CHAPTER 4
Equilibrium of Rigid Body &
Friction
Rigid Body
Definition of a rigid body:
• A rigid body is defined as a body on which the
distance between two points never changes
whatever be the force applied on it.
• Or you may say the body which does not deform
under the influence of forces is known as a rigid
body.
• But, in real life, there would be some force under
which the body starts to deform. For example, a
bridge does not deform under the weight of a
single man but it may deform under the load of a
truck or ten trucks. However, the deformation is
small.
Condition for rigid body equilibrium
• General equation of equilibrium for a rigid body
F=0
M=0
• Equilibrium in two dimension
Fx=0
Fy=0
M=0
• Equilibrium in three dimension
Fx=0
Fy=0
Fz=0
M=0
Free Body Diagram (2D)
•
Support Reactions
– there are various types of reaction that occur at
supports and point
– As a general rules,
• If a support prevents the translation of a body in
a given direction, then a force is developed on the
body in that direction. Likewise , if rotation is
prevented , a couple moment is exerted on the
body
Example :
(i) Prevents the beam from translating in the
vertical direction
Prevents the beam from
translating in the vertical
direction, the roller can only
exert a force on the
beam
The pin passes through
a hole in the beam and
two leaves which are
fixed to the ground. The
pin can prevent
translation of the beam
in any direction
This support prevent both
translation and rotation of
the beam, and so to do
this a force and couple
moment must be
developed on the beam at
its point of connection.
•
External and Internal Forces
– Internal Forces – not represented on Free
Body Diagram
– External Forces – do represented on Free
Body Diagram
• Weight and the center of Gravity
– When the body is uniform or made of
homogeneous material , the centre of gravity
will be located at the body’s geometric centre
or centroid
– the body nonhomogeneous or has an unusual
shape, then the location of its center of
gravity will be given
W=mg
• Idealized Model
APPLICATIONS
A 200 kg platform is suspended off an oil rig. How do we
determine the force reactions at the joints and the forces in the
cables?
How are the idealized model and the free body diagram used to do
this? Which diagram above is the idealized model?
Example 4.1
Draw the free body diagram of the uniform beam shown
in figure below. The beam has a mass of 100kg
Idealized model
Free body diagram
1. Draw an outlined shape. Imagine the body to be isolated
or cut “free” from its constraints and draw its outlined
shape.
2. Show all the external forces and couple moments.
These typically include: a) applied loads, b) support
reactions, and, c) the weight of the body.
Example 4.4
• The link shown in Figure below is pin-connected at A
and rests against a smooth support at B. Compute
the horizontal and vertical components of reaction at
the pin A.
Determine support reaction for beam
• In determining of support reactions, the
unknown reactions acted on the beam must be
identified. This can be done by referring to Table
of support reaction.
• By using the Equation of Equilibrium, the
support reactions can be solved.
For the point load:
• the distance is taken from the point of
force to moment about point.
For distributed load (DL),
• the distance is taken from the centroid of
DL to the moment about point.
• A point load is self-explanatory: a load that is
applied at one point
• A distributed load/uniform load, on the other hand,
is applied along an edge
Practical cases
• In buildings, point loads on beams is subjected to
loads from other beams or columns and uniformly
distributed loads are due to floors, walls and partitions
and the weights of the beams themselves
Example 1
Constraint for rigid body
• Type of constraint
– Redundant constraint
• When a body has redundant supports, that is
more supports than are necessary to hold it in
equilibrium, it become statically indeterminate
(more unknown loadings on the body than
equations of equilibrium).
– Improper constraint
• Resulting in stability
• axis is perpendicular to the plane of the forces
and therefore appears as a point, Hence, when
all the reactive forces are concurrent at this
point, the body is improperly constrained
Equilibrium of rigid body in 3 Dimension
Free body diagram of Support Reactions
• The magnitude of forces, F = Fx 2  Fy 2  Fz 2
• Force’s orientation defined by the coordinate
direction angles ,  and 
Example 4.3
A single bearing or hinge can prevent rotation by providing
a resistive couple moment. However, it is usually preferred
to use two or more properly aligned bearings or hinges.
Thus, in these cases, only force reactions are generated
and there are no moment reactions created.
Equation of equilibrium
• Vector Equations of Equilibrium ,
F = 0, vector sum of all the external forces acting on the
body
Mo = 0, is the sum of couple moments and the moments
of all the forces about any point O located either
on or off the body
• Scalar Equations of Equilibrium
If applies the external forces and couple moments in
Cartesian Vector.
F =
Fxi + Fyj + Fzk = 0
Mo =
Mxi + Myj + Mzk = 0
• Since the i, j and k components are independent from one
another, the above equations are satisfied provided
Fx = 0 , Fy = 0 and Fz = 0
Mx = 0 , My = 0 and Mz = 0
Example 4.14
• The homogeneous plate shown in figure below has a
mass of 100kg and is subjected to a force and
couple moment along it edges. If it is supported in
the horizontal plane by means of a roller at A, a ball
and socket joint at B and a cord at C, determine the
components of reaction at the supports.
Equations of Equilibrium:
Fx = 0;
Bx = 0
Fy = 0;
By = 0 and
Fz = 0;
Az + Bz + Tc - 300N - 981 N = 0
Moment at axis equation (right hand rules=define direction of moment):
Mx = 0; Tc ( 2m) - 981 ( 1m) + Bz (2m)
=0
My = 0; 300 N (1.5m) + 981N ( 1.5m) – Bz (3m) - Az(3m) – 200N.m = 0
The components of force at B can be eliminated if the x’, y’ and z’
axes are used. We obtain
Mx’ = 0; 981 N (1m) + 300 N (2m) - Az ( 2m)= 0
My’ = 0; -300 N (1.5m) - 981N (1.5m) – 200 N.m+Tc (3m) = 0
Results
Az = 790 N Bz = -217 N, negative sign indicates that Bz acts
downward
Tc = 707 N
Pulleys system
Example 4.7
• Determine the tension T in the cable of the pulley
system shown in Figure
The most solution
would be as follows:
Fy = 0
5T = 1500 N
T = 300 N
∑Fy = 0
T2 = 2T
T2 = 600 N
∑Fy = 0
T3 = 4T
T3 = 1200 N
∑Fy = 0
T + T4 = 1500 N
T = 300 N
Friction
Friction
• What is Friction?
– Friction is defined as a force of resistance acting on a
body which prevents or retards slipping of the body
relative to a second body
• Friction Law for dry surface
• Motion or impending motion of two forces in contact
causes a reaction force known friction force.
• This friction force is :
– Parallel to a flat surface or tangent to a curved surface
– Opposite in direction to the motion or impending
motion
– Dependent on the force pressing the surfaces together
– Generally independent of the area of surface of contact
– Independent of velocity, except for extreme cases not
to be considered in this module
– Dependent on the nature of the contacting surface
Friction
• How to define friction?
– Friction laws for dry surface
• On Horizontal plane
• On Inclined plane
Friction laws for dry surface on horizontal plane
– Condition 1: No applied load (Not moving)
• No friction
Weight(W)
F
Normal
force(N)
– Condition 2 : applied load P pushed the block and the block
has impending motion
• F acted as a reacting force an not be great enough to
balance P and subsequently the block will tend to slip.
• In other words P will slowly increases until the block is
on verge sliding, F can be to max value (Fmax) but the
block still in equilibrium
– At this state F is know as static friction force, FS
Weight(W)
Fs=μsN
P
F
Normal
force(N)
Coefficient of
static friction
Notes: if F=P, F< Fs, the equation of FS can not be use to determine the
friction since the friction force not achieved the maximum value
– Condition 3 : when the block is start moving (motion)
• F will reduced fastly because of the momentum influence
– At this state F is know as kinetic friction force, Fk
– At this condition Fk is less than Fs
Weight(W)
P
Fk=μkN
Fk
Normal
force(N)
– Condition 4 : the block is sliding/overturning
• When P is increased (P > Fk) until it make the block
sliding/overtuning at point B. Moment at B = zero.
+
ΣMB =P(h)-W(a)
Weight(W)
P
Fk
Normal
force(N)
•Total moment at B is +ve = the block is not sliding/overturning.
•Total moment at B is -ve = the block is sliding/overtuning.
• Resultant force
– Experimentally Those limiting static frictional force Fs
and Fk is directly proportional to the resultant normal
force (Fs) and to the magnitude of the resultant normal
force (Fk)
– Defining the angle of static friction, фs
W
P
P
F
N
Φ
R
tan Φs =Fs /N
= μs N / N
= μs
Φs = tan-1μs
W
N
Φs
Fs
R
– Defining the angle of kinetic friction, фk
W
P
P
W
Fk
F
N
Φ
R
tan Φk =Fk /N
= μk N / N
= μk
Φk = tan-1μk
N
Φk
R
Example 4.18
• The 70N force shown in figure causes impending motion to
the right. The block is not moving due to this force.
Determine the static coefficient of friction, μs. Given the
mass of block is 40kg.
70N
Solution;
Draw the free body diagram of block.
W
W = mg = 40(9.81)
= 392.4N
70N
W=N
Ncc
Fs
s 
F
70

 0.18
N 392 .4
Example 4.19
• The mass of block is 20000 kg is subjected to the applied
load as shown in figure. Determine the friction force if μs =
0.5.
80kN
Ө=20º
Solution;
Draw the free diagram of block.
80 sin 20º =27.4kN
80kN
W=mg
Ө=20º
80 cos 20º =75.2kN
=20000(9.81)
=196.2kN
N
F
ΣFy ↑=ΣFy↓
27.4 + N = 196.2 or can be write as 27.4 + N – 196.2 =0
N = 168.8 kN
Fx = 0 , 75.2 – F = 0
75.2 kN= F
• The friction force, Fs = μs N = 0.5(168.8) = 84.4kN.
• The value of Fs then compare to the F. Show that F < Fs.
The block is not moving yet. The equation of Fs can not be
used to determine the friction since the friction forces not
achieved the maximum value.
• The friction force is F = 75.2kN.
Example
• Te uniform crate shown in Figure has a mass of 20kg. If a
force P=80 N is applied to the crate, determine if it remains
in equilibrium. The coefficient of static friction is μs = 0.3
Solution
• Free body diagram
As shown in Figure, the
resultant normal force Nc must
act a distance x, from the
counteract the tipping effect
cause by P. there are three
unknowns F, Nc and x, which
can be determined strictly from
three equations of equilibrium



 FX  0; 80cos300  F  0
   Fy  0;  80 cos300  N c  196.2  0
  Mo  0; 80sin 300 (0.4)  80 cos300 (0.2)  N c ( x)  0
solving
F  69.3 N
N c 236N
x  0.00908m  9.08 m m
Friction laws for dry surfaces on an inclined plane
• Condition 1 : the block is not moving (no friction)
N can be determined using
ΣFx = 0,
F = W sin θ and
ΣFy = 0,
N = W cos θ.
The block tends to move down on
the inclined plane. The friction force
acted in the opposite direction.
F< Fs
• Condition 2 : the block is in the verge of
impending motion
N can be determined using
ΣFx = 0,
F = W sin θ and
ΣFy = 0,
N = W cos θ.
F = Fs, F = μs N
• Condition 3 : the block is moving
N can be determined
using ΣFy = 0, N = W
cos θ.
Fk = μk N
Fk = μk N= μk W cos θ
• Condition 4 : the block is sliding/overtuning
When P is increased (P > Fk) until it
make the block sliding/overtuning
at point A.
+
ΣMA = - W sin θ(h) + W cos θ(a)
Example 4.20
• A 100N force acts as shown on a 300N block placed on an
inclined plane. The coefficients of friction between the
block and plane are μs=0.25 and μk=0.2. determine whether
the block is in equilibrium and find the value of the friction
force.
• Free body diagram
Σfy = 0; - 240 + N = 0
N= 240N
Fs = μs N = 0.25(240) = 60N
Σfx = 0
100 - 180 – F = 0
F = -80N (the friction force is acted opposite direction, thus
directed up and to the right)
F = 80N (
)
F > Fs , 80N > 60N, the block is moving down due to this
condition. Use μk to determine the actual friction force.
The block is not in equilibrium condition.
The friction force; Fk = μk N = 0.2(240) = 48N
Tips Solving problems in dry friction
1) The first step is to draw a free body diagram of the body,
labeling and directing all forces involved at each surface
of contact.
2) The resultant, R exerted by a surface on a free body can
be resolved into a normal component N and a tangential
component F. F known as a friction force. When a body is
in contact with a fixed surface, the direction of the friction
force, F is opposite to that of the actual or impending
motion or applied load of the body.
3) No motion will occur as long as F does not exceed the
maximum value of Fs = μs N where μs is the coefficient of
static friction.
4) Motion will occur if a value of F larger than Fs is required
to maintain equilibrium. As motion takes place, the actual
force drops to Fk = μk N where μk is the coefficient of
kinetic friction.
Example 4.23
• The static coefficient μs is 0.4 between the point A and rough
surfaces meanwhile there are no friction occurred between
point B and contact surfaces. Indentify whether the wood is
in equilibrium or not. Assumed the weight of wood acts at
the centre of A-B.
Smooth
surface
w
Rough
surface