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Physics Matters for GCE ‘O’ Level (4 Edition): Full Solutions to Textbook Questions
Chapter 5
Chapter 5
Turning Effect of Forces
Quick Check (page 86)
The force that is applied on the longer lever arm produces a greater moment.
Test Yourself 5.1 (page 86)
1.
(Any two)
Using a spanner to turn a bolt or a nut.
Using a can opener.
Carrying a load in the hand and flexing the elbow of the load-carrying arm.
Stepping on the pedal of a bicycle to turn the wheel.
2.
A spoon is longer than a coin. When applied further from the pivot (i.e. contact point between the
spoon and the lid), a smaller force is required to produce a turning effect large enough to open the
lid.
Test Yourself 5.2 (page 91)
1.
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. The heavier
person has to sit nearer the pivot, while the lighter person has to sit further away on the other side
of the pivot. In this way, the moments of the two persons can be equal and the see-saw can be
balanced.
2.
Let d be the distance between Ali’s mother and the pivot.
Taking moments about the pivot, sum of anticlockwise moments = sum of clockwise moments
(400 N × 2 m) + (600 N × d) = 700 N × 2 m
∴d=1m
Ali’s mother should sit 1 m to the left of the pivot.
Test Yourself 5.3 & 5.4 (page 97)
1.
If the object is in a uniform gravitational field, then the centre of gravity remains the same point.
Near the surfaces of Earth and the Moon, the gravitational fields are relatively uniform. Hence, the
position of the centre of gravity of an object is the same near the surfaces of Earth and the Moon.
2.
Some possible advice:
• Grab the pole with the two hands positioned as far apart as possible.
• Hold the pole tilted such that the end opposite the hands is as high above the hands as
possible.
• Place the heavier clothes nearer the end of the pole being held by the hands.
3.
No. When a heavy load is placed on the roof of an empty minibus, the centre of gravity of the
minibus will be raised. This will cause the minibus to be less stable. Thus, it is more likely to
overturn.
4.
The base of the lamp should be broad and heavy. The lamp should not be too tall so as to ensure
that its centre of gravity is low.
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Physics Matters for GCE ‘O’ Level (4 Edition): Full Solutions to Textbook Questions
Chapter 5
IT Learning Room (page 99)
1.
The pivot of the set-up is where the nails rest on the supporting nail, and must lie at or directly
below the centre of gravity. This results in the stability of the nails and hence is critical to
balancing the nails successfully.
2.
A possible explanation for the nails not being balanced successfully is the addition of too many
nails to one side of the set-up. The excess weight causes the moments on both sides of the pivot
to become unequal, causing the set-up to become unstable and the nails to fall off.
Get It Right (page 99)
(a)
(b)
(c)
(d)
(e)
(f)
True
True
True
False
The centre of gravity of an object does not always lie within the object. For example, the centre of
gravity of a ring lies outside it.
True
False
If the centre of gravity of an object is lowered when tilted slightly, the object is in unstable
equilibrium.
Let's Review (pages 100–102)
Section A: Multiple-Choice Questions
1.
B
Moment about hinge = force × perpendicular distance of force from its line of action
32.5 N m = 50 N × X
X = 0.65 m
2.
C
Taking moments about the pivot,
sum of clockwise moments = sum of anticlockwise moments
(F)(5 cm) = (20 N)(40 cm)
F = 160 N
3.
B
Taking moments about the pivot,
sum of clockwise moments = sum of anticlockwise moments
(80 g)(20 cm) = (m)(30 cm – 14 cm)
m = 100 g
4.
A
Moment due to 60 N force about pivot = 60 N × 30 cm = 1800 N cm (anticlockwise moment)
To balance the beam, a 1800 N cm clockwise moment about the pivot is required.
Option (A): 30 N (downwards) × 60 cm (to right of midpoint) = 1800 N cm (clockwise moment)
Option (B): 30 N (upwards) × 60 cm (to right of midpoint) = 1800 N cm (anticlockwise moment)
Option (C): 50 N (downwards) × 40 cm (to right of midpoint) = 2000 N cm (clockwise moment)
Option (D): 50 N (upwards) × 40 cm (to left of midpoint) = 2000 N cm (clockwise moment)
Therefore, option (A) is the answer.
© 2013 Marshall Cavendish International (Singapore) Private Limited
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Physics Matters for GCE ‘O’ Level (4 Edition): Full Solutions to Textbook Questions
Chapter 5
5.
A
As the tank is emptied, the float moves downwards.
The diagrams below show the pivot at A and at a position further to the right of A, say D.
From the diagrams, it can be seen that when the tank is emptied, the right side of the pointer turns
through a larger angle when the pivot is positioned at A, as compared to when the pivot is
positioned at D.
Pivot positioned at A:
Pivot positioned to the right of A (e.g. D):
6.
A
An object is said to be in stable equilibrium when it is able to return to its original position after it
has been moved slightly.
7.
B
The object in option (B) has a wider base than the objects in options (A) and (D), and a lower
centre of gravity than the object in option (C) due to its shape. Thus, the object in option (B) is the
most stable.
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Physics Matters for GCE ‘O’ Level (4 Edition): Full Solutions to Textbook Questions
Chapter 5
Section B: Structured Questions
1.
(a)
(b)
(i)
The moment of a force is a measure of the turning effect of the force.
(ii)
The moment of a force is the product of the force and the perpendicular distance
from the pivot to the line of action of the force. The SI unit is the newton metre (N m).
(i)
Moment of 40 N force about hinge at A
= force × perpendicular distance from pivot to line of action of force
= 40 N × 1.8 m
= 72 N m (anticlockwise)
(ii)
Taking moments about point A,
sum of clockwise moments = sum of anticlockwise moments
(X)(1.2 m) = 72 N m
X = 60 N
A minimum force of 60 N should be applied at C in order to stop the door from
turning.
The Principle of Moments states that when an object is in equilibrium, the sum of
clockwise moments about any pivot is equal to the sum of anticlockwise moments
about the same pivot.
(iii)
2.
3.
(a)
Taking moments about the pivot,
sum of clockwise moments = sum of anticlockwise moments
(3000 N)(10 m) = (W)(4 m)
W = 7500 N
(b)
Let d be the distance of the block from the pivot.
Taking moments about the pivot,
sum of clockwise moments = sum of anticlockwise moments
(1800 N)(10 m) = (7500 N)(d)
d = 2.4 m
We can find the smallest force F needed to push the barrel over the step by applying the Principle
of Moments.
0.4 m
0.3 m
0.2 m
A
O
F
0.5 m
W
Taking moments about pivot A,
sum of clockwise moments = sum of anticlockwise moments
1500 N × 0.4 m = F × 0.3 m
1500 N × 0.4 m
F=
0.3 m
= 2000 N
Therefore, the smallest force is 2000 N.
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Physics Matters for GCE ‘O’ Level (4 Edition): Full Solutions to Textbook Questions
Chapter 5
Section C: Free-Response Questions
1.
(a)
(b)
(c)
Taking moments about the pivot,
sum of clockwise moments = sum of anticlockwise moments
W(9.0 cm – 5.0 cm) = (7 N)(5.0 cm – 1.4 cm)
W = 6.3 N
The weight is supported at the pivot, i.e. its moment is zero.
When the 7 N weight is replaced by a 4 N weight, the anticlockwise moment becomes
smaller than the clockwise moment. The rule will therefore rotate clockwise.
0 1.
0
2.
0
4
N
2.
(a)
(b)
(c)
4.
0
3.
0
pivo
t
5.
0
6.
0
7.
0
8.
0
9.
0
10.
0
6.3
N
Taking moments about the pivot,
sum of clockwise moments = sum of anticlockwise moments
–1
F (5 cm) = (1 kg × 10 N kg )(20 cm), where F is the force exerted by the beam on the spring.
F = 40 N
2
3
Volume of bucket = 0.012 m × 0.10 m = 0.0012 m
Mass of water = density × volume
–3
3
= 1000 kg m × 0.0012 m
= 1.2 kg
Shift the hook supporting the bucket nearer to the pivot. When the bucket is loaded, its
anticlockwise moment will be smaller than before. The spring will exert a smaller force on the
beam, causing the pointer to move by a smaller vertical distance.
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Physics Matters for GCE ‘O’ Level (4 Edition): Full Solutions to Textbook Questions
Chapter 5
3.
(a)
(b)
The Principle of Moments states that when an object is in equilibrium, the sum of clockwise
moments about any pivot is equal to the sum of anticlockwise moments about the same
pivot.
(i)
Forces acting on the ladder:
A
Rw: Normal reaction by the wall
W: Weight of the ladder
4m
Rg: Normal reaction by the ground
f : Friction between the ground and ladder
W
(ii)
f
To calculate Rw, find lengths OA and OB.
OB
4m
∴ OB = 2 m
A
cos 60° =
2
2
Rw
4m
Rg
m
2
AB = OA + OB
2
2
2
(4 m) = OA + (2 m)
∴ OA =
m
W
O
f
B
2 m (= 4 cos 60°)
Taking moments about B,
sum of clockwise moments = sum of anticlockwise moments
Rw ×
m=W×1m
–2
Rw ×
m = (60 kg × 10 m s )(1 m)
Rw = 173 N
4.
(a)
Frictional force F has no turning effect on the rod because its line of action passes through
the pivot A.
(b)
A
θ x
R2
B
y
R1
W F
C
Taking moments about A,
sum of anticlockwise moments = sum of clockwise moments
R1 × (x + y) = W × x sin θ
W × x sin θ
R1 =
(x + y)
Taking moments about C,
sum of clockwise moments = sum of anticlockwise moments
R2 × (x + y) cos θ = W × y sin θ
W × y tan θ
R2 =
(x + y)
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5.6