AS-Level Maths:
Mechanics 1
for Edexcel
M1.6 Statics of a
particle
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Types of force
Contents
Types of force
Resolving forces
Particles in equilibrium
Friction
Examination-style questions
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Types of force
There are many different types of force that may act on an
object.
The most common ones met in mechanics problems are:
Weight
Normal reaction forces
Tension
Thrust
Friction
When an object remains at rest it is said to be in static
equilibrium.
This state occurs when the net force acting on the object is zero.
If the net force is not zero, then the object will accelerate.
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Weight and acceleration due to gravity
The acceleration due to gravity, g, is the acceleration that
a body in free fall experiences if air resistance and other
forces are neglected.
On or near the surface of the earth g is taken to be
approximately 9.8 ms–2 whereas on the moon the acceleration
due to lunar gravity would be approximately 1.6 ms–2.
The weight, W, of a body is the downward force that the
earth exerts on the body due to its mass. This force, by
Newton’s Second Law, is equal to the mass multiplied by
the acceleration produced.
Therefore,
W = mg
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Friction
Friction is a very common force that acts on objects moving
relative to each other (for example a block sliding along a
table) to eventually slow them down.
Friction also acts to stop one object moving relative to the
other when it would otherwise do so because of a force
acting on it. So we need to consider friction when deciding
whether a system is in equilibrium.
Friction depends on the roughness of the bodies touching –
compare an iron bar sliding over grass to on an ice-hockey
puck sliding on ice.
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Thrust and Tension
A rod under compression produces an outward force at
each end. This is known as the thrust in the rod.
Note that a string cannot be compressed and so cannot
produce a thrust.
T
T
A taut string or rod under extension produces an inward
force at each end. This is known as the tension in the string
or rod.
T
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T
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Normal reaction force
When two surfaces are pressed against one another then a
force, called the normal reaction force or normal contact
force, acts between the two surfaces.
This force acts perpendicular to the area of the surfaces in
contact.
Normal reaction force
Weight
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Resolving forces
Contents
Types of force
Resolving forces
Particles in equilibrium
Friction
Examination-style questions
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Forces as vectors
A force is a vector quantity, so when we deal with forces,
we can use vector arithmetic.
One very important skill in mechanics is finding the
component of a force in a given direction.
This enables the resultant of several forces acting in
different directions to be calculated.
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Resolving forces
The component of a force in a given direction is equal to the
magnitude of the force multiplied by the cosine of the angle
between the force and the given direction.
C
F

A
B
The component of the force in the direction AB = F cos.
The component of the force in the direction AC = F cos(90 – )
= F sin.
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Resolving forces
A force of 7 N acts on a particle at
an angle of 30° to the horizontal.
7N
30°
Resolve this force into horizontal and vertical components:
7N
60°
30° b
a
a = 7 cos30° = 6.06 to 3 s.f.
b = 7 cos60° = 3.50
Therefore, 6.06 N act left and 3.50 N act upwards (to 3 s.f.).
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Resolving forces
A force of 5 N acts on a particle at
an angle of 55° to the horizontal.
5N
55°
Resolve this force into vertical and horizontal components:
Vertical component = 5 cos35° = 4.10 (to 3 s.f.)
Horizontal component = 5 cos55° = 2.87 (to 3 s.f.)
Therefore, 4.10 N act upwards and 2.88 N act left (to 3 s.f.).
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Resolving forces
A force of 10 N acts on a particle
at an angle of 40° to the vertical.
Calculate the horizontal and
vertical components of this force.
10 N 40°
Vertical component = 10 cos40° = 7.66 (to 3 s.f.)
Horizontal component = 10 cos50° = 6.43 (to 3 s.f.)
Therefore, 7.66 N act downwards and 6.43 N act to the left
(to 3 s.f.).
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Resultant forces
A particle is in equilibrium when acted on by the forces
 x   -6 
 1
 2  ,  y  and  -3 
   
 
 3   7 
 z
   
 
a) Find the values of x, y and z.
 7
A fourth force of  8  Newtons acts on the particle.
 
 1
 
b) Calculate the resultant force, F N, now acting on
this particle.
c) Find the magnitude of F.
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Resultant forces
 x   6   1 
     
a)  2    y    3   0
 3   7   z 
     
 x = 5, y = 1 and z = –4.
 5   6   1   7   7 
b)  2  +  1  +  3  +  8  =  8 
         
 3   7   4   1   1 
         
c) Magnitude = 72 +  8  +12 = 49 + 64 +1 = 114
2
Therefore the magnitude of the force is 114 N
= 10.7 N (to 3 s.f.)
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Resultant forces
A force of magnitude 39 N acts in the direction of –5i + 12j,
where i and j are the standard unit vectors.
a) Calculate the angle this force makes with the negative
i direction.
b) Find the force in the form ai + bj.
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Resultant forces
12
  67.4°
5
So the force acts at an angle of
67.4° (3 s.f.) with the horizontal.
a) tan 
12
5
b) 5i  12 j  52  122  13  the magnitude of the
direction vector is 13 N.
For a magnitude of 39 N, a force 3 times the direction vector
is needed (3 × 13 = 39). Therefore F = –15i + 36j.
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Resultant forces
Three coplanar forces act at a point in a vertical plane.
PN
60°
3N
6N
a) By resolving vertically, calculate the magnitude of P for
which the forces are in equilibrium.
b) If P = 7 and the 6 N force now acts at 50° to the horizontal,
find the magnitude and direction of the resultant force.
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Resultant forces
7
a) Resolving vertically, P = 6 cos30° = 5.20
b) Resultant vertical force  =
7 – 6 cos40° = 2.40 N
Resultant horizontal force  =
6 cos50° – 3 = 0.857
40°
6
Using the triangle law of addition:
R = 0.857  2.40  2.55
2
2
2.40
tan x 
 x  70.3°
0.857
2.40
2.40
0.857
3
50°
R
x°
0.857
Therefore the resultant force is 2.55 N (to 3 s.f.) at an
angle of 70.3° (to 3 s.f.) to the negative horizontal.
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Particles in equilibrium
Contents
Types of force
Resolving forces
Particles in equilibrium
Friction
Examination-style questions
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Static equilibrium
When an object remains at rest it is said to be in static
equilibrium.
This state occurs when an object remains still and the net
force acting on it is zero.
If the net force is not zero, then the object will accelerate.
To decide whether an object is in equilibrium, we must resolve
all the forces in all the dimensions we are considering to see
that no net force is acting.
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Example 1
An object is resting on the top of a vertical rod. Draw a
diagram showing the forces acting on the object.
T
mg
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Example 2
An object is hanging in equilibrium from the bottom of a
vertical string. Draw a diagram showing the forces acting on
the object.
T
mg
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Example 3
An object is resting on a smooth horizontal surface. Draw a
diagram showing the forces acting on the object.
R
mg
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Question 1
A particle is hanging in equilibrium at the end of a light
inextensible string. The mass of the particle is 2 kg.
Find the tension in the string.
T
2g
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T = 2g = 2 × 9.8 = 19.6
Therefore the tension in the
string is 19.6 N.
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Question 2
A particle of mass 3 kg rests on a smooth horizontal surface.
Find the force exerted on the particle by the table.
R
3g
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R = 3g = 3 × 9.8 = 29.4
Therefore the force exerted
on the particle by the table
is 29.4 N.
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Question 3
A particle of mass 5 kg is held in equilibrium by two light
inextensible strings suspended at angles of 15° and 20°
respectively to the horizontal.
Find the tension in each string.
T1
T2
15°
20°
5g
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Resolving vertically,
T1cos75° + T2cos70° = 5g
Resolving horizontally,
T1cos15° = T2cos20°
T1 
T2cos20°
cos15°
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Question 3 solution
T2cos20°
cos15°
cos75°  T cos70°  5 g
2
 cos20°

T2 
cos75°  cos70°   5 g
 cos15°

T2  82.5 (to 3 s.f.)
T2cos20°
T1 =
= 80.3 (to 3 s.f.)
cos15°
Therefore the tensions in the two strings
are 82.5 N and 80.3 N respectively.
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Friction
Contents
Types of force
Resolving forces
Particles in equilibrium
Friction
Examination-style questions
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Friction and the coefficient of friction
If two rough surfaces are in contact then a frictional force may
also act to prevent relative motion between the two surfaces.
The size of the frictional force depends on several things.
The maximum size of the frictional force depends on:
The surfaces of the objects involved, specifically how
rough they are, and
The normal contact force.
The frictional force is never higher than the force required to
keep equilibrium.
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Friction and the coefficient of friction
Friction needs to be taken into account when deciding
whether a system is in equilibrium.
The maximum frictional force that can act is proportional to
the normal contact force.
The constant of this proportionality for any two given surfaces
is called the coefficient of friction and is usually written as .
Therefore
Fmax = R
where R is the normal contact force.
Generally, the rougher the two surfaces, the closer  is to 1.
The smoother the surfaces, the closer  is to 0.
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Friction along a rough inclined plane
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Question 4
A particle of mass 0.15 kg is resting on a smooth plane
inclined at an angle of 30° to the horizontal. The particle is
held in equilibrium by a light inextensible string acting up the
line of greatest slope.
Find the tension in the string.
Resolving parallel to the plane:
T = 0.15g cos60°
T = 0.735 (to 3 s.f.)
Therefore the tension in the string is 0.735 N.
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Question 5
A particle of mass 2.5 kg is held in equilibrium on a smooth
plane inclined at an angle of 25° to the horizontal by means
of a horizontal force H.
Find the magnitude of this force.
Resolving parallel to the plane:
H cos25° = 2.5g cos65°
2.5 gcos65°
H=
=11.4 (to 3 s.f.)
cos25°
Therefore H has a magnitude of 11.4 N.
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Question 6
A particle of mass 2 kg is resting on a rough horizontal
surface. The coefficient of friction between the particle and
the surface is 0.1. A light inextensible string is attached to the
right of the particle at an angle of 30° to the horizontal.
If the particle is on the point of moving to the right, find the
tension in the string.
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Question 6 solution
R
F
T
30°
2g
Resolving vertically:
R + T cos60° = 2g
Resolving horizontally:
F = T cos30°
F = R
T cos30° = 0.1 × (2g – T cos60°)
T (cos30° + 0.1 cos60°) = 0.2g
T = 2.14 (to 3 s.f.)
Therefore the tension in the string is 2.14 N.
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Friction along a rough horizontal surface
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Question 7
A particle of mass 0.5 kg is resting on a rough horizontal
surface. A light inextensible string is attached to the particle
at an angle of 15° to the horizontal and exerts a force of 2 N.
If the particle is on the point of moving in the direction of the
force, find the coefficient of friction between the particle and
the surface.
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Question 7 solution
R
F
2
15°
0.5g
Resolving vertically:
R + 2 cos75° = 0.5g
R = 0.5g – 2 cos75° = 4.38 (to 3 s.f.)
Resolving horizontally:
F = 2cos15° = 1.93 (to 3 s.f.)
1.93
 = F/R =
= 0.441 (to 3 s.f.).
4.38
Therefore the coefficient of friction is 0.441
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Question 8
A particle of mass 1.7 kg rests on a rough plane inclined at
an angle of 20° to the horizontal. A horizontal force, H, is
applied to the particle. The coefficient of friction between the
particle and the plane is 0.15.
Find H if
a) The particle is on the point of sliding down the plane.
b) The particle is on the point of sliding up the plane.
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Question 8 solution a
Resolving perpendicular to the plane:
R = 1.7g cos20° + H cos70°
Resolving parallel to the plane:
F + Hcos20° = 1.7gcos70°
F = 1.7gcos70° – Hcos20°
F = R
 1.7gcos70° – Hcos20° = 0.15(1.7gcos20° + Hcos70°)
0.15Hcos70° + Hcos20° = 1.7gcos70° – 0.15 × 1.7gcos20°
H(0.15cos70° + cos20°) = 1.7gcos70° – 0.15 × 1.7gcos20°
1.7 gcos70°  0.15  1.7 gcos20°
H 
 3.38 (to 3 s.f.)
0.15cos70°  cos20°
Therefore the horizontal force is 3.38 N.
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Question 8 Solution b
Resolving perpendicular to the plane:
R = 1.7gcos20° + Hcos70°
Resolving parallel to the plane:
F + 1.7gcos70° = Hcos20°
F = Hcos20° – 1.7gcos70°
F = R
Hcos20° – 1.7gcos70° = 0.15(1.7gcos20° + Hcos70°)
Hcos20° – 0.15Hcos70° = 0.15 × 1.7gcos20° + 1.7gcos70°
H(cos20° – 0.15cos70°) = 0.15 × 1.7gcos20° + 1.7gcos70°
0.15  1.7 gcos20°  1.7 gcos70°
H 
 9.06 (to 3 s.f.)
cos20°  0.15cos70°
Therefore the horizontal force is 9.06 N.
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Examination-style questions
Contents
Types of force
Resolving forces
Particles in equilibrium
Friction
Examination-style questions
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Examination-style question 1
A particle of mass 10 kg is held in equilibrium on a smooth
plane inclined at an angle of 30o to the horizontal by means
of a light inextensible string acting at an angle of 10° to the
plane.
a) Draw a diagram showing all the forces acting on the
particle.
b) Calculate the tension in the string.
A horizontal force of 30 N is applied to the particle to try to
make it slide up the slope. The tension in the string is
adjusted so that the particle remains in equilibrium.
c) Calculate the new tension in the string.
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Solution 1
a)
b) Resolving parallel to the plane:
Tcos10° = 10gcos60°
T = 10 g cos60° = 49.8 (to 3sf)
cos10°
Therefore the tension in the string is 49.8 N.
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Solution 1
c) Resolving parallel to the plane:
Tcos10° + 30cos30° = 10gcos60°
10gcos60° – 30cos30°
T=
cos10°
T = 23.4 (to 3 s.f.)
Therefore the tension in the string is 23.4 N.
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Examination-style question 2
A and B are two fixed points on a horizontal line. Two light
inextensible strings are attached to A and B and the other
ends are attached to a particle C of mass 5 kg. AC = 3 cm,
BC = 4 cm and angle C = 90°. The strings are holding
particle C in equilibrium.
Find the tensions in the two strings in terms of g.
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Solution 2
By Pythagoras’ Theorem, AB = 5 cm.
The two angles marked  are alternate as are the two angles
marked .
Using trigonometry,
4 and cos = 3 ;
sin =
5
5
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3
4
sin =
and cos = .
5
5
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Solution 2
Resolving horizontally,
T1cos = T2cos
3
4
 T1  T2
5
5
4
 T1  T2
3
Resolving perpendicularly,
T1cos(90 – ) + T2cos(90 – ) = 5g
T1sin + T2sin = 5g
4
3
 T1  T2  5 g
5
5
44  3
  T   T2  5 g
53 2 5
25
 T2  5 g
15
Therefore T1 = 3g and T2 = 2g.
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Examination-style question 3
A particle of mass m kg lies on a rough plane inclined at an
angle of ° to the horizontal. The particle is held in
equilibrium by means of a light inextensible string held at an
angle of 30° to the plane.
The coefficient of friction between the plane and the particle
is 0.25 and the particle is about to slide up the plane.






÷
Show that the tension in the string is 2mg cosθ  4sinθ ÷÷
4 3 1 ÷
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Solution 3
Resolving perpendicular to the plane,
R + T cos60 = mgcos
1
2
R = mgcos – T
Using F = R,
F
1
= mgcos
4
–
1
T
8
Resolving parallel to the plane,
F + mgcos(90 – ) = Tcos30
1
mg cos  T  mg sin  T cos30
4
 4 3  1
1
3
1
T
mg cos  T + mg sin 
T + T =
8
4
2
8


 cos + 4 sin 
 T = 2mg 
 as required.
4 3 +1 

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