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: