# IGCSE PHYSICS EOCQ ans 4 ```Answers to end-of-chapter questions
Chapter 4
1 a increase
b increase
5 moment

6 no resultant force (forces balanced)


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





B
1.5 m
weight
of beam
centre
of
mass
unstable object
x
4 a
pivot
F
b moment = force &times; distance from pivot
c
Quantity
Unit
force
N
distance
m
moment of force
Nm
&copy; Cambridge University Press 2014 IGCSE&reg; Physics
a centre of mass correctly marked,
as in diagram

b arrows and labels added correctly

c moment of weight = force &times; distance
= 200 N &times; 0.5 m
= 100 N m
moment of force F is F &times; 1.0 = 100 N m
so F = 100 N





d upward contact force = sum of
downward forces
= 200 N + 100 N
= 300 N



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


b (1.5 + 0.6) &times; 10 = 21 N

c i stays in position

= 63 N m
v
make bar longer or move pivot/stone
to the left or move pivot to left or
increase mass of bar
&copy; Cambridge University Press 2014 IGCSE&reg; Physics

30
= 0.60 kg
iii moment of force = 40 &times; 1.2 = 48 N m

moment of weight = + 30 &times; 0.5 = 15 N m 
total clockwise moment = 48 + 15

iv F &times; 0.2 = 63
63
= 315 N
F=
0.2
10 a mass = (1 5 &times; 12 )


ii as the parrot is rotated, both distances
change in proportion
so clockwise moment = anticlockwise
moment

Answers to end-of-chapter questions: Chapter 4
2



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