Recap of M1
It is always best
to start with an example from the previous unit to remind you of the basics.
Example 1
A body of mass 5
kg is held in equilibrium under gravity by two inextensible light ropes. One
rope is horizontal and the other is inclined at an angle Ө to the horizontal, as shown in the diagram
below. The tension in the rope inclined at Ө to the horizontal is 72N.
Find
a) the angle Ө, giving your answer to the nearest degree.
b) the tension T in the horizontal rope,
giving your answer to the nearest N.
a) Most of the questions from the
moments section in M1 can be solved by simply resolving forces. The body is in
equilibrium hence the vertical component of the 72N force must equal 5g.
72SinӨ = 5g (1)
Ө = 42.88º
Ө = 43º
b) The horizontal component of the 72N
force must equal T.
72CosӨ = T
72
× Cos 43º = T
T
= 53N
The moment of a force about a particular
point is:
force × perpendicular distance.
Moment about Q = F × d
Remember that the unit is Nm.
Algebraic sum of moments
If a number of
coplanar forces act on a rigid body then their moments about a given point may
be added. Take great care with the
direction of rotation
Equilibrium of Rigid Bodies
A rigid body is said to be in
equilibrium if;
1.
the
vector sum of the forces acting is zero ( the sum of the components in any
direction is zero)
2.
the
algebraic sum of the moments of the force about a particular point is zero.
All problems within this unit need to be
attacked in the same fashion.
1.
Draw
a clearly labeled diagram showing all forces.
2.
Resolve
the forces perpendicularly and vertically
remembering to set them equal to zero.
On some occasions the forces will need to be resolved parallel and
perpendicular to a plane.
3.
Take
moments about a convenient point and equate the algebraic sum of the moments to
zero. The aim is to choose a point that has only a few unknowns. If you choose a point that makes things more
difficult simply select another point.
Example 2
A uniform rod AB of mass 12 kg and length 15 m is smoothly hinged at A and has a particle of mass 28 kg attached to it at B. A light inextensible string is attached to the rod at the point C where AC = 9m and to the point D vertically above A, keeping the rod in a horizontal position. The tension in the string is TN. If the angle between the rod and the string is 25°, calculate, in terms of T, the resultant moment about A of the forces acting on the rod.
So by taking moments about A, one needs to
consider, the weight of the rod, the particle placed at B and the tension in
the string acting at C.
(12g
× 7.5) + (28g × 15) – (9 ×
TSin25°)
(510g
– 3.80T)Nm
Example 3
A non-uniform rod PQ of mass 12kg and length 8m rests horizontally
in equilibrium, supported by two strings attached at the ends P and Q of the
rod. The strings make angles of 45° and 60° with the horizontal as shown in the
diagram.
(a) Obtain the tensions in each of the strings.
(b) Determine the position of the centre of mass of the rod.
a) Resolving horizontally gives:
Resolving vertically gives:
By substituting back in for T2
this gives:
Therefore T2 = 86.09N
b) Taking moments about Q gives:
One of the fundamental ideas to remember
with this type of problem is that surds must be used until the final
calculation. This example may, at first, appear tricky but there will be others
to practice on later.
Exam Question
A uniform rod AB, of length 8a and
weight W, is free to rotate in a vertical plane about a smooth pivot at A.
One end of a light inextensible string is attached to B. The other end
is attached to point C which is vertically above A, with AC
= 6a. The rod is in equilibrium with AB horizontal, as shown
below.
(a) By taking moments about A, or
otherwise, show that the tension in the string is 5/6W.
Add the forces to the diagram. 
Taking
moments about A gives:
4aW =
8aTSinB
By Pythagoras CB = 10a
Hence
(b) Calculate the magnitude of the horizontal
component of the force exerted by the pivot on the rod.
Resolving forces horizontally.
X =
TCosB
Therefore
Limiting Equilibrium
If a body is in limiting equilibrium then one of the forces acting
must be friction. The condition to remember is that F ≤ μR. At the
instant that motion is about to take place friction will have its highest value
of μR.
Example 4
A smooth horizontal rail is fixed at a
height of 3m above a horizontal playground whose surface is rough. A straight
uniform pole AB, of mass 20kg and length 6m, is placed to rest at point C on
the rail with the end A on the playground. The vertical plane containing the
pole is at right angles to the rail. The distance AC is 5m and the pole rests
in limiting equilibrium.
Calculate:
a) the magnitude of the force exerted by
the rail on the pole, giving your answer to 3 sig fig.
b) the coefficient of friction between
the pole and the playground, giving your answer to 2 decimal places.
c) the magnitude of the force exerted by
the playground on the pole, giving your answer to the nearest N.
a) Adding forces to the diagram:
Taking moments about A gives:
5S = 20g × 3 ×CosA
CosA = 0.8
Therefore S = 94N
b) Limiting equilibrium therefore F = μR.
Resolving horizontally gives:
S
× SinA = F
S × 0.6 = F
F = 56.4N
Resolving vertically gives:
S
× CosA + R = 20g
94
× 0.8 +R = 196
R
= 120.8N
F
= μR
Therefore μ
= 0.47
c) Magnitude of the force
exerted by the ground on the pole is given by:
Problems Involving ladders
Ladders will either be lent against a wall or horizontal. We
sometimes have to consider frictional forces on the ladder due to the floor or
wall (ladder is in contact with a 'rough' surface). Remember that the
friction F acts parallel to the surface in such a direction as to oppose the
motion.
Example 5
A uniform ladder of mass 30kg rests against a smooth vertical wall
with its lower end on rough ground (coefficient of friction 0.25), and its top against
a smooth vertical wall. The ladder rests at an angle of 60° to the horizontal.
Find the magnitude of the minimum horizontal force required at the base to
prevent slipping.
We need to find S, V, R and F.
The easiest one to find first is S. 
Taking moments about Q gives:
Resolving vertically gives:
R = 30g
Resolving
horizontally gives:
S = F + V
Given that F = μR
84.87
= 0.25 
30 g + V
84.87
= 73.5 + V
V = 11.4N
What is the
maximum horizontal force that could be applied at the base of the ladder
without slipping occurring?
In this situation
friction is acting in the opposite direction.
Therefore S + F = V
84.87 + 73.5 = V V = 158.4N
Example 6
The diagram shows
a ladder AB of mass 8kg and length 6m resting in equilibrium at an angle of 50° to the horizontal
with its upper end A against a smooth vertical wall and its lower end B on
rough horizontal ground, coefficient of friction μ. Find forces S, F and R and the least possible
value of μ if the centre of gravity of the ladder is 2m from B.
Taking moments
about B gives: 
S × 6Sin50° = 8g × 2Cos50°
S
= 21.9N
Resolving
horizontally gives:
S = F
F = 21.9N
Resolving
vertically gives:
R = 8g = 78.4N
Least value of μ occurs when F = μR
The following
example considers friction on the floor and the wall.
Example 7
A uniform ladder of mass 25kg and length L rests against a rough vertical
wall (coefficient of friction μ = 
) with its base on rough ground (coefficient of friction
α = 
) and it makes an angle of 61º with the ground. Find the magnitude
of the minimum horizontal force that must be applied to the base in order to
prevent slipping.
Taking moments
about Q gives: 
S × LSin61° + W × LCos61° = 12.5g × L × Cos61°
S × Sin61° + W × Cos61° = 12.5g × Cos61° (1)
Using W = μS equation (1) becomes:
Resolving vertically gives:
W
+ R = 25g
μS
+ R = 25g
R
= 25g – (
)
R
= 225.9N
Using F = αR F
= 0.2 × 225.9
F
= 45.18N
Finally, resolving horizontally gives:
F
+ V = S
45.18
+ V = 57.31
V
= 12.1N
Therefore the minimum horizontal force to prevent slipping is
12.1N
Climbing ladders
Obviously safety is the ultimate concern when climbing a
ladder. In deciding whether it is safe
to climb to the top of a ladder one has to consider the magnitude of the
frictional force acting on the ladder. This in itself is dependent on the
roughness of the ground. If a MAN is already on a ladder and the system
is in limiting equilibrium then any further movement up the ladder will cause
it to slip. The example below considers such a situation.
Example 8
A uniform ladder of mass 30kg and length 10m rests against a
smooth vertical wall with its lower end on rough ground. The coefficient of
friction between the ground and the ladder is 0.3. The ladder is inclined at an
angle θ to the horizontal where tan θ = 2. Find how far a boy of mass 30 kg can ascend the
ladder without it slipping.
Assume that the
boy can climb a height y m
up the ladder.
Taking moments
about B gives: 
Resolving
vertically gives:
R = 60g
Given that F = μR, F = 0.3 × 60g = 18g
Resolving
horizontally gives:
F = S
Therefore S = 18g
Using
Therefore the boy can climb 7m up the
ladder.
Questions
1 A uniform rod of mass M rests in limiting
equilibrium with the end A standing on rough horizontal ground and the end B
resting against a smooth vertical wall. The vertical plane containing AB is
perpendicular to the wall. The
coefficient of friction between the rod and the ground is 0.2.
Find, to the
nearest degree, the angle at which the rod is inclined to the vertical.
2 A uniform rod of mass M rests in
limiting equilibrium with the end A standing on rough horizontal ground and the
end B resting against a smooth vertical wall. The vertical plane containing AB
is perpendicular to the wall. The coefficient of friction between the rod and
the ground is 0.75. Given that the ladder makes an angle α with the floor
show that
3 A non uniform ladder AB of length 15m
and mass 40kg has its centre of gravity at a point 5m from A. The ladder rests
with end A on rough horizontal ground (coefficient of friction 0.25) and end B
against a rough vertical wall (coefficient of friction 0.2). The ladder makes
an angle α with the horizontal such that 
. A straight horizontal string connects A to a point at the
base of the wall directly below B. A man of mass 80kg begins to climb the
ladder. How far up the ladder can the man climb without causing tension in the
string? What tension must the string be capable of withstanding if the man is
to climb to the top of the ladder?
4 The diagram below shows a uniform ladder
AB of length 2a and mass m, with the
end A resting on a rough horizontal floor.
The ladder is held
at an angle θ to the vertical by means of a light inextensible rope
attached to the point N, where AN = 1.5a. The other end of the rope is attached
to a point c, which is at a height 3a vertically above the end A of the ladder.
By taking moments about C find the magnitude of the force of friction acting on
the ladder at A. Also calculate the magnitude of the vertical component of the
reaction at A.
Given that the
coefficient between the floor and the ladder is 
, show that when the ladder is on the point of slipping at A
its inclination to the vertical is given by 
.
5 The figure shows a uniform rod AB of weight
W resting with one end A against a rough vertical wall. One end of a light
inextensible string is attached at B and the other end is attached at a point
C, vertically above A. The points A, B
and C lie in the same vertical plane with AB = BC = 4a and
AC = a. If the system is in limiting equilibrium,
calculate:
a) the tension in the string
b) the angle that the rod makes with the
horizontal.
c) the magnitude of the resultant force acting
at A.
6 The diagram below shows a uniform rod
AB, of weight W being held in limiting equilibrium at an angle of 30º to a
rough plane by a vertical string at the point B. The end A is in contact with a
rough surface inclined at an angle of 30º to the horizontal. Find T in terms of
W and calculate the coefficient of friction between the rod and the inclined
plane.
7 The diagram below shows a uniform rod at
rest in limiting equilibrium on a rough peg at A and a smooth peg at C. Given
that MC = CB, f
Extension
1 The diagram below shows a heavy uniform
rod of length 2a and mass m, resting
in equilibrium with its two ends on two smooth surfaces. The normal reactions
at the ends of the rods have magnitudes R and S. The inclinations of the planes
to the horizontal are 
and the rod lies in
the vertical plane
containing lines of greatest slope of both planes.
a) Show that 
.
b) By taking moments about the centre of the
rod, prove that the inclination of the rod to the horizontal is 
. The identities for