Active Physics Full Solutions to Textbook Exercises
Chapter 7
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Checkpoint 3 (p.143)
Checkpoint
Checkpoint 1 (p.135)
1.
B The tension in thread L balances the total weight
of the 3 shells beneath it, and this is equal to 30 N.
2.
The spring in the spring balance is stretched
until it reads 100 N. The spring balance experiences
two horizontal forces of 100 N, i.e. a leftward and
a rightward pulling force. This is why it remains
stationary. Nonetheless, the tension in the spring
must be exactly 100 N in order to balance the 100 N
weight on the right.
1. dummy
B
When you weigh an object vertically, you have to hold the
spring balance still, so that the tension in the spring shows the
force exerted on its hook.
Resultant force = 2.41 N (N48.4°W)
2. (a) No
(b) Yes
(c) Yes
(d) No
Checkpoint 2 (p.138)
1.
y -component = 8 cos 30° = 6.93 N
Take the direction to the right as positive.
Consider the two blocks as a single system. By
F net = ma , we get
A
3 = (1 + 2)a ⇒ a = 1 m s−2
So the acceleration of the 1 kg block is 1 m s−2 . .
2.
3. (a) x -component = 8 sin 30° = 4 N
(b) x ′ -component = 5 cos 50° = 3.21 N
y ′ -component = 5 sin 50° = 3.83 N
Checkpoint 4 (p.146)
1.
The force F does not act on B , and hence option
A is incorrect.
C
As A remains at rest, the friction acting on A by B
points to the right in order to balance the leftward
force F . By Newton’s third law, the friction acting
on B by A , denoted by f A , points to the left. Hence,
option B is incorrect.
Similarly, as B remains at rest, the friction acting
on B by the ground, denoted by f g , points to the left
in order to balance the rightward force f A . Hence,
option C is correct.
C
√
Adding the forces algebraically, we get
42 + (3 − 2)2 N.
p
2. Sum of x -components = 2 cos 30° = 3 N
Sum of y -components = 2 sin 30° − 1 = 0
Resultant F =
p
3N
(towards the +x -direction)
Checkpoint 5 (p.155)
1.
A As the cart moves at a constant speed, there is
no net force acting on it.
2.
B Since the string is held still, all the horizontal
force components balance each other. Hence we get
F = T cos θ + T cos θ = 2T cos θ
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3. (a)
Active Physics Full Solutions to Textbook Exercises
sum of clockwise moment = sum of anticlockwise moment.
C
Option A is incorrect because the normal
reaction R should be perpendicular to the
supporting surface (i.e. the water chute).
(600)(1.2) = (550)(0.6 + 0.4) + W (0.4)
W=
Option B is incorrect because the direction
of the friction f should be opposite to the
direction of motion of the girl.
(b)
720 − 550
= 425 N
0.4
(b) In equilibrium, the net force acting on the
seesaw is zero.
B Adding up all the forces acting on the girl
along her direction of motion, we get
R = 550 + 425 + 100 + 600 = 1675 N
F net = mg sin θ − f .
Exercise
Checkpoint 6 (p.164)
1. (a) Net moment = (5)(0.2) = 1 N m (clockwise)
(b) Net moment = (5 sin 30°)(0.2) = 0.5 N m
(anticlockwise)
Exercise 7.1 (p.138)
1.
Consider the boxes as a single system. By
F net = ma , we have
B
(c) Net moment = (5)(0.1) − (2)(0.1 + 0.1) = 0.1 N m
(anticlockwise)
2. Since the lever is balanced, its net moment is zero.
6 − 3 = (2 + 1)a ⇒ a = 1 m s−2
Taking moments about O , we get
2.
A Let T A and TB be the tensions in strings A and B
respectively.
sum of clockwise moments = sum of anticlockwise
moments
Consider the blocks as a single system. By Fnet = ma ,
we have
TB = (2 + 1)a ⇒ a =
(1)(9.81)(0.4) + m(9.81)(0.4 + 0.4) = (2)(9.81)(0.8)
m=
TB
3
Consider the 1 kg block alone. By Fnet = ma , we have
1.6 − 0.4
0.8
T A = (1)(a) =
= 1.5 kg
TB
3
Therefore T A : TB = 1 : 3.
Checkpoint 7 (p.173)
1. (a)
(b)
OR: Let a be the common acceleration of the blocks.
For the 1 kg block, T A = (1)(a) = a .
For the 2 kg block, TB − T A = (2)(a) = 2T A ⇒ TB = 3a .
T
The c.g. of a metal ring lies at its centre,
which is outside the ring itself.
F
(c)
T
(d)
An object experiencing a zero net force
may undergoing uniform motion. Similarly,
an object experiencing a zero net moment may
undergoing uniform rotational motion.
F
3.
Consider the boxes as a single system. By
F net = ma , we have
B
3 = (1 + 1 + 1)a ⇒ a = 1 m s−2
Consider A alone.
2. Bus A will topple, because its c.g. lies outside of its
supporting base.
3. (a) In equilibrium, the net moment on the seesaw
is zero.
Taking moments about the pivot O , we get
By Fnet = ma , we have
f B = (1)(1) = 1 N
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Active Physics Full Solutions to Textbook Exercises
The new acceleration is 2.5 m s−2 towards the
right.
Consider B alone.
7. (a) Take the direction to the right as positive.
Consider the blocks as a single system. By
F net = ma , we have
We have f A = f B = 1 N because the forces form an
action–reaction pair. By Fnet = ma , we have
fC − f A = (1)(1) ⇒ fC = 2 N
4 = (2 + 1 + 1)a ⇒ a = 1 m s−2
The common acceleration is 1 m s−2 towards
the right.
(b) (i) Consider P alone. By Fnet = ma , we have
So the ratio is 1 : 2.
4.
In Fig. a, we have T = W , where W is the weight
of the block. In Fig. b, we have 2T ′ = W ⇒ T ′ = T2 .
TPQ = (2)(1) = 2 N
B
5. (a) Take the direction to the right as positive.
Consider the trolleys as a single system. By
F net = ma , we have
1 = (0.2 + 0.2)a ⇒ a = 2.5 m s−2
The common acceleration of the trolleys is
2.5 m s−2 towards the right.
The tension in the string connecting P and
Q is 2 N .
(ii) Consider P and Q as a single system. By
F net = ma , we have
TQR = (2 + 1)(1) = 3 N
The tension in the string connecting Q and
R is 3 N .
(c) Consider the blocks and the plasticine as a
single system. By Fnet = ma , we have
(b) dummy
4 = (2 + 1 + 1 + 3)a ⇒ a = 0.5714 m s−2
Consider trolley A alone. By Fnet = ma , we have
1 − T = (0.2)(2.5) ⇒ T = 0.5 N
TPQ = (2)(0.5714) ≈ 1.14 N
The tension in the string connecting P and Q is
1.14 N .
The tension in the string is 0.5 N .
6. (a) Take the direction to the right as positive.
Consider the blocks as a single system. By
F net = ma , we have
15 = (4 + 1)a ⇒ a = 3 m s
Consider P alone. By Fnet = ma , we have
Consider P and Q as a single system. By
F net = ma , we have
TQR = (2 + 1 + 3)(0.5714) ≈ 3.43 N
−2
Consider B alone. By Fnet = ma , we have
F = (1)(3) = 3 N
The force acting on B by A is 3 N towards the
right.
(b) Consider the three blocks as a whole system.
By Fnet = ma , we have
15 = (1 + 1 + 4)a ⇒ a = 2.5 m s−2
The tension in the string connecting Q and R is
3.43 N .
8. Consider the mass and the block as a single system.
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Active Physics Full Solutions to Textbook Exercises
Take the direction of motion of the mass as positive.
(b) dummy
By Fnet = ma , we have
mg − f = (M + m)a
(0.25)(9.81) − f = (0.5 + 0.25)(2)
∴ f = 0.9525 ≈ 0.953 N
The friction acting on the block is 0.953 N towards
the left.
Consider C alone. By Fnet = ma , we have
9. (a) The 0.4 kg mass will fall.
T 1 − mC g = mC a
(b) Take the direction of motion of the 0.4 kg mass
as positive.
T1 − (0.4)(9.81) = (0.4)(0.481)
∴ T1 ≈ 4.12 N
Consider the masses as a single system. By
F net = ma , we have
Therefore the tension in the string connecting
B and C is 4.12 N .
M g − mg = (M + m)a
Consider A alone. By Fnet = ma , we have
(0.4)(9.81) − (0.2)(9.81) = (0.4 + 0.2)a
m A g − T2 = m A a
∴ a = 3.27 m s−2
(0.6)(9.81) − T2 = (0.6)(0.481)
Consider the 0.4 kg mass alone.
∴ T2 ≈ 5.60 N
Therefore the tension in the string connecting
A and B is 5.60 N .
By Fnet = ma , we have
Exercise 7.2 (p.147)
(0.4)(9.81) − T = (0.4)(3.27) ⇒ T ≈ 2.62 N
1. (a) dummy
(b) dummy
The tension in the string is 2.62 N .
10. (a) Take the direction to the right as positive.
Consider the blocks as a single system.
.
2. (a) dummy
(b) dummy
By Fnet = ma , we have
m A g − mC g − f = (m A + m B + mC )a
m A g − mC g − f
m A + m B + mC
(0.6 − 0.4)(9.81) − 1
=
0.6 + 1 + 0.4
∴a=
= 0.481 m s
The acceleration of B is 0.481 m s
right.
−2
−2
3. (a) x -component: T cos 30°
y -component: T sin 30°
towards the
(b) x ′ -component: mg sin 30°
y ′ -component: mg cos 30°
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Active Physics Full Solutions to Textbook Exercises
4. (a) dummy
Hence, the resultant is 1 N upwards.
(b) dummy
(c) dummy
Hence, the resultant is
0
.
5. (a) If θ = 0°, then the magnitude of the resultant
#»
force F is F = 1 + 1 = 2 N .
(b) If θ = 45°, we have
#»
x -component of F :
F x = 1 + (1)(cos 45°) = 1.707 N
#»
y -component of F :
F y = (1)(sin 45°) = 0.7071 N
#»
Magnitude of F :
F=
Hence, the resultant is 1 N downwards.
√
√
F x 2 + F y 2 = 1.7072 + 0.70712 ≈ 1.85 N
(c) If θ = 90°, by Pythagoras’ theorem, the
#»
magnitude of F
F=
√
12 + 12 ≈ 1.41 N
(d) If θ = 135°, we have
#»
x -component of F
F x = 1 − (1)(cos 45°) = 0.2929 N
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Active Physics Full Solutions to Textbook Exercises
#»
y -component of F
F y = (1)(sin 45°) = 0.7071 N
8. By symmetry, we may only consider the nail in the
top left-hand corner. The magnitude of the resultant
force acting on the nail is
#»
Magnitude of F
F=
√
F=
Fx 2 + F y 2 =
√
0.29292 + 0.70712 ≈ 0.765 N
√
0.52 + 0.52 ≈ 0.707 N
Let θ be the angle between the resultant force and
the horizontal.
6. De ine the x and y -direction as shown.
tan θ =
0.5
⇒ θ = 45°
0.5
Due to symmetry, the resultant forces on the four
nails are 0.707 N and they all point to the centre of
the board.
x -component of the resultant force:
9. Let the direction of F1 be the +x direction as shown.
F x = 1500 + 1200 cos 75° = 1811 N
y -component of the resultant force:
F y = 1200 sin 75° = 1159 N
Magnitude of the resultant force:
F=
√
Fx 2 + F y 2 =
√
18112 + 11592 ≈ 2150 N
Since F1 has no y -component, F1x = 12 N and F1y = 0.
The y -component of F must be solely provided by
F 2 . Hence, we have
Let θ be the angle between the resultant force and
the +x direction. The direction of the resultant force
is given by
tan θ =
Fy
Fx
=
1159
⇒ θ ≈ 32.6°
1811
The resultant force is 2150 N and makes an angle
of 32.6° with the 1500 N force.
7. (a) The forces are opposite to each other. Hence,
F = F 2 − F 1 = 15 − 8 = 7 N
(b) The forces are perpendicular to each other.
Hence,
√
√
F = F 1 2 + F 2 2 = 82 + 152 = 17 N
F 2y = F sin 30° = 20 sin 30° = 10 N
F 2x = F cos 30° − F 1x = 20 cos 30° − 12 = 5.321 N
Magnitude of F2 :
F2 =
√
F 2x 2 + F 2y 2 =
F = F 1 + F 2 = 8 + 15 = 23 N
5.3212 + 102 ≈ 11.3 N
Direction of F2 :
tan θ =
(c) The forces are acting in the same direction.
Hence,
√
F 2y
F 2x
=
10
⇒ θ ≈ 62.0°
5.321
So F2 has a magnitude of
angle of 62.0° with F1 .
11.3 N
and makes an
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(b) dummy
Exercise 7.3 (p.156)
1. (a) dummy
x ′ -direction: mg sin 30° − f = ma ;
y ′ -direction: R = mg cos 30°
(b) dummy
3. (a) F increases ⇒ a increases
(b) θ increases ⇒ a decreases
(c) m increases ⇒ a decreases
θ
F cos θ = ma ⇒ a = F cos
m
4.
A The friction in region 1 is zero because the block
moves together with the belt without slipping. The
friction in region 2 is non-zero because it balances
the component of the weight of the block along the
slope.
5.
D For the bee to be at rest or moving at a constant
speed, all the forces acting on it must be balanced,
and this is possible in the given situation.
(c) dummy
(d) dummy
6. dummy
2. (a) dummy
.
Take the direction to the right as positive.
Consider the horizontal forces acting on the
suitcase. By Fnet = ma , we have
100 cos 40° − f = (20)(1) ⇒ f ≈ 56.6 N
x -direction: R sin 30° − f cos 30° = ma cos 30°;
y -direction: mg − R cos 30° − f sin 30° = ma sin 30°
So the friction acting on the suitcase is
towards the left.
56.6 N
7. Consider the vertical forces acting on the painting.
By Fnet = ma , we have
2T sin θ − mg = 0
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Active Physics Full Solutions to Textbook Exercises
Rearranging the equation, we have
sin θ =
mg (1)(9.81)
=
⇒ θ ≈ 37.8°
2T
(2)(8)
8. (a) Since the normal reaction R balances out the
component of the weight perpendicular to the
ramp, it is given by
R = mg cos θ
Since the T-shirt is at rest, the forces acting on
it are balanced.
Considering the horizontal components of the
forces, we have
T A cos 45° − TB cos 30° = 0
p
p
2T A
3TB
−
=0
2
2
p
p
2T A − 3TB = 0
(1)
= (3000)(9.81) cos 6°
Considering the vertical components of the
forces, we have
= 2.93 × 104 N
(b) dummy
T A sin 45° + TB sin 30° = (1.2)(9.81)
p
2T A TB
+
= 11.772
2
2
p
2T A + TB = 23.544
(2)
(2) − (1), we obtain
p
(1 + 3)TB = 23.544 ⇒ TB = 8.618 ≈ 8.62 N
Take the direction up the ramp as positive.
Consider the forces along the ramp when the
block moves at a constant speed. By Fnet = ma ,
we have
F − mg sin θ − f = ma
18 000 − (3000)(9.81) sin 6° − f = 0
∴ f = 14 920 ≈ 14 900 N
The friction acting on the block is 14 900 N.
Consider the forces along the ramp when the
block accelerates. By Fnet = ma , we have
F − mg sin θ − f = ma
F − (3000)(9.81) sin 6° − 14 920 = (3000)(0.2)
∴ F ≈ 18 600 N
Therefore, the pulling force required is
18 600 N up the ramp.
9. (a) dummy
Putting TB = 8.618 N into (1), we get
p
p
2T A − 3(8.618) = 0 ⇒ T A = 10.6 N
So the tensions in strings A and B are
and 8.62 N respectively.
10.6 N
(b) If the weight of the T-shirt decreases, both T A
and TB decrease.
10. The angle θ between the track and the horizontal
decreases along the track. Besides, the only force
acting on the block is its weight.
As the block slides down the track, the component
of its weight along the direction of motion (i.e.
mg sin θ ) decreases. Hence, the component of its
acceleration in that direction also decreases.
11. Weigh yourself on a scale placed on level ground.
Record your weight W .
Weigh yourself again on a scale placed on a slope of
inclination θ . Record the normal reaction R exerted
on you by the ground.
When you are at rest, the normal reaction R is equal
to the component of your weight perpendicular to
the slope, i.e. R = W cos θ , and hence we can ind θ .
More about Force | 9
Active Physics Full Solutions to Textbook Exercises
12. (a) The doll is swinging to the right in the car due
to inertia, and so the car is travelling to the left.
Exercise 7.4 (p.173)
1. (a) F increases ⇒ τ increases
(b) dummy
(b) d increases ⇒ τ increases
2. (a) d = 600×2
300 = 4 m
(b) W = 500×4
= 400 N
5
3.
A
The moment about the hinge is given by τ =
F sin θ , and so τ increases with increasing θ .
4.
Considering the vertical components of the
forces acting on the doll, we have
A
Taking moments about the pivot, we get
(30g )(x) + (60g )(2x) = (30g )y
150x = 30y
T cos θ = mg
(1)
Considering the horizontal components of the
forces acting on the doll, we have
T sin θ = ma
(2)
(2)
(1) , we have
tan θ =
a
3
=
⇒ θ ≈ 17.0°
g 9.81
(c) The following shows the free body diagram
of the doll if air resistance is taken into the
account.
x : y =1:5
5.
A Increasing the load increases the clockwise moment on the crane about the pivot. To maintain the
equilibrium of the crane, the counterweight should
be moved to the left to increase the anticlockwise
moment on the crane about the pivot.
6. The moment of the couple is given by
( )
( )
ℓ
ℓ
τ=F
sin θ + F
sin θ = F ℓ sin θ
2
2
7. (a) The vertical net force on the plate = 2 − 2 = 0.
The horizontal net force on the plate = 1 N
towards the right.
So the net force acting on the plate is 1 N
towards the right.
(b) The sum of the clockwise moment about O
= 2 × 0.5 = 1 N m.
The sum of the anticlockwise moment about O
Considering the horizontal direction, we have
T ′ sin θ ′ − f = ma ⇒ T ′ sin θ ′ > T sin θ
Hence, the horizontal component of the
tension increases.
On the other hand, the vertical component of
the tension remains unchanged:
= 1 × 0.5 = 0.5 N m.
The net moment about O is 0.5 N m
(clockwise).
(c) The sum of the clockwise moment about P
= 2 × 0.5 = 1 N m.
The sum of the anticlockwise moment about P
= 1 × 1 = 1 N m.
The net moment about P is
′
0
.
′
T cos θ = T cos θ
Combining the above two conditions, we have
tan θ ′ > tan θ
Hence, the value of θ would increase.
8. (a) Simon is incorrect .
From the information given, we can only
conclude that the c.g. of the stick must be
located along the dashed line between points X
and C . The c.g. of the stick may not necessarily
be located at C .
10 | More about Force
Active Physics Full Solutions to Textbook Exercises
(b) The stick swings back and forth about point X
until it comes to rest again.
11. (a) dummy
9. dummy
In equilibrium, the net moment on the plank
about any point is zero. Taking moments about
trestle A , we have
In equilibrium, the net moment about the left end of
the platform is zero. Hence we have
m K g (2.5) + m p g (2) = F B (3)
(50)(9.81)(2.5) + (20)(9.81)(2) = 3F B
∴ FB = 539.55 ≈ 540 N
T2 (3) = mg (1) + M g (1.5)
3T2 = (60 × 9.81)(1) + (100 × 9.81)(1.5)
In equilibrium, the net force on the plank is
zero. Hence, we have
∴ T2 = 686.7 N
In equilibrium, the net force acting on the platform
is zero. Hence we have
F A + FB = mK g + mp g
F A + 539.55 = (50)(9.81) + (20)(9.81)
T1 + T2 = mg + M g
∴ F A = 147.15 ≈ 147 N
T1 + 686.7 = (60)(9.81) + (100)(9.81)
So the forces F A and FB are
respectively.
∴ T1 = 882.9 N
So the tensions in the left cable and the right cable
are 882.9 N and 686.7 N respectively.
10. Let W be the weight of Horace. In equilibrium, the
net moment on Horace about any point is zero.
147 N
and
(b) When she walks towards the right, F A
decreases. Suppose, after she walks for a
distance x , the plank topples as F A = 0.
Taking moments about balance A , we have
W x = (250)(2)
W x = 500
(1)
Taking moments about balance B , we have
W (2 − x) = (550)(2)
W (2 − x) = 1100
(2)
(1) , we have
(2)
Taking moments about trestle B , we have
2 − x 1100
=
x
500
∴ x = 0.625 m
m p g (1) = m K g (x − 0.5)
20 = 50(x − 0.5)
∴ x = 0.9 m
540 N
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Active Physics Full Solutions to Textbook Exercises
She can walk for a distance of
toppling the plank.
0.9 m
.
without
2.
12. (a) Let T be the tension in the rope. The c.g.
of the plank is 1 m away from the hinge. In
equilibrium, the net moment on the plank
about any point is zero. Taking moments about
the hinge, we have
The parcel moves at a constant velocity from B to
C , and so the net force acting on the parcel is zero.
The belt does not exert any friction on the parcel,
because they move together without any sliding
motion between their surfaces.
(40 × 9.81)(1) = (T sin 45°)(2)
∴ T ≈ 277 N
3.
The tension in the rope is
277 N
.
180° − 45°
= 67.5°
2
Taking moments about the hinge, we have
4.
(40 × 9.81 × sin 45°)(1) = (T sin 67.5°)(2)
150 N
C Statement (1) is correct. The block moves at a
constant velocity, and so the net force acting on it is
zero.
Statement (2) is correct. Lubricant decreases the
friction acting on the block. Therefore, the block will
accelerate down the inclined plane.
∴ T ≈ 150 N
The tension in the rope is
B For the object to remain at rest, the three forces
must balance each other.
The resultant force of the two forces shown is 1 N to
the −y -direction. Hence, the third force is 1 N to the
+y -direction.
(b) The angle between the rope and the plank is
θ=
The parcel accelerates while moving from A to B ,
and the only force acting on it is the friction exerted
by the belt. Hence, the friction points to the right.
C
.
Statement (3) is incorrect. If the angle of inclination
increases, the block accelerates down the inclined
plane.
13. (a) The force can produce the largest moment
when the force is applied in a direction
perpendicular to the handle.
(b) In equilibrium, the net moment on the hammer
about any point is zero. Taking moments about
P , we have
5.
D
F d Y P = 600d X P
)
dY P
F
= 600
dX P
(
F (15) = 600
∴ F = 40 N
So the minimum pulling force is
40 N
.
Chapter Exercise
Since the block is at rest, the net force acting on the
block is zero.
Multiple-choice Questions (p.178)
Forces on the block perpendicular to the force F :
1.
In Fig. Q1a, the force exerted on Q by P is the
only force acting to accelerate the less massive Q .
This force has the same magnitude as the force
exerted on P by Q .
A
In Fig. Q1b, the force exerted on P by Q is the only
force acting to accelerate the more massive P .
Hence, this force is greater than that in Fig. Q1a.
T = W cos 30° < W
Forces on the block along the force F :
F = W sin 30°
As cos 30° < sin 30°, we get F < T .
12 | More about Force
6.
Active Physics Full Solutions to Textbook Exercises
If X moves towards the right end of the plank,
the supporting force exerted by X must increase to
produce the same clockwise moment about the right
end in order to maintain the equilibrium.
Subsequently, ℓ decreases and θ increases, and
both of them lead to the decrease of F , and so
F < 0.94M g .
C
11.
If the supporting force exerted by X increases, the
supporting force exerted by Y must decrease, such
that the net force acting on the plank is zero.
7.
8.
A Since the dumbbell falls at a constant speed,
the net force acting on the dumbbell is zero. As the
dumbbell remains horizontal, the net torque on the
dumbbell is also zero. Therefore, the dumbbell is in
equilibrium.
Consider the blocks and the spring balance as a
single system.
C
By Fnet = ma , we get
F 1 − F 2 = (3m)a ⇒ a =
The net force acting on the block is zero because
it is moving with a constant velocity.
D
F1 − F2
3m
Consider P alone.
Considering the vertical forces, we have
R + F sin θ = W ⇒ R < W
R is smaller than W .
9.
By Fnet = ma , we get
Let ℓ be the required length. To balance the bar,
the net moment on the bar about any point must be
zero. Taking moments about the left hand end, we
have
C
F 1 − T = ma
Solving for T , we have
)
(
2F 1 + F 2
F1 − F2
=
T = F 1 −
m
m
3
3
(3ℓ) = (10 − ℓ) +
(5 − ℓ) ⇒ ℓ = 3 units
mg
mg
mg
10.
D
Consider the forces acting on the gangplank.
The reading of the spring balance =
12.
.
Taking moments about P , we get
M g × ℓ = F sin θ × 1.5
∴F =
Mgℓ
1.5 sin θ ◦
Initially, ℓ = 1 m and θ = 45°. Hence, we get
F=
M g (1)
≈ 0.94M g
1.5 sin 45◦
D
See the following steps.
2F 1 + F 2
.
3
More about Force | 13
Active Physics Full Solutions to Textbook Exercises
13.
second compartment. By Fnet = ma , we get
In equilibrium, all the forces acting on the
weight are balanced. Considering the vertical forces
acting on the weight, we have
A
T2 − f 2 = m 2 a
) (
)
T2 − 1.8 × 105 = 6 × 104 (1.167)
(
W = 30 sin θ + 20 sin ϕ
< 30 + 20 sin ϕ
(∵ sin θ < 1 for 0 < θ < 90°)
< 30 + 20
(∵ sin ϕ < 1 for 0 < ϕ < 90°)
(1M)
T2 = 2.5 × 105 N
So the tensions in the irst and the second chains are 5 × 105 N and 2.5 × 105 N
respectively.
Hence, W < 50 N.
(1A)
(c) If the chain connecting the irst and the
second compartments is suddenly broken, the
mass of the system decreases, and hence the
acceleration of the system increases.
(1A)
Structured Questions (p.180)
14. (a) dummy
Considering the horizontal forces acting on the
locomotive, we have
F − T1 − f 1 = ma → T1 = F − f 1 − ma
Therefore, the tension in the chain connecting
the locomotive and the irst compartment will
decrease.
(1A)
.
Consider the locomotive and the compartments as a single system.
15. Put ive weights on the weight rack of the rider.
Release the system. Record its acceleration a and
the net force Fnet acting on it (i.e. the weight of the
weight holder).
(1A)
The total mass of the locomotive and the
compartments is
)
)
(
(
M = 1.2 × 105 + 2 × 6 × 104 = 2.4 × 105 kg
Transfer one weight from the rider to the weight
holder. Release the system again and record a and
F net (i.e. the new weight of the weight holder).
(1A)
Take the direction to the right as positive. By
F net = ma , we get
Repeat the above steps until all the weights are
transferred.
F − f1 − 2 f2 = M a
(
)
(
) (
)
6
5
10 − 3.6 × 10 − 2 1.8 × 105 = 2.4 × 105 a
(1M)
∴ a = 1.167
≈ 1.17 m s−2
The acceleration of the train is
the left.
1.17 m s−2
(1A)
16. (a) Take the direction of motion of the blocks as
positive.
(b) Let T1 and T2 be the tensions in the irst and
the second chains respectively.
Consider the blocks as a single system. By
F net = ma , we have
Consider the horizontal forces acting on the
locomotive. By Fnet = ma , we get
T1 = 5 × 105 N
Consider the horizontal forces acting on the
Plot a graph of Fnet against a . Newton’s second
law is veri ied if a straight line passing through the
origin is obtained.
(1A)
In the experiment, the weights are transferred within the
system so as to keep the total mass m of the system unchanged.
to
F − T1 − f 1 = m 1 a
(
) (
)
6
10 − T1 − 3.6 × 105 = 1.2 × 105 (1.167)
(1A)
m B g + mC g − f = (m A + m B + mC )a
(1M)
(1)(9.81) + (1)(9.81) − 4 = (1 + 1 + 1)a
∴ a = 5.207 ≈ 5.21 m s
The common acceleration of the blocks is
5.21 m s−2 .
(1M)
−2
(1A)
14 | More about Force
Active Physics Full Solutions to Textbook Exercises
(b) Consider A alone. By Fnet = ma , we have
Considering the horizontal forces acting on the
helicopter, we have
T AB − f = m A a
T AB − 4 = (1)(5.207)
L sin θ − T sin 10° = M a
(1M)
∴ T AB ≈ 9.21 N
Rearranging the equation,
Consider C alone. By Fnet = ma , we have
L sin θ = M a + T sin 10°
)
(
= (2500)(1.730) + 1.992 × 104 sin 10°
mC g − TBC = mC a
(1)(9.81) − TBC = (1)(5.207)
= 7784 N
(1M)
∴ TBC ≈ 4.60 N
Therefore, the tension in the string connecting
A and B is 9.21 N , and the tension in the
string connecting B and C is 4.60 N .
(1A)
Considering the vertical forces acting on the
helicopter, we have
L cos θ = M g + T cos 10°
= (2500)(9.81) + 19 923 cos 10°
= 4.415 × 104 N
17. (a) dummy
Set up the equations: 1M
Magnitude of L :
L=
√
)2
(
77842 + 4.415 × 104 ≈ 4.483 × 104 N
Direction of L :
By Newton’s second law, we have
tan θ =
T cos 10° = mg
T cos 10° = (2000)(9.81)
(1M)
∴ T = 1.992 × 10 ≈ 1.99 × 10 N
4
4
The tension in the cord is
1.99 × 104 N
.
7784
4.415 × 104
⇒ θ ≈ 10.0°
Combine the equations to ind the unknowns: 1M
The lifting force has a magnitude of 44 800 N
and makes an angle of 10.0° with the vertical.
(1A)
(b) Considering the horizontal forces acting on the
bucket, we have
T sin 10° = ma
(
)
1.992 × 104 sin 10° = 2000a
(1A)
18. (a) (i) dummy
(1M)
−2
∴ a = 1.730 ≈ 1.73 m s
Since the bucket and the helicopter have the
same acceleration, the acceleration of the
(1A)
helicopter is 1.73 m s−2 forwards.
When the box is raised steadily, the
magnitude of the pulling force balances
the weight of the box, i.e. F = mg .
(1M)
Since that the pulley remains stationary,
all the forces acting on the pulley must
be balanced. By symmetry, we know that
(1A)
θ = 30° .
(c) dummy
(ii) The tension in the string is
T = 2(100 cos 30°) ≈ 173 N
(b) No.
(1M+1A)
(1A)
More about Force | 15
Active Physics Full Solutions to Textbook Exercises
If θ = φ, then the tension T and the pulling
force F will be parallel to each other, like this:
From the graph, when θ = 90°, F = 700 N.
Hence,
700
= 700 N
sin 90°
W=
.
The net force acting on the pulley is no longer
zero because no force can balance the weight
mg of the box.
(1A)
19. (a) The normal reaction acting on A is
R A = mg cos θ
= (0.5)(9.81) cos 15°
= 4.737 ≈ 4.74 N
(1A)
(b) (i) As the player has given two incorrect
answers, we have θ = 20°.
From the graph, when θ = 20°, f > F .
Thus the frictional force acting by the track
f ′ = 240 N , and the net force acting on the
cart is 0 .
(2A)
(ii) By Newton’s third law, the normal reaction
R acting on the cart by the track is equal
to the force acting on the track by the cart,
which is W cos θ .
(1M)
Therefore,
R = W cos θ
= 700 cos 20°
= 657.8 ≈ 658 N
The friction acting on A is
f A = mg sin θ
The normal reaction R is
= (0.5)(9.81) sin 15°
.
(1A)
(1A)
Hence, the player can only give 3 incorrect
answers before he slides into the pool.
(1A)
(2A)
(b) (i) Consider the books as a single system. By
F net = ma , we have
f = M g sin 45°
= (0.5 + 1)(9.81) sin 45°
658 N
(c) From the graph, if θ > 30°, F will be greater
than f .
= 1.270 ≈ 1.27 N
So the normal reaction and the friction
acting on A by B are 4.74 N and 1.27 N
respectively.
(1M)
(1M)
≈ 10.4 N
The limiting friction between book B and
the inclined plane is 10.4 N .
(1A)
(ii) Consider book A alone. By Fnet = ma , we
have
f = mg sin 60°
= (0.5)(9.81) sin 60°
≈ 4.25 N
The limiting friction between the books is
(1A)
4.25 N .
20. (a) Since F is the component of W parallel to the
track, we get F = W sin θ .
(1A)
21. (a) Let F A and FB be the supporting forces
provided by workers A and B respectively.
In equilibrium, the net moment on the pipe
about any point is zero. Taking moments about
the point where worker B supports the pipe,
we have
( )
L
= F A d AB
2
5
(80 × 9.81) × = F A × 4.2
2
mg
(1M)
∴ F A = 467.1 ≈ 467 N
In equilibrium, the net force acting on the pipe
is zero. Considering the vertical forces only, we
have
F A + F B = mg
467.1 + F B = (80)(9.81)
(1M)
∴ FB ≈ 318 N
The forces exerted on the pipe by A and B are
(1A)
467 N and 318 N respectively.
16 | More about Force
Active Physics Full Solutions to Textbook Exercises
(b) Worker A should hold the pipe farther from
worker B .
Taking moments about B ′ , we have
(1A)
(mg cos θ)(1.5) = (R A ′ sin θ)(3)
Consider the moment about the point where
worker B supports the pipe. Worker A could
exert a smaller supporting force but still
produce the moment for the pipe remains in
equilibrium.
(1A)
Rearranging the equation, we have
R A′ =
1.5mg cos θ
mg
=
3 sin θ
2 tan θ
(1M)
If θ decreases, then tan θ decreases and hence
22. (a) dummy
R A ′ increases. Therefore, both the answers in
(b) and (c) will increase.
(1A)
.
23. (a) In equilibrium, the net moment on his spine
about any point is zero. Taking moments about
point J , we have
(F sin 10°) (0.25) = (W sin 60°) (0.25)+(L sin 60°) (0.4)
In equilibrium, the net force acting on the
board is zero. Considering the vertical forces
only, we have
Rearranging the equation, we have
F=
R B = W = 100 N
The normal reaction exerted by the ground on
the board is 100 N .
(1A)
(b) In equilibrium, the net moment on the board
about any point is zero. Taking moments about
B , we have
≈ 2790 N
(1A)
(b) In equilibrium, the net moment on his spine
about any point is zero. Taking moments about
point J , we have
Rearranging the equation, we have
(1M)
∴ R A = 50 N
F=
The normal reaction exerted by the wall on the
board is 50 N .
(1A)
p
Note that cos 45° = sin 45° = 22 .
(c) In equilibrium, the net force acting on the
board is zero. Considering the horizontal
forces only, we have
f = R A = 50 N
The friction between the board and the ground
is 50 N .
(1A)
(d) dummy
(1M)
(F sin 10°) (0.25) = (W sin 30°) (0.25)+(L sin 30°) (0.4)
°)(1.5) = (R 45
cos
(mg
A sin 45°)(3)
(100)(1.5) = (R A )(3)
(400 sin 60°) (0.25) + (100 sin 60°) (0.4)
(sin 10°) (0.25)
(400 sin 30°) (0.25) + (100 sin 30°) (0.4)
(sin 10°) (0.25)
≈ 1610 N
(1M)
(1A)
(c) Standing as close to the object as possible
while lifting reduces the angle between your
spine and the vertical.
(1A)
Lifting with such posture requires less force
and hence reduces the burden of the back
muscle.
(1A)
24. (a) For an object that does not rotate, we get
sum of clockwise moments = sum of anticlockwise moments.
(1A)
More about Force | 17
Active Physics Full Solutions to Textbook Exercises
(b) (i) dummy
(c) Taking moments about P , we have
F A × d AP = W × dC P
F A × 0.10 = 160 × 0.40
F A = 640 N
(d) The minimum force becomes smaller.
mg d b = W d t
(
)
m(9.80)(2.6 − 0.4) = 2 × 104 (1.25)
m=
25 000
21.56
(1A)
(1M)
(1M)
≈ 1160 kg
.
The mass of the block is
1160 kg
.
(1A)
(iii) Yes, it would be possible.
(1A)
The distances of the centres of gravity of
both the concrete block and the truck from
P are the same as in (a)(ii). Hence, the
block can be raised to that point without
toppling the truck.
(1A)
(iv) Any
of the following:
(1A)
(1A)
The distance of A from the pivot (Q ) is longer
than before.
(1A)
Correct label: 1A
(ii) In equilibrium, the net moment on the
forklift truck about any point is zero.
Taking moment about P , we have
(1M)
By exerting a smaller force at A , the clockwise
moment it produced can still balance the
anticlockwise moment produced by the
weight.
(2A)
26. (a) (i) The negative sign indicates that the force
acts upwards.
(1A)
Under normal working conditions, the
scale measures the weight of the object
to ind its mass. But since the scale reading
shows the tension in the string, we have
T = mg
= (0.0094)(9.81) = 0.092 21 ≈ 0.1 N
(1A)
(ii) dummy
(1A)
• Increase the mass of the truck.
• Reduce the length of the forklift arm.
• Move the centre of gravity of the truck
backward.
25. (a) The point at which the weight of an object
appears to act
• Tension in the string: 1A
• Weight of the balloon: 1A
(2A)
Since the net force acting on the balance is
zero, we have
The c.g. of an object may lie outside its body.
(b) (i) In equilibrium, the net moment on the
trolley about any point is zero. Taking
moments about Q , we have
W = F −T
= 0.16 − 0.092 21 = 0.067 79 ≈ 0.07 N
F P × d PQ = W × dCQ
F P × 0.90 = 160 × 0.50
∴ FP = 88.89 ≈ 88.9 N
(b) (i) dummy
(1M)
(1A)
(ii) In equilibrium, the net force on the trolley
is zero. Considering the vertical forces
only, we have
FQ = W − F P = 160 − 88.89 ≈ 71.1 N
(1A)
• Weight of the balloon: 1A
• Air resistance acting on the balloon: 1A
(1A)
18 | More about Force
Active Physics Full Solutions to Textbook Exercises
(ii) T cos 43°
(1A)
(iii) Since the balloon is moving at a constant
speed, the net force acting on it is zero.
Hence, we have
T cos 43° = 0.16 − 0.068
(iii) Applying the equation of uniformly
accelerated motion, we have
*+0 1 at 2 ⇒ s = 1 at 2 ⇒ t 2 = 2 s
s =
ut
2
2
a
So the slope of the graph is a2 .
Solving for the acceleration a , we get
(1M)
∴ T ≈ 0.126 N
The tension in the string is
0.126 N
.
2
= 1.556 ⇒ a = 1.286 ≈ 1.29 m s−1
a
(1A)
(c) When the child walks faster, the air resistance
exerted on the balloon increases.
(1A)
(1M)
(1A)
(b) dummy
Therefore, the horizontal component of the
tension has to increase in order to overcome
the increase in air resistance.
(1A)
27. (a) (i) dummy
D /m
t /s
t 2 / s2
0.4
0.79
0.62
• Correct arrows: 1A
0.8
1.12
1.25
• Correct labels: 1A
1.2
1.37
1.88
1.6
1.59
2.53
(c) Take the direction down the slope as positive.
Considering the forces acting along the slope,
we have
Refer to the igure below.
mg sin θ − f = ma
(1M)
Solving for f , we have
∴ f = mg sin θ − ma
= m(g sin θ − a)
= (0.178)(10 sin 25° − 1.285)
= 0.524 N
Hence, the friction acting on the block is
0.524 N (up the slope).
(1A)
28. (a) Without a ixed base, the model will topple in
the anticlockwise direction,
(1A)
due to the net moment on the model about the
base.
(1A)
• Correct axes and labels: 1A
• Correct scale: 1A
(b) Considering the moments produced by the
weights of the counterweight and arm C D
about O , we have
• Correct data points: 1A
• Correct best- it line: 1A
(ii) The slope of the graph is
{
1.4 − 0
slope =
0.9 − 0
= 1.556 ≈ 1.56 m s−1
moment = 4 × 10 × 0.2 = 8 N m
moment = 50 × 10 × 0.5 = 250 N m
(1A)
So the anticlockwise moment is much greater
than the clockwise moment. However, in
(1A)
More about Force | 19
Active Physics Full Solutions to Textbook Exercises
equilibrium, the net moment about any point
on the steel rod must be zero, and so wire
1 has to be taut in order to provide an extra
clockwise moment.
(1A)
The normal reaction R acting on the arm is
shown below.
Shoot-the-stars Questions (p.185)
1.
B The lifting force F balances the weight of
the plate. (Note that the mass of the strings is
negligible.)
2. (a) The mass of Q is
4
mQ = 10 × π(4)3 = 2681 g = 2.681 kg
3
Free body diagram of Q :
mg
we have F cos θ = mg ⇒ F = cos θ .
(1M)
Considering the following right-angled triangle,
Correct arrow: 1A
R balances the other forces acting on the arm.
(c) Taking moments about O , we have
(1M)
5
we have sin θ = 10
12 = 6 ⇒ cos θ =
moment = (40 sin 29°)(0.5) + (4 × 10)(0.2)
= 314.7 N m
F=
moment = (T sin 17°)(0.9) + (50 × 10)(0.5)
The force exerted on Q by P is
(1M)
∴ T ≈ 246 N
.
47.6 N
.
(b) If the length of the container decreases, the
value of θ decreases.
Since the rod is balanced, we have
0.2631T + 250 = 314.7
mg
(2.681)(9.81)
=
≈ 47.6 N
p
11
cos θ
6
= (0.2631T + 250) N m
246 N
(1M)
Hence,
= + (33 × 10)(0.9)
The tension in wire 2 is
p
11
6 .
Therefore, the answer in (a) will decrease.
3. (a) (i) The c.g. of X must lie on top of Y .
(1A)
(ii) The c.g. of the system of X and Y as a
whole must lie on top of the table.
(b) First consider brick X .
(1A)
(1A)
(1A)
(1A)
(1A)
20 | More about Force
Active Physics Full Solutions to Textbook Exercises
If the c.g. of X is not supported by Y , its weight
will produce a moment about P (the edge of
Y ), but there is no force acting on X to balance
the moment. Hence, the farthest position
which X can be placed is where its c.g. is just
above P . Therefore,
a1 =
24
= 12 cm
2
of the table.
(1M)
Taking moments about Q , we have
a 2 = (12 − a2 ) ⇒ a2 = 6 cm
mg
mg
(1A)
(c) Consider brick Z .
(1A)
Then consider brick Y .
If the three bricks do not topple over the table,
the overall c.g. of X , Y and Z is located just
above Q .
Similarly, if the bricks do not topple over the
table, the farthest position is where the overall
c.g. of X and Y is located just above Q , the edge
Taking moments about Q , we have
a 3 = (12 − a3 ) ⇒ a3 = 4 cm
2
mg
mg
.
(1A)