P.1
PHYSICS
STATICS
3.1 Moment of a force
Definition –
the moment or torque of a force about a point in the product of the force
and the perpendicular ar distance from the point to the of action of the force.
T=rxF
T = rF sinθ
T = torque of the force
F = force applied
r = displacement
– the moment or a force which applied to a hinged body determines the turning
effect of the body about the hinge.
e.g.
Torque / moment of F = rF sinθ (it provides an anticlockwise turning effect to the
rod)
3.2 Couples
– A couple consists of two equal and opposite parallel forces whose lines of action do
not coincide.
tap handle
moment of a couple = F x OA + F x OB (both are clockwise)
= F x AB
moment of couple = one force x perpendicular distance between forces
3.3 Displacement and rotation
– any change in the position of a rigid body can be resolved into a linear displacement
and a single rotation.
–
Any set of coplanar forces can be reduced to a single resultant force and a single
resultant couple.
P.2
3.4 Conditions for equilibrium
If a body is acted on by a number of coplanar forces and is in equilibrium, then
(i) the components of the forces in both of any two directions (usually taken at right angle)
must balance, and
(ii) the sum of the clockwise moments about any point equals the sum of the anticlockwise
moments about the same point.
** In brief, if a body is in equilibrium, the forces and the moments must balance.
3.5 Centre of gravity (e.g.)
– every particle is attracted towards the centre of the earth by the force of gravity.
– the centre of gravity of a body is the point where the resultant force of attraction or
weight of the body acts.
3.6 Centre of mass (CM)
– the centre of mass of an object may be defined as the point at which an applied force
produces acceleration but no rotation.
** Usually, the centre of gravity and centre of mass of a body are coincide. (exceptional
case can see PHYSICS – RESNICK & HALLIDAY P.325)
3.7 Stability of equilibrium
– It is vlassified into 3 types. Consider that a sphere is displaced slightly and then
released.
a)
stable equilibrium – the sphere will move back to the equilibrium position.
b)
neutral equilibrium – the sphere will stay in any new position it moved.
c)
Unstable equilibrium – the sphere will roll along the convex surface.
P.3
3.8 Worked Examples
e.g.1 A uniform ladder 4m long, of mass 25 kg, rests with its upper end against a smooth
vertical wall and with its lower end on rough ground. Also it inclines at 60∘ with
the horizontal without slipping. (g=10 ms-2)
Find (i) the magnitude and direction of the reaction exerted on the lower end by the
ground.
(ii) the coefficient of static friction between the ground and ladder.
(the normal reaction and
frictional force contribute
the resultant reaction)
Since the ladder is in equilibrium.
resolving vertically,
N = 250 newtons
resolving horizontally f =  N = S
….. (1)
….. (2)
Taking moments about A,
S = BC = W x AD
S x 4 sin 60∘= 250 x 1
S = 72.17 newtons ….. (3)
f = 72.17 newtons
from (2)
a)
R=
f 2  N 2 = 260.2 newtons
tanθ= N / f
θ= 73.9∘
b)
sub. (1), (3) into (2),
 x 250 = 72.17

= 0.29
e.g.2 A sign of mass 5kg is hung from the end B of a uniform bar AB of mass 2kg. The bar
is hinged to a wall at A and held horizontal by a wire joining B to a point C which is on
the wall vertically above A. If angle ABC = 30∘, find the force in the wire and that
exerted by the hinge. (take g = 10 ms-2)
(let the length of the bar be 2  )
P.4
Solution:
Take moment about A
20  + 50 x 2  = P2  sin30∘
P = 120N
F + 120 sin 30∘= 20 + 50
N = cos30∘
f 2  N 2 = 104N
R=
tanθ=
θ
f
N
= tan
f
= 5.5
N