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Maths Quest 12 Specialist Mathematics
b The magnitude and direction, anticlockwise from the i direction, respectively of a
˜
third force for the resultant force to be zero is nearest to:
A 182 N, 49° B 15.3 N, 41° C 13.5 N, 41° D 182 N, 229° E 13.5 N, 227°
Spreadshe
et
EXCEL
3 multiple
ultiple choice
A (5 newtons)
~
The three coplanar forces, A , B and C act in such a way that the
˜ ˜a triangle
˜ of forces to assist in
resultant force is zero. Use
C
~
answering the following questions.
a The magnitude of C is:
B (4 newtons)
A 15 N B 21 ˜N C 3 N
D −21 N E 0 N
~
b The angle between the forces A and C , to the nearest degree, is:
˜ C 143°
˜
A 37°
B 53°
D 137°
E 127°
Vector
addition
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Example
xample
2
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Example
xample
3
4 Two forces — F 1 and F 2 — act on an object. They are described by the vectors:
˜
˜
i F 1 = 13i – 5 j, F 2 = –4i + 9 j
˜
˜
˜
˜ ˜
˜
ii F 1 = 8i + 6 j , F 2 = –14i – 9 j
˜
˜
˜
˜ ˜
˜
iii F 1 = 2 2i – 3 j, F 2 = – 3 5i + 2 j .
˜
˜
˜
˜ ˜
˜
For each of the above find:
a the resultant force, R , acting on the object
b the magnitude of the˜ resultant force, R
˜ that the resultant force is equal to zero.
c a third force, F 3 , applied to the body so
˜
5 Three forces — F 1 , F 2 and F 3 — act on an object. They are described by the vectors:
˜
˜
˜
i F 1 = 3i – 5 j, F 2 = 4i + 9 j and F 3 = –6i – 2 j
˜
˜
˜
˜
˜
˜
˜
˜
˜
ii F 1 = i – j, F 2 = – 2 i + 3 j and F 3 = 2i – 2 j .
˜
˜
˜ ˜ ˜
˜
˜
˜
˜
For each of the above find:
a the resultant force, R , acting on the object
b the magnitude of the˜ resultant force, R
˜ i vector
c the angle that the vector R makes to the
˜
d a fourth force, F 4 , applied to the body so˜ that the resultant force is equal to zero.
˜
6 In each of the following cases:
i find the resultant force, R , for the specified forces acting on a body
˜
ii describe the additional force
needed to attain equilibrium; that is, R = 0.
˜
a F = 25 N at N30°E and G = 45 N at N30°W
˜
˜
b W = – 80 j and F = 50i + 75 j
˜
˜
˜
˜
˜
c A = 26i – 80 j and B = –30i + 92 j
˜
˜
˜
˜
˜
˜
d F = 10 N at S50°W and G = 7.0 N at N45°W and H = 12 N at N70°E
˜
˜
˜
7 Three forces — X , Y and Z — act on an object such that the resultant force R = 0.
˜ angle˜of 120° to the force X and has the same magnitude˜as the
The force Y acts ˜at an
˜
˜
force X which is 10 N.
˜
a Determine the magnitude of Z .
b Find the angle that the force Z˜ makes to X .
˜
˜
8 Three coplanar forces — A , B and C — have magnitudes of 1000 N, 1200 N and
˜ a body
˜ such that the resultant force is zero. Find the
1700 N respectively. They ˜act on
angle between A and B .
˜
˜
Chapter 11 Mechanics
545
686 N or 70 g N downwards. That is, the weight W of a mass m in a gravitational field
˜
g is given by˜ the equation:
˜
W = mg
˜ is used
˜ universally in examples and problems
Note: The value for g of 9.8 N/kg down
˜
in this textbook. Sometimes the unit for g is quoted as m/s2. This is the acceleration
˜
due to gravity. For a mass of 1 kg, it becomes
equal to the weight force when all other
forces are ignored.
Resolving a force into its components
In many cases the size of a force acting in a particular direction
j
~
needs to be calculated. For example, when a jet aircraft is taking
~i
off from a runway the engines provide a thrust. Part of the thrust
~F
assists in lifting the plane up into the air (the vertical component)
~Fy
θ
and the other component exerts a horizontal push making the
F
plane move forward. In the diagram at right, the combined thrust Aircraft ~ x
of the jet’s engines is represented by a single vector F acting on
˜
a particle representing the aircraft.
If F is the applied force acting on the plane due to the action of the engines then we
˜
can write
the force vector as the sum of two vectors F x and F y which are parallel to
˜
˜
the unit vectors i and j respectively.
˜
˜
F = Fx + Fy
˜
˜
˜
F = Fx i + Fy j
˜
˜ ˜
˜ ˜
where
F x = F cos θ
˜
˜
and
F y = F sin θ.
˜
˜
The quantities F x and F y are referred to as the components or resolutes of the
˜ i and ˜j directions respectively.
vector force F in the
˜
˜
˜
A force F can be resolved into perpendicular components:
˜
F = F cos θ i + F sin θ j
˜
˜
˜
where F = F and θ is the angle between F and i .
˜
˜
˜
For example, if the combined thrust of the engines is 2 × 106 N and the plane during
take off had a 15° elevation angle then the vertical component would be:
Thus
F y = F sin θ
˜
˜
= 2 × 106 × sin 15°
≈ 5.18 × 105 N
while the horizontal component would be:
F x = F cos θ
˜
˜
= 2 × 106 × cos 15°
≈ 1.93 × 106 N
or F ≈ (1.93 × 106) i + (5.18 × 105) j
˜
˜
˜
Note: The i and j components are usually horizontal
˜
and vertical˜ components
respectively but they can be
rotated to suit a particular problem.
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Maths Quest 12 Specialist Mathematics
11B
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Example
xample
4
Newton’s First Law of
Motion
Note: Use g = 9.8 m/s2 down for all problems involving weight.
˜
1 A girl is pulling along her baby brother in a cart attached to a rope. The cart is on a
horizontal path. The rope makes an angle of 20° to the horizontal and exerts a force on
the cart of magnitude 25 N.
a Calculate the horizontal component of the force exerted on the cart by the rope.
b Calculate the vertical component of the force exerted on the cart by the rope.
The cart moves along at a constant velocity; that is, the acceleration of the cart is zero.
c Calculate the size of the horizontal friction force acting on the cart.
2 multiple
ultiple choice
A child of mass 40 kg is held on a ‘swinging rope’ at an angle of 37.43° to the vertical by
a horizontal force of 300 N. If T is the tension force of the rope acting on the child, then:
˜
a the force vector diagram which
best represents this situation is:
A
B
52.57°
~T
37.43°
300 N
~T
300 N
40g N
C
40g N
D
37.43°
~T
300 N
37.43°
40g N
E
37.43°
~T
~T
300 N
40g N
40g N
300 N
b the horizontal component of T is:
A 40 g N
B 300 N ˜ C 127 N
c the vertical component of T is:
A 300 N
B 127 N ˜
C 40 g N
d the magnitude of T is nearest to:
A 494 N
B˜ 440 N
C 477 N
D 272 N
E 200 N
D 200 N
E 272 N
D 92 N
E 300 N
3 A ship of mass m is being pulled by two tugboats. It glides
Tugboat
10°
Ship
through the water at constant velocity. The angle between the
two ropes connecting the two tugboats to the front of the ship is
10°
Tugboat
20° and they each support a tension of magnitude 20 000 N.
a Draw a force vector diagram showing all three coplanar forces which act on the ship.
b What is the magnitude of the resultant force acting on the ship? (Be careful; use
Newton’s First Law of Motion.)
c Calculate the magnitude of the force of friction due to the water acting on the ship.
Chapter 11 Mechanics
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Example
xample
5
551
4 A swing chair of mass 8 kg is suspended
by two taut ropes as shown at right. The
tension forces in rope 1 and rope 2 are T 1
˜
60°
30°
and T 2 respectively.
˜
a Draw a force vector diagram showing
Rope 2
Rope 1
all three forces which act on the swing.
b By resolving vectors into i – j com˜ ˜ of T and T .
ponents find the exact magnitudes
˜1
˜2
5 A speaker in an auditorium has a mass of 50 kg and is suspended
6.0 m
from the ceiling by two 4.0-m ropes. The ropes are attached to
A
B
the ceiling at points A and B whose separation is 6.0 m as shown
at right.
4.0 m
a Draw a force vector diagram showing all forces which act on
j
~
the speaker.
b Calculate the angle that each rope makes with the ceiling.
~i
c Determine the vertical component of the tension force in each rope.
d Give the vectors for the tensions in the left and right rope respectively using
i – j notation.
˜ ˜
e Calculate
the magnitude of the tension in each rope.
f The speaker is to be raised by increasing the separation between the points A and B,
but the ropes will break if the tension exceeds 4000 N. Find the maximum possible
separation between A and B; that is, when the tensions in the ropes are equal to
4000 N.
6 Sam earns some extra pocket money by mowing his neighbour’s front lawn.
When he pushes the lawn-mower at a constant velocity he applies a
force of 120 N down the shaft of the mower which is angled at
120 N
40° to the vertical. The lawn-mower has a mass of 40 kg.
40°
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Maths Quest 12 Specialist Mathematics
a Draw a force vector diagram illustrating all four forces acting on the lawn-mower.
(Treat the lawn-mower as a particle.)
b Calculate both the vertical and horizontal components of the force that Sam applies
to the lawn-mower.
c Using i – j notation, write a vector equation for the resultant force, R , acting on the
˜ ˜ in terms of the four forces acting on the lawn-mower. ˜
lawn-mower
d Determine the magnitude of the force of friction acting on the lawn-mower as it
moves across the lawn at constant velocity.
e Determine the magnitude of the normal contact force.
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Example
xample
6
7 A 1.5-kg mass is placed on a smooth inclined plane angled at
j
~
30° to the horizontal. To stop it from sliding down the plane a
string is attached to the upper side as shown at right. The unit
vectors i and j are also shown.
˜ force diagram showing the forces which act on the
a Draw˜ a vector
forces N for the normal contact force, T for the tension force and
˜
˜
arising from
the effect of gravity.
b What is the magnitude of the resultant force, R ?
˜
c Determine the weight vector, W , using i – j notation.
˜
˜
d Determine the tension force, T , using i – j ˜ notation.
˜ normal˜ contact
˜
e Determine the magnitude of the
force, N.
8 A 1.5-kg mass is placed on a smooth inclined plane angled at
30° to the horizontal. To stop it from sliding down the plane a
horizontal force is applied to the mass as shown at right. The
unit vectors i and j are also shown.
˜
˜
a Draw a vector
force
diagram showing the forces which act
on the mass. Label the forces N for the normal contact
˜ W for the force due to
force, H for the horizontal force and
˜
˜
gravity.
b Determine the weight vector, W , using i – j notation.
c Determine the horizontal force,˜ H , using˜ i˜– j notation.
˜ N , using
˜ ˜ i – j notation.
d Determine the normal contact force,
˜ ˜
˜
3W
W
e Show that H = ------˜ .
˜ - = --------------3
˜
3
~i
30°
mass. Label the
W for the force
˜
j
~
Horizontal
force, ~
H
~i
30°
9 A ball of mass 0.20 kg is shot vertically in the air. It decelerates under the action of two
forces, namely the weight force and the force of air resistance. When the ball moves
with a speed of 40 m/s it has an air resistance of 1.0 N. When the ball is stationary the
air resistance force is equal to zero.
a Draw a force diagram to show all the forces acting on the ball as it is moving
upwards.
b Determine the magnitude and direction of the resultant force acting on the ball
when it is moving upwards at a speed of 40 m/s.
c Determine the magnitude and direction of the resultant force acting on the ball
when it is at its maximum height above the ground.
Later, the ball is travelling toward the ground at 40 m/s.
d Draw a force diagram to show all the forces acting on the ball as it is moving downwards.
e Determine the magnitude and direction of the resultant force acting on the ball
when it is moving downwards at a speed of 40 m/s.
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Maths Quest 12 Specialist Mathematics
11F
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Example
xample
17
Variable forces
1 A particle of mass 1 kg is acted on by a force of magnitude (4t − 3t2) where 0 ≤ t ≤ 5.
If the particle is initially at rest at the origin, find:
a the velocity at any time t
b the displacement at any time t.
t
2 An object of mass 3 kg is acted on by a force of magnitude t − cos --- where
2
0 ≤ t ≤ π. If the object is initially at rest at the origin, find:
a the velocity at any time t
b the displacement at any time t.
3 An object of mass 4 kg, initially at rest, is acted on by a force F = (2 − t)2 N where
0 ≤ t ≤ 7 seconds. Find:
a the speed of the object after 5 seconds
b the distance travelled after 5 seconds.
4 A body of mass 6 kg, initially at rest, is acted on by a force F = 1 + sin 2t N where
0 ≤ t ≤ 4 seconds. Find:
π
a the speed of the body after --- seconds
2
π
b the distance travelled after --- seconds.
2
40
5 A 2-kg mass initially travelling at 2 m/s is brought to rest by a force F = -----------------2- N
(
t
+
5)
where t ≥ 0.
a Calculate the time at which the body comes to rest.
b Calculate the distance travelled in this time.
6 A car of mass 1 tonne starts from rest and moves against a constant resistance force of
400 N. The driving force decreases uniformly from 2400 N so that after 10 seconds it
has a value of 2000 N. Find:
a the resultant force at any time t
b i the velocity and
ii distance travelled
after 5 seconds.
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Example
xample
18
7 A resultant force of 2x + 3 acts on a particle of mass 4 kg which is moving in a
straight line, where x is the displacement from a fixed origin. If the velocity v = 2
when x = 0, find v when x = 2.
8 A resultant force of x2 + 1 acts on a body of mass 2 kg which is moving in a straight
line, where x is the displacement from a fixed origin. If the velocity v = 4 when x = 3,
find v when x = 1.
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Maths Quest 12 Specialist Mathematics
CHAPTER
review
Multiple choice
11A
1 A body is in equilibrium under the action of three forces f 1 , f 2 and f 3 .
˜ ˜
˜
f 1 = 3i + 7 j and f 2 = – 5 i + 2 j .
˜
˜
˜
˜
˜
˜
The magnitude of the vector f 3 is:
˜
85
A 2i – 9 j
B
C 77
D – 85
E
˜
˜
17
11B
2 An object experiences three forces of equal magnitude, 80 N. The angle between all
adjacent pairs of forces is 120°. The magnitude of the resultant force is:
A 240 N
B 3 × 80 cos 120° N
C 3 × 80 sin 120° N
D 3 × 80 N
E 0N
11C
3 An 800-kg car has two horizontal opposing forces acting on it as it travels along a horizontal
road. The first is a retarding force of 400 N; the second is the applied force of the engine
which is 800 N in magnitude. The acceleration of the car, in m/s2, is:
A 1.5
B 1.0
C 0.50
D 0
E −0.50
11C
4 A ball is thrown vertically up into the air. At a particular instant it is accelerating downwards
at 8.4 m/s2. Which one of the following statements is true?
A The ball is moving downward.
B The ball is moving upward.
C The ball is at the top of its flight.
D The ball may be moving upwards or downwards since the acceleration due to gravity is
a constant for the motion.
E There is insufficient information to identify the direction of movement.
11D
5 Two opposing forces act on a 2.0-kg body to produce an acceleration of 3.0 m/s2. If the
magnitude of one force is 8.0 N, the magnitude of the second force could be:
A 12.0 N
B 6.0 N
C 10 N
D 14 N
E 16 Ν
11D
6 A particle of mass 2 kg rests on a rough plane inclined at 60° to the horizontal. The friction
force which prevents the mass slipping down the plane is:
A 4g N
B gN
C 2g N
D 3g N
E 2 3g Ν
11D
7 A 2.0-kg mass whose initial speed is 5.0 m/s slides to rest along a rough floor over 5.0 m.
The coefficient of friction μ is (to 2 decimal places):
A 0.24
B 0.48
C 0.26
D 0.52
E 0.60
11D
8 The acceleration of a 5.0-kg mass down an inclined plane elevated at 10° to the horizontal is
1.50 m/s2. The magnitude of the force of friction is:
A 1.0 N
B 1.2 N
C 1.4 N
D 1.6 N
E 1.8 Ν