AS-Level Maths:
Mechanics 2
for Edexcel
M2.6 Statics of Rigid
Bodies
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Moments and non-parallel forces
So far, all questions requiring the use of moments have
involved parallel forces acting perpendicular to a body.
This is not always the case. Where forces are not acting
perpendicular to a body they can be resolved into two
components – perpendicular and parallel to the body.
2
The moment is 2 × 10 cos θ Nm ↻.
θ
10
10 cos θ
Alternatively, we can multiply the magnitude of the force by the
perpendicular distance from the pivot to the line of direction of
the force. 2 cos θ
The moment is 2 cos θ × 10 Nm ↻.
θ
2
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10
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Equilibrium and rigid bodies
A rigid body is one that does not bend or change shape
when forces are applied to it. Possible examples are rods
and ladders.
For a rigid body to be in equilibrium two conditions need to be
satisfied:
i. The vector sum of the forces acting on the body must be
zero.
ii. The sum of the moments about any point must be zero.
Rods are sometimes said to be freely hinged or smoothly
hinged. In this case there will be a reaction at the hinge but
the direction of this reaction is not known. Hence horizontal
and vertical components of the reaction force are used in the
solution.
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Non-parallel forces
Question 1: A uniform rod AB of length 6 m and mass 5 kg
is resting on a smooth pivot 1 m from A. It is held in
equilibrium by means of a light inextensible string attached
to the rod 5 m from A. The other end of the string is fixed to
a point vertically above A so that the angle it makes with the
rod is 30°.
Find the tension in the string if the rod is in equilibrium.
T sin 30°
T
R
A 1
30
2
2
5g
1
B
Take moments about the pivot:
5g × 2 = T sin 30° × 4
10g = 2T
T = 5g
T = 49 N
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Non-parallel forces
Question 2: A uniform rod AB of length 4 m and mass 4 kg is
resting in equilibrium. The forces acting on the rod are shown
in the diagram. Find X, Y and Z.
X
B
Y
60
1
First we need to find the vertical
and horizontal components of X.
X sin 30°
1
X
X cos 30°
30°
60°
30
2 4g
Z
A
Resolving the forces vertically gives:
X sin 30° = 4g
X = 78.4 N
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Non-parallel forces
Resolving the forces horizontally gives:
Y = Z + X cos 30°
Now take moments about B using the components of Y, 4g
and Z perpendicular to the rod:
(1 × Y cos 30°) + (2 × 4g cos 60°) = (4 × Z cos 30°)
We can substitute (Z + X cos 30°) for Y:
cos 30°(Z + X cos 30°) + (2 × 4g cos 60°) = 4 × Z cos 30°
Z cos 30° + 78.4 cos2 30° + 8g cos 60° = 4Z cos 30°
78.4 cos2 30° + 8g cos 60° = 3Z cos 30°
2 30°)
(8g
cos
60°
+
78.4
cos
Z=
= 113 N (3 s.f.)
3 cos 30°
 Y = 113.16 + 78.4 cos 30° = 181 N (3 s.f.)
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Non-parallel forces
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Equilibrium and rigid bodies
Question 1: A uniform ladder of length 4 m and weight 6 kg
rests on rough horizontal ground against a smooth vertical
wall. The ladder is inclined at an angle of 70° to the vertical
when it is on the point of slipping. Calculate the coefficient
of friction between the ladder and the ground.
B
S
2
R
2
6g
70 F
A
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We first need to draw a diagram showing
the forces acting on the ladder.
Since the ladder is uniform, the weight will
act through its mid-point.
There is a normal contact force from the
smooth wall.
The total reaction force on the ladder
from the rough ground has normal and
frictional components, R and F.
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Equilibrium and rigid bodies
Resolving vertically gives:
S sin 70°
S
2
6g cos 70°
R
2 6g
70 F
A
R = 6g
Resolving horizontally gives: F = S
B
In order to take moments we need to use
the components of S and 6g perpendicular
to the ladder.
Take moments about A:
(6g cos 70° × 2) = (S sin 70° × 4)
S=
1 2 g co s7 0
4 sin 7 0
= 3g cot 70°
The ladder is on the point of slipping, so F = µR.
(F = S and R = 6g:) 3g cot 70° = 6µg
 µ = ½ cot 70° = 0.182 (3 s.f.)
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Equilibrium and rigid bodies
Question 2: A uniform ladder AB of mass 20 kg stands on
rough horizontal ground and leans against a smooth vertical
wall.
A mass of 10 kg is attached to the ladder ¾ of the way up. The
coefficient of friction between the ladder and the ground is ½.
If the ladder is on the point of slipping, find the angle it makes
with the ground.
Resolving vertically, R = 30g
S
B
Resolving horizontally, S = F
10g
20g
F
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R

At the point of slipping F = R:
F = ½ × 30g = 15g
A
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Equilibrium and rigid bodies
Let the length of the ladder be a.
Take moments about A:
( 21 a × 20g cos ) + ( 34 a × 10g cos ) = (a × S sin )
Cancel the a’s. S = F = 15g, therefore:
17.5g cos  = 15g sin 
sin 
co s 
=
tan =
1 7 .5 g
15 g
7
6
 = 49.4 (3 s.f.)
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Examination-style question 1
Question 1: A ladder AB of mass 15 kg and length 4 m is
resting on rough horizontal ground and is leaning against a
smooth vertical wall. The ladder is modelled as a uniform
rod.
a) When the ladder is inclined at an angle of 30° to the
horizontal it is on the point of slipping. Find the
coefficient of friction between the ladder and the ground.
The ladder is moved so that it now makes an angle of 40
with the horizontal.
b) A boy of mass 40 kg climbs the ladder. How far can he
climb up the ladder before it starts to slip?
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Examination-style question 1
B
S
Resolving : R = 15g
Resolving : S = F
2
R
2
15g
30 F
A
Take moments about A:
4 × S sin 30° = 2 × 15g cos 30°
S =
3 0 g co s3 0
4 sin 3 0
S = 127.31 (5 s.f.) = F
When the ladder is on the point of slipping, F = R:
127.31 = 15g
 = 0.866 (3 s.f.)
Therefore the coefficient of friction between the ladder
and the ground is 0.866 (3 s.f.).
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Examination-style question 1
S
B
2–x
x
R
2
Let x be the distance from the midpoint of
AB to the point at which the boy makes
the ladder slip.
40g
Resolving , R = 55g
15g
Resolving , S = F
F
At the point of slipping, F = R:
40
A
F = 0.86603 × 55g
F = 466.79 (5 s.f.) = S
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Examination-style question 1
Taking moments about A:
(2 × 15g cos 40°) + ((2 + x) × 40g cos 40°) = (S sin 40° × 4)
30g cos 40° + 80g cos 40° + 40gx cos 40° = 466.79 sin 40° × 4
x(40g cos 40°) = 1967.2 sin 40° – 110g cos 40°
x=
1 9 6 7 .2 sin 4 0  1 1 0 g co s4 0
4 0 g co s4 0
x = 1.25 (3 s.f.)
The boy can climb 1.25 m past the midpoint of the ladder.
Therefore he can climb 3.25 m up the ladder before it
starts to slip.
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Examination-style question 2
Question 2: A uniform ladder AB of mass m kg and length 2a
rests on smooth horizontal ground and leans against a
smooth vertical wall.
The ladder is attached to the wall by means of a horizontal
light inextensible string secured at the mid-point of the
ladder.
The ladder is inclined at an angle ° to the horizontal where
tan = ¾.
If the maximum tension possible in the string before it breaks
is 10mg N, find, in terms of m, the maximum mass of a
person that can reach the top of the ladder.
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Examination-style question 2
Let the maximum mass be M kg. We need
to find M at the top of the ladder.
S
B
Mg
a
a
T
mg

If tan =
R
A
3
4
: sin =
3
and cos =
5
4
5
Resolving , R = mg + Mg
Resolving , S = T
Take moments about B: (Tsin × a) + (mgcos × a) = Rcos × 2a
R = mg + Mg, and for the maximum mass, T = 10mg:
3
4
5
5
 (10mg( ) +
The a’s cancel to give:
8
5
Mg = 6mg +
M=
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mg)a =
26
8
m
4
5
4
5
(mg + Mg) × 2a
mg –
8
5
mg =
26
5
mg
M = 3.25m
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Examination-style question 3
Question 3: A uniform rod AB of weight W N and length 2 m is
freely hinged at A and is held in a horizontal position by
means of a light inextensible string inclined at an angle of  
to the horizontal and attached to B as shown in the diagram.
a) Find, in terms of W and , the tension in the string.
b) Find X and Y in terms of W and  only.
X
A
T

Y
1
1
B
W
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Examination-style question 3
Resolving  gives: X + T sin  = W
Resolving  gives: Y = T cos 
Take moments about B:
W×1=X×2
X
T

Y
A
1
1
W
 X = ½W
We can now use the equation above to find T in terms of W
and  :
½W + T sin  = W
T sin  = ½W
T=
W
2 sin θ
Now we can use T to find Y in terms of W and  :
W
 Y = 2 sin
cos = ½W cot 
θ
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B