MODULE 3
INTRODUCTION
When one body moves over another body (say stationary body), then stationary body
offers a resistance to the motion of moving body at the surface of contact. Thus, the property of
bodies by virtue of which a force is exerted to resist the motion of one body upon the other is
called FRICTION and the force exerted is called FORCE OF FRICTION. Frictional force
always acts parallel to the surface of contact and depends upon the roughness of surfaces.
LIMITING FORCE OF FRICTION
Consider a block resting on horizontal surface as shown in Figure. A force P is applied
through c.g. of the block and parallel to the surface of contact. When P is small, block does not
move and it is not on the point of motion. Such a condition of the block is known as non-limiting
equilibrium. The force of friction F becomes equal to P in magnitude and opposite to the
direction of P.
Now, when P is increased further, then block is on the point of motion. Such a state of
the block is called in limiting equilibrium. Force of friction at this stage is called as limiting force
of friction. Limiting force of friction is maximum amount of friction that can be developed.
 Flimiting = P and R = W are conditions of equilibrium, limiting and such conditions are
known as “static frictional conditions”. If P is further increased, block begins to move. Force of
friction that acts when the body is in motion is known as “dynamic friction” or “kinetic friction.”
TYPES OF FRICTION
As described above, the friction is divided into two types:
1. Static friction
2. Dynamic friction
The force of friction developed between the surfaces of contact of two bodies in the
position of rest, is known as static friction.
The force of friction developed between the surfaces of contact of two bodies in motion
is known as dynamic friction.
If the surfaces are dry (not lubricated), then friction between surfaces is known as solid
friction.
LAWS OF SOLID FRICTION: (STATIC OR DYNAMIC)
The following are the laws of solid friction between two surfaces A and B in contact
with each other. When A is moving or has the tendency to move relative to B, then (ref. Figure)
1. The direction of force of friction on A is opposite to that in which A is moving or
has the tendency to move relative to B.
2. So long as A is at rest, the force of friction is a self adjusting force, and is equal to
P.
3. When A is on the point of motion, the force of friction is maximum and known as
limiting force of friction.
4. The limiting force of friction bears a constant ratio to the normal reaction between
two surfaces.
5. The limiting force of friction does not depend upon the shape and areas of the
surfaces in contact.
6. The ratio between limiting friction and normal reaction is slightly less, when two
surfaces are in motion.
7. The force of friction is independent of the velocity of sliding.
COEFFICIENT OF FRICTION
It is defined as the ratio of limiting force of friction (F) to the normal reaction (R)
between two bodies. It is denoted by symbol  . Thus,

Flim iting
R

F
R
F  Flim  R
In most cases, the value of  is less than one.
There are two values of  . When there is limiting equilibrium, the coefficient of friction
is called static and is denoted by s .
When the bodies are in relative motion, the coefficient of friction is called kinetic and is
denoted by  k . From laws of solid friction s   k .
ANGLE OF FRICTION (  )
It is defined as the angle between the normal reaction R, and the resultant S of limiting
force of friction ( R ) and normal reaction R. It is denoted by  . In the Figure, A is a body
resting on horizontal surface. P is a force which tends to move A. Limiting force of friction F =
R is developed. Resultant of F and normal reaction R is denoted by S. Then angle of friction 
is the angle between R and S.
Mathematically, tan  

F R


R R
  tan 
i.e. tangent of angle of friction is equal to coefficient of friction.
In Figure
Resolving forces horizontally:Pcos   F  R
Resolving forces vertically:P sin   R  W
CONE OF FRICTION
When F and R meet at point O, then S is the resultant, which makes an angle  with R.
Thus O is vertex of a cone whose axis is R and semi-vertical angle is  , then such a right
circular cone is called as cone of friction, as shown in figure.
ANGLE OF REPOSE
Consider the Figure 9.5 in which OA is an inclined plane making an angle  with
horizontal. Let a body represented by a point, having weight W rests on this plane. Now the
angle of inclination  is increased gradually, such that the body is just on the point of sliding.
The forces keeping the body in equilibrium are:-
(1) Weight W of the body acting vertically downward.
(2) Normal reaction R acting at right angle to inclined plane.
(3) Limiting force of friction R acting up the plane as the body is on the point of
motion down the plane.
Resolving all forces parallel and perpendicular to the plane, we have
R  Wsin 
...(i)
R  W cos 
...(ii)
Dividing equn.(i) by equn.(ii), we get
  tan 
...(i)
Since from eq. 9.2,   tan  , we get
tan   tan 
or

Angle of friction = angle of repose
Hence, angle of repose is the maximum angle of inclination of the plane at which a
body remains in equilibrium under the action of friction only.
Angle of repose for the different materials, i.e. coal, wheat etc. are different, depending
upon the angle of friction.
Example
A body of weight 90 N is placed on a rough horizontal plane. Determine the coefficient
of friction if a horizontal force of 63 N just cause the body to slide over the horizontal plane.
Solution
Since there is no motion in the vertical direction, net force in the vertical direction is
zero.
RN  W  0
R N  W  90N
Since P just causes motion,
R N  P  63N
 
63 63

R N 90
= 0.7
Example
A block of weight 200 N is placed on a rough horizontal floor. If   0.25,
pull P required to move the block if P is inclined upwards at 300 to the horizontal.
find
the
Solution
Since there is no motion in the vertical direction, net force in the vertical direction is
zero.
R N  Psin 30  200  0
R N  200  Psin 30
Since P is the limiting force causing motion of the block.
Pcos30  R N  0
Pcos30  0.25  200  Psin30
 50  0.25Psin 30
P  cos30  0.25sin30  50
P =50.45 N
Example
Block A in fig 2.132. weights 200 N and block B weights 300 N. Find the force P
required to move block B. Assume the coefficient of friction for all surface as 0.3.
Solution.
Consider the upper block A, let T be the tension in the string.
For
F
V
0
R NI  W  0
R NI  W  200N
For
F
H
0
R NI  T  0
0.3 × 200 = T
T = 60 N
Consider the lower block B.
Since
F
V
0
R N2  W  R N1  0
R N2  W  R N1
= 300 + 200 = 500 N
For
F
H
0
P  R N2  R N1  0
P  R N2  R N1
= 0.3 × 500 + 0.3 × 200
= 210 N
Wedge friction
Wedges are small pieces of materials with triangular or trapezoidal cross section. They
are generally used for lifting heavy weights, for slight adjustments in the position of a body etc.
The weight of the wedge is very small compared to the weight lifted. Hence generally the weight
of wedge will be neglected. The problems on wedges are generally the problems of equilibrium
of bodies on inclined planes.
Fig. 2.149 shows the forces acting on the body.
Consider the equilibrium of the body.
Resolving the forces vertically,
R2 cos  – 2R2 sin  – 1R1 – W = 0
Resolving forces horizontally:
R1 – 2R2 cos  –R2 sin  = 0
For equilibrium of the wedge
Resolving the forces horizontally,

3R3 + 2R2 cos  + R2 sin  – P = 0
Resolving the forces vertically,
R3 + 2R2 sin  – R2 cos  = 0
Example 2.63
Find the horizontal force P on the 10o wedge shown in fig. 2.151 to raise the 1500 N
load. The coefficient of friction is 0.3 at all contact surfaces.
Solution
Consider the equilibrium of the load of 1500 N.
Resolving the forces horizontally
R1 – 0.3 R2 cos 10 – R2 sin 10 = 0
R1
=
R2 (0.3 cos 10 + sin 10)
=
0.47 R2
.............. (i)
Resolving the forces vertically
R2 cos 10 – 0.3 R2 sin 10 – 1500 – 0.3 R1 = 0
R2 cos 10 – 0.3 R2 sin 10 – 0.3 (0.47 R2) = 1500
R2 (cos 10 – 0.3 sin 10 – 0.3 × 0.47) = 1500
R2
=
1894.63 N
Consider the equilibrium of wedge.
Resolving the forces vertically
R3 + 0.3 R2 sin 10 – R2 cos 10 = 0
R3
=
R2 (cos 10 – 0.3 sin 10)
=
1894.63 (cos 10 – 0.3 sin 10)
=
1767.15 N
Resolving the forces horizontally
0.3 R2 cos 10 + R2 sin 10 + 0.3 R3 – P = 0
P
=
1894.63 (0.3 cos 10 + sin 10) + 0.3 × 1767.15
=
1418.90 N
Ladder Friction
Fig. 2.162 shows a ladder, AB, with the end A on the ground and end B on the wall. The
ladder exerts a force on the wall and RW is its reaction on the ladder. Similarly the ladder exerts a
force on the ground and Rf is its reaction on the ladder. The upper end B of the ladder tends to
slip downwards and hence the force of friction will be vertically upwards. The lower end tends to
move away from the wall and hence the direction of friction force will be towards the wall. For
equilibrium of ladder the algebraic sum of vertical forces and algebraic sum of horizontal forces
must be zero. Also the sum of moments of all the forces about any point must be zero.
For limiting equilibrium,
For
 FH  0,
f R f  R W  0
For
.................(i)
 FV  0,
R f  W  W  0
For
.................(ii)
 M  0, taking moments of all the forces about the lower end A, W × AC
× cos  – WRW × OA – RW × OB = 0
l
W  cos    W R W  l cos   R W  l sin   0
2
.................(iii)
Using the above three equations the limiting inclination of ladder with the floor can be
calculated.
Example 2.66
A uniform ladder 5m long, weighing 250N, is placed against a smooth vertical wall with
its lower end 2m from the wall. The coefficient of friction between the ladder and floor is 0.25.
Show that the ladder will remain in equilibrium in this position.
Solution
Since the wall is smooth, the frictional force at wall is zero.
Consider the free-body diagram of ladder AB.
BC
=
AB2  AC2  52  22
= 4.58 m
For
 FV
= 0,
Rf – 250
Rf
= 0
= 250 N
The limiting frictional force is RN.
= 0.25 × 250
= 62.5 N
Taking moments about B,
Rf × AC – 250 × 1 – F × BC
= 0
250 × 2 – 250 × 1
= F × 4.58
F
= 54.59 N
Since the frictional force at A is less than the
limiting frictional force of 62.5N,
the ladder will remain in equilibrium.
FRAMES AND TRUSSES
INTRODUCTION
A frame is an assemblage of a number of members, which resist geometrical distortion
under any applied system of loading. Frames are used in the roofs of sheds at railway platforms,
work shops, and in industrial buildings, bridges etc. For a number of given loading forces, the
members of the frame are determined and then the members are designed to carry the required
forces. Frames are classified into (1) Statically determinate frames and (2) Statically
indeterminate frames. Statically determinate frames are those frames which can be analysed with
the help of equations of statics alone. In this chapter, we will learn about statically determinate
perfect frames only.
TYPES OF FRAMES
Following are the types of frames:(a) Perfect frame
(b) Imperfect frame
(a) Perfect frames:- Simplest perfect frame is a
triangular assemblage of three member AB, BC, CA
C
meeting at joints B, C and A as shown in Figure
Mathematically,
B
A
m = number of members
j = number of joints
then
m = 2j – 3 is the condition for the frame to be a perfect frame.
For the figure,
m = 3, j = 3
m = 2j – 3
3=2×3–3
3 = 3 condition is satisfied.
E
Now consider the frame in Figure,
Here
D
m= 7
j = 5
A
B
C
m = 2j – 3
7 =2×5–3
7 = 10 – 3 = 7
Hence the frame in Figure is perfect frame.
(b) Imperfect frames:- (a) When the numbers are less than that required by equation m
= 2j – 3 then frame is called as imperfect frame. Such frame, can not resist geometrical distortion
under the action of loads.
(c) Redundant frames: If the number of members are more than that required by
equation m = 2j – 3, then such frames will be called as redundant frames.
.
A
In Figure,
.F
E
D
.B
.C
E
m = 12, j = 7
.
m = 2j – 3
12 = 2 × 7 – 3
12 = 14 – 3 = 11
Hence, the frame is redundant to a single degree, because one member is more.
REACTIONS AT SUPPORT
There are three types of supports.
(a) Roller support
(b) Hinged support
(c) Fixed support
(a) Roller supports. Figure consists of a support which is known as Roller support. It
is always a normal reaction R perpendicular to surface of rolling. Hence, Roller supports always
give one reaction component and in perpendicular direction.
(b) Hinged support. It is support at which inclined reaction R is developed. It has two
components, one is in vertical direction i.e. V and other is horizontal direction i.e. H. Hence a
hinged support always offers two reaction components V and H, i.e. one in vertical direction and
other in horizontal direction. This is shown in Figure.
(c) Fixed Support. Figure shows a fixed support at which three reaction components
are developed. First in vertical direction, i.e. V second is horizontal direction i.e. H and third is a
moment M.
We shall consider only (a) Roller support and (b) Hinged support in this chapter.
ASSUMPTION
Following assumptions are made in finding out the forces in a frame.
(a) Frame is a perfect frame.
(b) Load is applied at joints only.
(c) All members are hinged or pin-jointed. (It means members will have only axial
force and there will be no moment due to pin, because at a pin moment becomes
zero.)
ANALYSIS OF A FRAME
(a) Reaction at the supports are first determined due to applied external loads at joints
of a perfect frame, considering conditions of equilibrium.
(b) Forces in the members of the frame are determined by considering conditions of
equilibrium.
CONDITIONS OF EQUILIBRIUM
There are three conditions of equilibrium
(a) V  0 (i.e. Algebraic sum of all the forces in a vertical direction must be equal to
zero).
(b) H  0 (i.e. Algebraic sum of all the forces in a horizontal direction must be equal
to zero).
(c) M  0 (i.e. Algebraic sum of moments of all the forces about a point must be equal
to zero).
METHODS OF ANALYSIS
(a) Method of joints
(b) Method of sections
(c) Graphical methods
METHOD OF JOINTS
After determining the reactions at the supports, the equilibrium of each joint is
considered one by one. Each joint will be in equilibrium if V  0 and H  0 , these two
conditions are satisfied. Forces in the members will either be tensile in nature or compressive in
nature. The joint is selected in such a way that at any time there are not more than two
unknowns.
The direction of an unknown force is assumed. If the magnitude of force comes out to
be positive than assumed direction will be correct. If the magnitude of force comes out to be
negative than assumed direction will be incorrect. Now the method will be illustrated with the
help of some examples.
Example: Find th forces in the members AB, BC, AC of the Truss shown below in
Figure. End A is hinged and B is supported on rollers.
Sol. A roller offers a reaction perpendicular to plane of rolling. Let RB is reaction at B.
A hinge offers two reaction components one in vertical direction and other in horizontal
direction. Since the load of 40 kN acts vertically downward, therefore only vertical reaction RA is
developed and no horizontal reaction.
From the geometry of the figure, the distance of 40 kN load from A in horizontal
direction along AB is AC cos 60o.
In ΔACB,
ACB = 90o
1
AC  ABcos60o ,  5   2.5m
2
BC  ABsin 60o  5 
3
m
2
Distance of 40 kN Load from A  ACcos60o  2.5 
1
2
= 1.25m
Taking moments about A, we have
RB × 5 = 40 × 1.25
RB 
For equilibrium
40 1.25
 10.00kN
5
V  0, i.e.
RA + RB = 40
RA + 10 = 40
RA = 30 kN
Considering equilibrium of joint A, first, because RA is known and only two unknown
forces F1 and F2 are there. At each joint two equation of statical equilibrium are available i.e.
V  0 and H  0 .
Let F1 is force produced in the member AC and F2 is the force produced in the member
AB as shown in Figure. Joint A has to be in equilibrium. Component of force F1 in vertical
direction will balance vertical reaction RA. Therefore, the arrow is marked in member (1) in
down direction. Applying condition V  0 at joint A.
F1 sin 60o = 30
F1  30 
2
3

60
3
= 20 3 kN (+ve sign indicates that as assumed
direction is correct)
= 34.62 kN
As the force F1 is pushing joint A, therefore F1 is compressive force. Mark arrow at joint
C as pushing it to show that member AC is compression member.
Now applying condition H  0 at joint A,
F1 cos 60o = F2
1
F2  20 3   10 3 kN
2
(Again +ve sign indicated that the arrow marked in member AB is correct). As the force
F2 is pulling the joint A, therefore F2 is a tensile force. At B, place the arrow marking away from
B to show that member AB is a tension member.
Next consider joint B, shown in Figure.
Let F3 is the force produced in the member BC. The joint B has to be in equilibrium. It
must satisfy the two conditions of statical equilibrium viz. V  0 , and H  0 . Let us assume
the direction of arrow towards B. Applying V  0 at joint B.
F3 sin 30o = 10
F3 = +20 kN (+ sign indicates that direction of F3
is correct).
As F3 pushes the joint B, therefore it is a compressive force. Now applying second
condition H  0 at the joint B,
F3 cos 30o = F2
20 
3
 10 3
2
10 3 = 10 3
Now the forces in the various members are tabulated in the table.
Member
AB
Force
10 3 kN tensile =
17.32 kN
BC
20 kN compressive
AC
20 3 kN compressive
= 34.64 kN
Forces are marked in the truss, as shown in Figure. If we take tensile forces as +ve, then
compressive forces will be –ve or vice-versa.
METHOD OF SECTIONS
This method is the powerful method of determining the forces in desired members
directly, without determining the forces in the pervious members. Thus this method is quick.
Both the methods, i.e. method of joints and method of sections can be applied for analysis of
truss simultaneously. For members near to supports can be analysed with the method of joints
and for members remote from supports can be quickly analysed with the help of method of
section.
In this method a section line is passed through the members, in which forces are to be
determined in such a way that not more than three members are cut. Any of the cut part is then
considered for equilibrium under the action of internal forces developed in the cut members and
external forces on the cut part of the truss. The conditions of equilibrium, i.e. V  0 , H  0 ,
M  0 are applied to the cut part of the truss under consideration. As three equations are
available, therefore three unknown forces in the three members can be determined. Unknown
forces in the members can be assumed to act in any direction. If the magnitude of a force comes
out to be positive then the assumed direction is correct. If magnitude of a force is negative than
reverse the direction of that force.
Method will now be illustrated with the help of some examples.
Example: Find the forces in the members AC and AB of the truss loaded as shown
in Figure, using method of section.
Sol.
AC = 5 cos 60o = 2.5m,
BC  5sin 60o  5
3
 4.33m
2
(a) Determination of Reactions.
Let RA and RB be the reactions at A and B.
Taking moments about A,
RB × 5 = 40 × AC. cos 60o
5R B  40  2.5 
1
2
RB = 10 kN
RA + RB = 40,
RA = 40 –10 = 30 kN
(b) Passing a section 1–1, thereby cutting the truss in two parts. Considering
equilibrium of left part. The left part of the truss is shown in Figure. This part is in equilibrium
under the action of one external force RA = 30 kN and other two internal unknown forces AC
and AB in the members AC and AB respectively. The directions of AC and AB both are
considered as tensile as marked in Figure.
(c) Determination of AC ,
Taking moments of all forces about point B.
The moment of force AB about pt. B is zero.
R A  5  (AC  BC)  0 (
BC is perpendicular distance between force AC and pt.
B, i.e. 4.33m).
(30  5)  AC  5sin 60o  0
AC 
150
5sin 60
o

150
 34.64kN
4.33
-ve sign indicates that the assumed direction is wrong. This force is actually
compression force.
Hence
AC = 34.64 kN (compressive)
(d) Determination of AB ,
If we take the moments of all forces about point C, then AC will be eliminated and
there will be only one unknown force AB .
Hence, taking moments about pt. C
(AB  ACsin 60o )  (R A  AC.cos60o )  0.
(M  0)
or (AB  2.5  0.866)  (30  2.5  0.5)  0
or
2.165 AB = 37.5
or
AB = +17.32 kN
The positive sign indicates that the assumed direction is correct. This force is tensile
force.
(b) Rsult.
AC = 34.64 kN (Comp.)
AB = 17.32 kN (Tensile)
Those results are same as obtained in Example.
NOTE: If the force in member BC is also to be determined then we will have to take
another section 2-2, so as to cut the members BC and AB, as shown in Figure. Now considering
equilibrium of right part of the truss, under the action of two internal forces BC , AB and one
external forces RB = 10 kN, we can apply condition M  0 . If we take moments about A, then
force AB will be eliminated and only one unknown force BC will remain. Hence by taking
moments about A, we get (BC  2.5)  (10  5)  0
or
BC  
50
 20kN
2.5
Negative sign indicates that assumed direction is wrong. This force is actually
compressive. Similarly if we take moments about C, force BC is eliminated and AB = 17.32 kN
(Tensile).