Answers to end-of-chapter questions
Chapter 4
1 a increase
b increase
5 moment
[1]
6 no resultant force (forces balanced)
[1]
[1]
no resultant moment
2 a resultant
7 See Activity 4.3.
b zero
Make three small pinholes around the
edge of the lamina.
Suspend the lamina freely from a pin
through one hole.
Mark a vertical line below the pin
using a plumb-line.
Repeat this process for the other
two pinholes.
The centre of mass is where the
three lines intersect.
3 a, b for example
centre
of
mass
stable object
8
contact
force
A
for example
1m
0.9 m
centre
of mass
[1]
[1]
[1]
[1]
[1]
B
1.5 m
weight
of beam
centre
of
mass
unstable object
x
4 a
pivot
F
b moment = force × distance from pivot
c
Quantity
Unit
force
N
distance
m
moment of force
Nm
© Cambridge University Press 2014 IGCSE Physics
a centre of mass correctly marked,
as in diagram
[1]
b arrows and labels added correctly
[2]
c moment of weight = force × distance
= 200 N × 0.5 m
= 100 N m
moment of force F is F × 1.0 = 100 N m
so F = 100 N
[1]
[1]
[1]
[1]
[1]
d upward contact force = sum of
downward forces
= 200 N + 100 N
= 300 N
[1]
[1]
[1]
Answers to end-of-chapter questions: Chapter 4
1
9 a force and perpendicular distance (of force)
from the point
b i
downward force arrow at centre of bar
ii 0.50 m or 50 cm
[1]
© Cambridge University Press 2014 IGCSE Physics
[1]
[1]
30
= 0.60 kg
[1]
b (1.5 + 0.6) × 10 = 21 N
[1]
c i stays in position
[1]
iii moment of force = 40 × 1.2 = 48 N m
[1]
moment of weight = + 30 × 0.5 = 15 N m [1]
total clockwise moment = 48 + 15
= 63 N m
[1]
iv F × 0.2 = 63
63
F=
= 315 N
0.2
v make bar longer or move pivot/stone
to the left or move pivot to left or
increase mass of bar
10 a mass = 1 5 × 12
[1]
[1]
ii as the parrot is rotated, both distances
change in proportion
so clockwise moment = anticlockwise
moment
[1]
Answers to end-of-chapter questions: Chapter 4
2
[1]
[1]
[1]