of gravity or
mass.
2. For irregular shaped object:
For an irregular shaped object, centre of gravity/mass lies in the massive part.

Plumb line

At first three holes are drilled at the edge of this lamina far away from each other.
This lamina is hung through one of these holes using pin and retort Stand. A plumb
line is suspended from the pin. When this plumb line become steady, a line is drawn
on the lamina along the plumb line. This process, is repeated for another two holes.
It would be found that these three lines drawn on the lamina passes through a point.
This point represents the centre of gravity of this irregular shaped object.
An experiment to verify principle of moment:

Centre of gravity

M2
M1

d1 d2

Procedure: The metre rule is balanced. The location at which it is balanced is called the
centre of gravity. Two known masses, m1 and m2 are then placed on each side of the ruler.
The positions of the masses are altered until the metre rule is balanced again. The distance
of the masses from their centers and the pivot are recorded from the metre rule as d 1 and
d2 respectively.
If it is observed that m1 x d1 = m2 x d2, then the principle of moment is verified.
The procedure should be repeated for different values of masses, verifying each time the
principle of moment is accurate.
Precaution:
1. The mass values are checked with electronic mass balance with high precision.
2. The length readings are to be taken perpendicularly to avoid parallax error.
3. The experiment should be done in a closed room to avoid any problem caused by wind.
4. The pivot edge should be made rough to avoid slipping of the metre rule.
5. It is difficult to measure the perpendicular distance from the centre of the mass to the
pivot. To locate the centre of the mass, the metre rule readings should be taken from
each side of the slotted mass and average of the readings would give us the location of
the centre of the mass.
Experiment: To find the mass of a metre rule using principle of moment.

Procedure: The ruler is placed on the pivot and the position of the pivot under the ruler
is slowly changed until it is balanced. Position of the pivot is marked on the ruler. The
weight of this ruler acts through this point. There isn’t any turning effect about this point.

Now a known load is suspended at one end of the ruler and again this ruler is moved until
it is balanced on the pivot. The perpendicular distance from the centre of the suspended
load to the pivot is measured from this metre rule. The perpendicular distance from the
centre of gravity to the pivot is also measured.
m2

x2
x1
W2
W 1=m1 g

Clockwise moment (due to the weight of the metre rule) is found using = W 1 x X1
Anti-Clockwise moment (due to the known mass) is found using = W2 x X2
Since the metre rule is balanced, according to principle of moment: W 1X1 = W2X2
Using this method, we can determine the mass of meter rule.
M1gX1 = M2gX2
M2 X 2
M1 = X
1

Precaution:

1) The experiment should be done in a closed room to avoid any problem caused by wind.
2) The pivot edge should be made rough to avoid slipping of the metre rule.
3) It is difficult to measure the perpendicular distance from the centre of the mass to the
pivot. To locate the centre of the mass, the metre rule readings should be taken from
 each side of the slotted mass and average of the readings would give us the location of
the centre of the mass.
4) The mass values are checked with electronic mass balance with high precision.
5) The length readings are to be taken perpendicularly to avoid parallax error.

Determination of moment of a force at an angle:

Fsinθ

Since force F makes an angle θ, it has two components. Component of force F H is parallel to
X, there will be no turning effect of this force. Another component of force F v is
perpendicular to X, so the moment of this force about P (pivot) can be determined using
Moment = F v × x = Fsinθ × x

A disc of radius r is rotating about its centre by applying two forces as shown below:
According to this above diagram,
Total clockwise moment = (F x r) + (F x r) =2Fr
Since F1 makes an angle θ to the horizontal, the vertical component of this force gives
anticlockwise moment.
r
∴ Anti-clockwise moment = F1Sinθ x 2
r
 2Fr = F1Sinθ x 2

Question:
A uniform ladder of mass 1.2 kg is balanced by a hinge and a string. The length of the ladder is
1.5m, another mass of 0.5 Kg is hung from the ladder 0.95m away from the hinge.

a) Using Principle of Moment, calculate the tension acting in the string.

b) Calculate the force exerted by the ladder on the hinge and also state the magnitude of
force that is exerted by the hinge on the ladder. Give your reason.
Answer:
a) (1.2×9.81×0.75) + (0.5×9.81×0.95) = 1.5×Tsin30
⇒13.48875 = 1.5×Tsin30
⇒ T = 17.99N

b) Tcos30 = 17.99cos30 = 15.58N

15.58N is exerted by the hinge on the ladder as well because according to Newton’s
third
law, every action has an equal and opposite reaction.

Net Moment and rotation:

When this object is freely suspended from a point, due to anti- clockwise moment on this
object, it turns in anticlockwise direction. When centre of gravity of this object comes
down below the pivot, perpendicular distance between centre of gravity and pivot
becomes zero and there will be no net turning effect due to the weight and the object will
be in rotational equilibrium.

Stability: It is the capability of an object to return to its original position when it is

displaced slightly. There are three types of stability depending on the position of centre of
gravity and the shape of the object.

They are:
1. Stable equilibrium
2. Unstable equilibrium
3. Neutral equilibrium
1. Stable equilibrium

When an object is tilted slightly by applying force, the centre of gravity rises and the line
of action of weight through the centre of gravity falls inside the base. Due to the turning
effect, the object returns back to its original position when the force is removed. This is
called stable equilibrium.
Lower the centre of gravity and wider the base of an object, the greater the stability.
2. Unstable equilibrium

When an object is tilted slightly by applying a force, if the centre of gravity falls and the
line of action of weight through the centre of gravity falls outside the base and due to the
turning effect, it doesn't return back to its original position when the force is removed, this
is called unstable equilibrium.
Higher the centre of gravity and the narrow the base of an object, the greater the instability.
3. Neutral equilibrium

When a spherical shaped object is rolled along a horizontal surface, the position of centre
of gravity does not rise nor fall and the line of action of weight through the centre of
gravity always falls inside the base.