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2 Force and Motion
5
Chapter 5 Moment of a Force
Moment of a Force
8
(a)
normal reaction N
by plank
Practice 5.1 (p.185)
1
P
A
Moment about X
weight W
= Fd = 30 (0.7) sin 15 = 5.44 N
2
D
normal reaction
from pivot
Maximum magnitude of moment
= FD = 5  (0.2  2) = 2 N m
3
A
4
D
force FQ by Q
force FP by P
clockwise moment
(b)
= anticlockwise moment
Normal reaction N acting on P by the
plank and the force FP acting on the
100  0.3 = f  0.05
plank by P
f = 600 N
(c)
5
D
6
The moment arms are the same for the three
Since P is stationary,
N=W
By Newton’s third law,
forces.
N = FP
 They produce moments of the same
 FP = W
magnitude.
(a)
2d
O
Just before the nail moves,
7
d
plank
(d)
Moment about O
Clockwise moment
= anticlockwise moment
= 5  0.25 cos 35
M  2d = m  d
m
M=
2
= 1.02 N m (anticlockwise)
(b)
 The mass of Q is
X
m
.
2
Practice 5.2 (p.198)
1
1 .2
= cos 30
OX
1.2
OX =
= 1.386 m
cos 30 
B
Take moment about the pivot.
Sum of clockwise moment
= sum of anticlockwise moment
600  3 = 400  (1 + 2) + W  2
Moment about O
W = 300 N
= 50  OX  sin 65
= 50  1.386  sin 65
= 62.8 N m (anticlockwise)
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2
D
Distance of c.g. from pivot =
50
 10 = 15 cm
2
1
2 Force and Motion
Chapter 5 Moment of a Force
Distance of P from pivot
Revision exercise 5
= 50  10  15 = 25 cm
Cocept traps (p.200)
Take moment about the pivot.
1
Sum of clockwise moment
A couple, which consists of two equal and
= sum of anticlockwise moment
20  15 = F  25 sin 40
F
opposite forces, can make an object rotate.
2
T
F = 18.7 N
3
C
Multiple-choice questions (p.200)
4
A
3
A
5
After releasing, the c.g. of the ruler is outside
4
B
6
the base of support (the table). Therefore, the
Magnitude of moment
ruler loses balance and falls.
= FD
No, he cannot. When he stands against a wall,
= 8  0.4 sin 130
he cannot move backwards and his centre of
= 2.45 N m
gravity would be outside the base of support,
5
C
i.e. his feet, if he bends over.
20 N
7
0.5 m
0.5 m
X
c.g.
weight of rod W
weight
contact point
The c.g. of the rod is at its middle.
When the roly-poly is tilted as shown, its
Take moment about X.
weight produces a clockwise moment about
Clockwise moment = anticlockwise moment
W  0.5 = 20  1
the contact point and makes it return to the
W = 40 N
original position. Therefore, it stays upright.
8
(a)
(b)
The c.g. of the bus will be higher when
6
B
passengers stand on the upper deck. The
7
D
c.g. may be shifted outside the base of
8
B
support even if the bus is only tilted
The force acting on the rod by the pivot at O is
slightly. This produces a net moment
not zero.
and makes the bus topple over.
 (1) is incorrect.
If the c.g. is not vertically below the
Take moment about O. Also, take clockwise
point of suspension, the weight of the
moment as positive.
object will produce a moment that makes
Net moment = 6  0.07  10  0.02  4  0.02
the object rotate. This moment will be
zero only if the c.g. is vertically below
= 0.14 N m
 (2) is correct.
the point of suspension.
2
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2 Force and Motion
Chapter 5 Moment of a Force
The rod is not in equilibrium. The net moment
Since m =
about different points may be different.
 (3) is incorrect.
2.4d1
,
d2
1A
the answer to (a) will be smaller.
15
(a)
(i)
1A
Take moment about the contact
9
(HKALE 2008 Paper 2 Q1)
10
(HKALE 2009 Paper 2 Q2)
point.
11
(HKDSE 2012 Paper 1A Q6)
Anticlockwise moment
12
(HKDSE 2014 Paper 1A Q3)
= clockwise moment
F(5) = 1.5(9.81)15 + 5(9.81)30
Conventional quesionts (p.201)
13
1M
= 338 N
normal reaction
from Y
(a)
1A
The force exerted by the biceps is
338 N.
friction
(ii) No,
weight
there is an inclined force acting on
the forearm at the contact point.1A
(1 correct force with correct name) 1A
(b)
14
(a)
1A
(All correct)
1A
No, it cannot.
1A
(b)
The weight is further away from the
pivot (the shoulder).
1A
If the c.g. is not vertically above Y, it is
Therefore, the clockwise moment
outside the base of support and X will
produced by the weight about the pivot
lose balance.
is larger.
1A
1A
Just before toppling, the force by the
He must exert a larger force on the arm
table will act on the stand at A.
to produce a larger anticlockwise
Take moment about A.
moment.
Clockwise moment
As a result, he feels more tired.
= anticlockwise moment
2.4  9.81  3 = m  9.81  10
m = 0.72 kg
16
(a)
1A
tension T
tension T
1M
50
50
1A
The maximum mass is 0.72 kg.
(b)
Shorten the distance between the clamp
and the stand.
1A
(1 correct force with correct name) 1A
Put a weight on the stand.
1A
(All correct)
(Or other reasonable answers)
(c)
weight = 10 N
(b)
Consider the vertical direction.
When the system is put on an inclined
Since the board is stationary, by
plane, the horizontal distance d1 between
Newton’s first law,
X and A will be shorter and the
2T sin 50 = 10
horizontal distance d2 between the
weight and A will be longer.
New Senior Secondary Physics at Work (Second Edition)
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T = 6.53 N
1A
1A
1M
1A
The tension in each string is 6.53 N.
3
2 Force and Motion
(c)
(i)
Chapter 5 Moment of a Force
The total force acting on her shoulder
tension T1 tension T2
would become larger.
50
P
50
Therefore, it would be harder if the two
Q
loads are put together.
X
S
R
18
(a)
TB = 90 N
length of SX be c and the weight of
the mass be W.
Take moment about the contact point
between A and the plank.
1A
Anticlockwise moment
= clockwise moment
 15  65  20

 15 
90  65 = 50d + 150 
2


Take moment about Q.
W(2d  c) + 10d = T1(2d) sin 50
W (2d  c)  10 d
T1 =
2d sin 50 
 The left string breaks.
1M
d = 12 cm
1A
The distance between the robot and
1A
string A is 12 cm.
(c)
M
1A
The tension is 90 N.
(b)
Take moment about P.
> T2
Since the plank does not move, by
TB + 110 = 50 + 150
Let the length of PQ be 2d, the
Wc + 10d = T2(2d) sin 50
Wc  10 d
T2 =
2d sin 50 
1A
Newton’s first law,
W 10 N
(ii)
1A
(i)
1A
Take moment about the contact
point between B and the plank.
Anticlockwise moment
= clockwise moment
50(65  d) + 150(50  20)
C
= 115  65
1M
d = 5.5 cm
(C vertically below M)
17
(a)
No,
1A
1A
this is because the two forces point in the
same direction.
(b)
The force acting on the woman’ shoulder
= 200 + 200 = 400 N
1A
(c)
It would be the same.
1A
(d)
The heavier load should be put closer to
her.
(e)
1A
She has to apply a downward force on
the pole to keep it in equilibrium.
4
1A
1A
1A
The minimum distance is 5.5 cm.
(ii) No, it cannot.
1A
Take moment about the contact
point between string B and the
plank. The anticlockwise moment
will be larger when the robot is on
the left of A than when it is on the
right.
1A
Therefore, the tension needed by A
will be even larger
1A
and A will break.
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
2 Force and Motion
19
(a)
Chapter 5 Moment of a Force
Take moment about the pivot.
Clockwise moment
= anticlockwise moment
500(9.81)12 = (m + 300)(9.81)6 1M
m = 700 g
50
1A
40
The mass of the object is 700 g.
(b)
(c)
20
(a)
Since the pan and the 500-g mass have
the same acceleration,
1A
the result will remain unchanged.
1A
weight = 400 N
Clockwise moment
= anticlockwise moment
Use a heavier mass to replace the 500-g
mass.
1A
Move the pivot closer to the pan.
1A
400(0.9) sin 40 = T(0.6) sin 70
1M
T = 410.4 N
Let T be the tension in the cable and N
 410 N
be the normal reaction.
The tension in the rope is 410 N.
Take moment about Q.
(ii)
Clockwise moment
= anticlockwise moment
410.4 N
(T sin 60)(1.6 + 0.7)
= 9000  0.7 + (T cos 60)0.4
20
T = 3516 N
400 N
1A
Consider the vertical direction.
3516 sin 60 + N = 9000
N = 5960 N
(b)
(a)
400 = 410.4 sin 20 + F sin  1M
1M
1A
(c)
 F sin  = 259.6
F cos  = 410.4 cos 20
(1)  (2),
(i)
No
1A
1A
The centre of gravity of a rigid body is
(1)
Consider the horizontal direction.
the normal reaction is 5960 N.
the position that the weight of the body
(b)
Consider the vertical direction.
The tension in the cable is 3520 N and
(ii) Yes
21
force on pole
by ground F

1M
 3520 N
1A
(2) 1M
tan  = 0.673
 = 33.94
 33.9
seems to act.
1A
No net force acts on it.
1A
From (2),
410 .4 cos 20 
F=
= 465 N
cos 33 .94 
No net moment acts on it.
1A
The required force is 465 N
(i)
Take moment about X.
1A
1A
(towards right at 33.9 above
horizontal).
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
5
2 Force and Motion
Chapter 5 Moment of a Force
22
(HKALE 2009 Paper 1 Q2(a))
Moment produced by W = Wr
23
(HKALE 2012 Paper 1 Q2)
To avoid toppling,
Wr  2Fr sin 
W

1
1A
2F sin 
W
is independent of the
2F sin 
Experiment questions (p.205)
24
Set up the apparatus as shown.
string
ruler
string
protractor
radius r and increases when 

decreases.
string
weight
1A
Therefore, decreasing  helps
stabilize the bowl while increasing
spring
balance
r does not.
(b)
Zero
1A
F
(Correct set-up)
1A
Pull the spring balance as shown so that the
ruler remains horizontal.
1A
Measure the force F with the spring balance
and the angle  with the protractor.
1A
Record several sets of F and  while keeping
the ruler horizontal.
1A
1
Plot a graph of F against
. If the graph
sin 
is a straight line passing through the origin, F
is inversely proportional to sin .
1A
Physics in article (p.205)
25
(a)
(i)
Moment about T
= 3(0.09  2) sin 45
1M
= 0.382 N m (clockwise)
1A
(ii) Let W be the bowl’s weight, r be
the bowl’s radius, F be the force
acting along the slanted surface and
 be the slanting angle.
Take moment about T.
Moment produced by F
= F(2r) sin  = 2Fr sin 
6
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015