Engineering Mechanics 1st Year

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P

INTRODUCTION
We have so far dealt with smooth surfaces, which offer a single reaction force
R. In this chapter w-e deal with rough surfaces which offer an additional
reaction known as friction force. Friction force is developed whenever there is
a motion or tendency of motion of one body with respect to the other body
involving rubbing of the surfaces of contact. We will understand the concept
of friction and also present the laws of friction. Application of friction to the
problems of blocks, ladder, wedges, square threaded screw, and rope friction
will be dealt with in this chapter.


FRICTIONAL FORCE
Friction is of two types i) Dry Fiction ii) Fluid Fiction. Dry friction also known
as Coulomb friction involves friction due to rubbing of rigid bodies, for
example a block tending to move on table or a wheel rolling on the ground.
Fluid friction is developed between layers of fluids as they move with different
velocities inside a pipe, bodies moving over lubricated surface, etc. Our study
would be limited to Dry Friction.
Frictional force is generated whenever a body moves or tends to move over
another surface. This is best illustrated by the following discussion. Consider
a block of weight W resting on a rough surface. Let the normal reaction be N.
The block is in equilibrium under the action of two forces. Fig. (a)

Now if an attempt is made to disturb the equilibrium by applying a force P as
shown in Fig. (b), the rough surface generates a friction force F to maintain
the equilibrium. The force F is equal to P and the block is maintained
equilibrium. If P is now increased the friction force F also increases {refer
shown in Fig. (c)}. However the surface can generate a maximum friction
Friction

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known as limiting fiction force Fmax. If P exceeds Fmax, the body would be set
into motion and the block is said to be in a dynamic state. In dynamic state,
the same surfaces develop a lower friction force known as kinetic frictional
force Fk.
Friction force generated is generally due to presence of hills and valleys on
any surface and to smaller extent due to molecular attraction. Fig. Shows a
macroscopic view of two contacting surfaces enlarged 1000 times.

When the body is in static condition, there is greater interlocking between the
hills and the valleys of the two surfaces. As the body becomes dynamics, the
interlocking reduces resulting in lowering the friction force to a new value Fk.


COEFFICIENT OF FRICTION
It has been experimentally found that the ratio of the limiting frictional forces
Fmax and the normal reaction N is a constant. This constant is referred to as
coefficient of static friction, denoted as s
μ . max
s
F
μ=
N

…..4.2
Since k
F is less than max
F , we have ks
μ < μ
Coefficient of friction s
μ or k
μ depend on the nature of surfaces of contact and
have value less than 1. For example the approximate value of coefficient of
static friction (s
μ ) between wood on glass ranges between 0.2 to 0.6. The
corresponding coefficient of kinetic friction (k
μ ) is around 25 % lower.

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LAWS OF FRICTION
1. The frictional force is always tangential to the contact surface and acts
opposite to the direction of impending motion.
2. The value of frictional force F increases as the applied disturbing forces
increases till it reaches the limiting value Fmax. At this limiting stage the
body is on the verge of motion.
3. The ratio of limiting frictional force (Fmax). and the normal reaction N is a
constant and it is referred to as coefficient of static friction (s
μ ).
4. For bodies in motion, frictional force developed (k
F ) is less than the limiting
frictional force (max
F ). The ratio of k
F and the normal reaction N is a constant
and is referred to as coefficient of kinetic friction (k
μ ).
5. The frictional force F generated between the two rubbing surfaces is
independent of the area of contact.


ANGLE OF FRICTION, CONE OF FRICTION AND ANGLE OF REPOSE
Angle of Friction:
"It is the angle made by the resultant of the limiting frictional forcemax
F and
the normal reaction N with the normal reaction". Fig. shows a block of weight
W under the action of the applied force P. Let N be the normal reaction. If the
block is on the verge of impending motion, the frictional force max
F . Would be
developed as shown.
Let R be the resultant of max
F , and N, making an angle  with the normal
reaction. Here  is known as the angle of fiction.
Here 22
max
R = F + N

2
2
s
=μ N + N
Also max s
F μN
tan = =
NN

s
tan =μ
1
s
tanμ

 .....4.3

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CONE OF FRICTION
Fig. shows a block of Weight W on the verge of motion acted upon by force P.
Let R be the resultant reaction at the contact surface acting at an angle of
friction . If the direction of force is changed by rotating it through 360° in a
plane parallel to the contact surface, the force R also rotates and generates a
right circular cone of semi-central angle equal to . This right circular cone is
known as the cone of friction.

The significance of cone of friction is that for static body the resultant reaction
force at the rough contact surface lies within the cone, while for a static body
on the verge of motion, the reaction force lies on the surface of the cone of
friction.


ANGLE OF RESPONSE
It is defined as the minimum angle of inclination of a plane with the
horizontal for which a body kept on it will just slide down on it without
application of any external force.
Consider a block of weight W resting on a
rough horizontal plane. The plane is slowly
tilted till the block is just on the verge of
sliding down the plane, Fig. The angle of
inclination of the plane at this position is
known as the angle of repose. It is denoted
by letter α. Angle of repose is independent
of the weight of the body and depends only
on the coefficient of static friction. Let us

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derive the relation between angle of repose α and coefficient of static frictionμ
s .
Applying COE
F0
x
 + ve
max
F -Wsinα=0 μsN-W sinα=0
----(1)
F0
y
 + ve
N-W cos =0
N=Wcosα ----(2)
Substituting (2) in (1) μ
s
(W cos α) – W sin α = 0
tan α = μ
or α = tan
-1
μ
s …..(3)
but we have seen that
1
tanμ
s




i.e. Angle of Response = Angle of Friction
Though magnitude wise both angle of repose and angle of friction have the
same value, their meaning and application are different, as we have seen.


PROBLEM ON BLOCKS
We will encounter dimensionless blocks acted upon by forces tending to cause
motion. The blocks can be modelled as a particle forming a concurrent system
of forces. The following steps are adopted in the solution of problems on
blocks.
Step 1: Draw the FBD of the block showing the weight W, the normal reaction
N, the friction force F and other applied forces. The direction of F is always
directed opposite to the direction of impending motion. When the block is on
verge of motion, F = μ
s N should be taken.
Step 2: Since we have a concurrent system of forces in equilibrium we apply
two conditions of equilibrium viz.
x
F =0
y
F =0

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EXERCISE 1
1. For the following cases find force P needed to just impend the motion of the
block. Take weight of block to be 100 N and coefficient of static friction at
the contact Surface to be 0.4.

2. A block weighing 800 N has to rest on an incline of 30°. If the angle of
limiting friction is 18°, Find the least and greatest force that need to be
applied on the block, parallel to the plane so as to keep the block in
equilibrium.

3. A support block is acted upon by two forces
as shown. Knowing μ
s = 0.35 and μ
k = 0.25,
determine the force P required.
a) to start the block moving up the incline.
b) to keep it moving up.
c) to prevent it from sliding down.
4. A block of weight 200 N rests on a horizontal surface. The co-efficient of
friction between the block and the horizontal surface is 0.4. Find the
frictional force acting on the block if a horizontal force of 40 N is applied to
the block.

5. A horizontal force P is applied to the block A of mass
50 kg kept on an inclined plane as shown. If P = 200
N, find whether the block is in equilibrium. Also find
the magnitude and direction of frictional force.
Take μ
s = 0.25.

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6. A car of 1000 kg mass is to be parked on the same 10° incline year around.
The static coefficient of friction between the tires and the road varies
between the extremes of 0.05 and 0.9. Is it always possible to park the car
at this place? Assume that the car can be modelled as a particle.
7. Block A of weight 2000 N is kept on a plane inclined at 35°. It is connected
to weight B by an in extensible string passing over a smooth pulley.
Determine weight of B so that B just moves down.
(a)  = 0 (b) 0 = 20°
Takeμ=
0.2



8. Two blocks A and B of weight 500 N and 750 N respectively are connected
by a cord that passes over a frictionless pulley as shown.μ between the
block A and plane is 0.4 and between the block B and plane is 0.3.
Determine the force P to be applied to block B to produce the impending
motion of block B down the plane.

9. Determine the force P to cause motion to impend. Take masses A and B as
9 kg and 4 kg respectively and coefficient of static friction as 0.25. The force
P and rope are parallel to the inclined plane. Assume smooth pulley.
b) Solve taking mA = 8 kg, B
m 4 kg, μ 0.3.
s


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10. What is the minimum value of mass of block B required to maintain the
equilibrium? The rope connection A and B passes over a frictionless pulley.

11. Determine the least and greatest value of W for the equilibrium of the whole
system.

12. Three blocks are placed on the surface one above
the other as shown.
The coefficient of friction between the surfaces is
shown. Determine the maximum value of P that can
be applied before any slipping takes place.
Hint: check for three possibilities.
13. Block A weighing 1000 N rests over a block B
weighing 2000 N as shown. Block A is tied to a
wall with a horizontal string.
If p = o.25 between blocks A and B and μ = l/3
between block B and floor, determine the force P
needed to move the block if (a) P is horizontal, P acts at 30° upwards to
horizontal.
14. Block A of mass 27 kg rests on block B of mass
36kg. Block A restrained from moving by a
horizontal rope to the wall at C. What force P,
parallel to the plane inclined at 30° with the
horizontal is necessary to start B down the plane?
Assume p for all surfaces = 0.33.

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15. What should be the value of ' ' so that the motion of
block A impends down the plane? Take m A = 40 kg
and mB = 13.5 kg.
μ = 1 / 3 for all surfaces.

16. Block A has a mass of 25 kg and block B has a mass of 15 kg. Knowing s
μ
= 0.2 for all surfaces, determine value of 0 for which motion impends.
Assume frictionless pulley.

17. A block A of mass 10 kg rests on a rough inclined plane as shown. This
block is connected to another block B of mass 30 kg resting' on a rough
horizontal plane by a rigid bar inclined at an angle of 30°. Find the
horizontal P required to be applied to block B to just move block A in the
upward direction. Take μ = 0.25 for all contact surfaces.

18. Find the maximum value of W B for the rod AB to remain horizontal. Also
find the corresponding axial force in the rod.
Take μ = O.2 for all contact surfaces.

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19. Two blocks connected by a horizontal link
AB are supported on two rough planes, μ
between block A, and horizontal surface is
0.4. The limiting angle of friction between
block B and inclined plane is 20°. What is
the smallest weight W of the block A for
which equilibrium of the system can exist, if the weight of block B is 5 kN?

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WEDGES
Wedges are tapering shaped devices which are used for lifting or shifting to
little extent heavy blocks, machinery, pre-cast beams and columns etc.
Usually two or more wedges are used in combination. Fig. shows two
arrangements of wedges. In Fig. (a) Two wedges are used for imparting a small
vertical movement to heavy machinery of weight W. Effort P is applied to the
wedge B which is driven inside causing the vertical movement. In Fig. (b) Two
wedges are used in combination to cause a small horizontal movement to the
heavy block of weight W. Effort P is applied to wedge B, driving it down,
thereby causing horizontal movement of the heavy block.

For solving problems on wedges, we require to isolate the wedges and then
apply COE to each wedge separately. Since the dimensions are neglected,
wedges are treated as articles forming a concurrent force system. Only two
COE viz.x
F = 0 and F
y = 0 are therefore useful.
Solving equations (1) and (2), we get
N1 = 537.25 N and N3 = 587.8 N
Applying COE to Wedge A F0
y

2 3 3
N - 0.2N sin75-N cos75-800 = 0

Substituting N3 = 587.8 N, we get N2 = 1065.7 N
F0
x

2 3 3
P+0.3N + 0.2 N cos 75-N sin 75 = 0
Substituting N2 = 1065.7 N and N3 = 587.8 N, we get
P = 217.6 N

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EXERCISE 2

1. The horizontal position of the 1000 kg block is adjusted by 6° wedges. If
coefficient of friction for all surfaces is 0.6, determine the least value of
force P required to move the block.

2. A block weighing 1000 N is raised against a surface inclined at 60° to the
horizontal by mean of a 15° wedge as shown in figure. Find the horizontal
force P which will just start the block to move if the coefficient of friction
between all the surfaces of contact be 0.2 Assume the wedge to be of
negligible weight.

3. The vertical position of the 200 kg mass I section is being adjusted by two
15° Wedges as shown. Find force P to just raise the mass.
Take μ = 0.2 for all surfaces.

4. Wedge A supports a load of W = 5000 N which is to be raised by forcing the
wedge B under it. The angle of friction for all surfaces is 15°. Determine the
necessary force P to initiate upward
motion of the load Neglect the weight of
the wedges. Hint: Wedge A comes in
contact with the right wall, as the load
gets lifted.

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5. Two blocks A and B are resting against the wall and floor as shown in
figure. Find minimum value of P that will hold the system in equilibrium. μ
= 0.25 at the floor, μ = 0.3 at the wall and μ = 0.2 between the blocks.


PROBLEMS ON LADDERS
We have all made use of ladders at one time or the other to reach for things
kept high. But every time while climbing the ladder we fear that the ladder
might slip. All this is because the friction force at the floor and at the wall is
responsible for preventing the slip. If these surfaces do not produce sufficient
frictional force the ladder would slip. Invariably we adjust the slope of the
ladder and even ask someone to hold the ladder as we use it. The mechanics
involving ladders is dealt in this section
Forces acting on ladder viz. the normal reactions and friction-force at the
floor and wall contact points, the weight of the ladder, the weight of the
person climbing the ladder and any other force, form a non-concurrent force
system. It will therefore require all the three COE viz. F
x = 0, F0
y
 and M=0
for analyzing the equilibrium of a ladder.
EXERCISE 3

1. A person of weight P = 600 N ascends the 5 m
ladder of weight 4OO N as shown. How far up the
ladder may the person climb before sliding motion
of ladder takes place.

2. A 1OO N uniform ladder AB is held in equilibrium
as shown. If μ = 0.15 at A and B, calculate the range
of values of P for which equilibrium is maintained.

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3. A 150 N uniform ladder 4 m long supports a 500 N weight person at its
top. Assuming the wall to be smooth, find the minimum coefficient of
friction which is required at the bottom rough surface to prevent the
ladder from slipping.

4. The ladder shown is 6m long and is supported by a horizontal floor and a
vertical wall.μ Between floor and ladder is 0.4 and between wall and ladder
is 0.25. The weight of ladder is 200 N. The ladder also supports a vertical
load of 900 N at C. Determine the greatest value of 0 for which the ladder
may be placed without slipping.

5. A 6.5m ladder AB of mass 10kg leans against a wall as shown. Assuming
that the coefficient of static friction μ is the same at both surface of
contact, determine the smallest value of p for which equilibrium can be
maintained.

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6. A ladder shown is 4 m long and is supported by a horizontal floor and
vertical wall. The coefficient of friction at the wall is 0.25 and that at the
floor is 0.5. The weight of ladder is 30 N. It also supports a vertical load of
150 N at 'C'. Determine the least value of 'α' at which the ladder may be
placed without slipping to the left.

7. A uniform ladder of length 2.6m and weight 24O N is placed against a
smooth vertical wall at A, with its lower end 1 m from the wall on the floor
at B. The coefficient of friction between the ladder and the floor is 0.3.
Find the frictional force acting on the ladder at the point of contact
between the ladder and the floor. Will the ladder remain in equilibrium in
this position? Give reason.