Unit - 2

Distributed Forces and Friction

2.1 Moment of area

• Gravity axis :

Line of action of the gravitational force acting on the body

• Centre of gravity :

The point through which the whole mass or weight of the body acts, irrespective of its position is known as centre of gravity

• Centre of mass :

The point where the entire mass of a body is supposed to be concentrated

2.2 Centroid of plane lamina and wire bends

• Centroid : (geometries centre of plane areas )

The plane figures like triangle, quadrilateral, circle have only areas but no mass the centre of area of such figures is known as centroid.

• Procedure to find centroid of composite areas and line :

Select suitable coordinate axes if not given

Divide given area into different parts having known length & area & centroid distanced

If the section is symmetrical about X axis, then we can directly find ȳ axis, then we can directly find X

Let x= x co-ordinate of centre of gravity or distance of C.G. of the area from y axis

y = y co- ordinate of centre of gravity or distance of C.G. of the area from X axis

let a1, a2, a3……… are the areas into which whole figure is divided ,

thenx =

y =

where,

x1, x2, x3 ....... = distances of centroid of part 1,2,3 … from y axis and

x1, x2, x3 ....... = distances of centroid of part x axis

For line segment & wire bents

Replace the area (A) by length (L)

• Problems on the centroid of areas

1. Locate the centroid of area of the following figure w.r.t. given axis

Solution:

Let us divide the given area as shown above into 3 simple parts.

Consider area ① i.e. Square abcd,

of □ avcd = A1 = = 36 cm2

X1 = = 3 cm

y1 = = 3 cm

consider area ② i.e.le cfd,

A2 = area of cfd =

A2 = 9 cm2

X2 = 6 + = 6 + = 7 cm

Y2 = = = 2 cm

Consider area ③ i.e. ade,

area of ade =

A3 =

A3 = 27 cm2

X3 = = 6 cm

y3 = = -2 cm

now

X =

=

X = 4.625 cm

Y =

=

Y = 1 cm

co-ordinate of centroid of the given area are

Y = 1 cm (from x axis)

2.     Determine the Y co-ordinate of the centroid of the shaded area as shown in figure

Consider trapezoid as area ① and semicircle as area ②

For trapezoid and semicircle, shaded area is symmetrical about Y axis and thus as Y axis passing through symmetry line, X = 0

1)     For trapezoid :

A1 = h

A1 = 324 mm2

Y1

Y1 = 10 mm

2)     For semicircle :

A2 = = = = 56.55 mm2

Y2 = 18 - = 18 - = 15.45 mm

   Y = = =

y = 8.84 mm

Q. Locate the centroid of shaded lamina as shown in figure

Let us split given composite

Area into simple parts

Area ① - square

Area ② - triangle

Area ③ - quarter circle

1.

A1 = 2002 = 40000 mm2

X1 = Y1 = = 100 mm

2.

A2 =

=

= 11250 mm2

X2 = 200 –

X2 = 200 – = 150 mm

y2 = 200 – = 200 – = 150 mm

3.

A3 = = 2

A3 =  17671.46 mm2

x3 = y3 = =

x3 = y3 = 63.66 mm

X = = 107.91 mm

Y = = 107.91 mm

Q: Locate the centroid of shaded lamina as shown in fig.

Consider the composite area

Shown in figure:

Let us split thus composite area into simple parts thus

Area ① - quarter circle (shaded)

Area ② - square

Area ③ - quarter circle (non-shaded)

• Area ①

A1 = = 2 = 7854 mm2

X1 = = = 42.44 mm

Y1 = = 42.44 mm

• Area ②

A2 = 100 100 = 10000 mm2

X2 = = -50 mm

Y2 = = 50 mm

• Area ③

A3 = = 2 = 7854 mm2

X3 = = 57.56 mm

Y3 = = 57.56 mm

Centroid of shaded portion will be

X =

=

=

X = 28.54 mm

Y =

=

Y= 38.12 mm

Ans:

Centroid of shaded lamina will be at (28.54 mm, 38.12 mm)

• Determine the X co-ordinate of the centroid of shaded area with respect to origin 0 as shown in figure.

1. Consider rectangle O A B C

area = A1 = 400 250 = 100000 mm2

X1 = 200 mm

2.     Consider right angle B C D

area = A2 =

mm2

X2 = 400 +

= 400 +

X2 = 450 mm

3.     Consider circle of 120 mm

Area = A3 =

A3 = = 11309.73 mm2

X3 = 200 mm

Thus X =

=

X = 242.52 mm

4.     Determine the co-ordinate of centroid of the shaded area as shown in figure

The given figure and Shadedarea is Symmetricalabout y axis Shaded area,

X = 300 mm

Let us divide the given diagram into 3 parts

① rectangle O A C D , ② A B C , ③ O E D

① for rectangle O A C D

A1 = 600 mm2

Y1 = = 150 mm

② for A B C

A2 =

=

= 45000 mm2

Y2 = 300 +

Y2 = 300 + (150/3) = 350 mm

③ For O E D

A3 = = 45000 mm2

Y3 = = 150/3 = 50 mm

Y =

=

Y = 225 mm

• Determine the y coordinate of centroid of the shaded area as shown in figure:

① consider  Rectangle O A C D

A1 = 200 100 200000 mm2

X1 = 100 mm

Y1 = 50 mm

② consider A B C

A2 = AC height

  = 200 100

A2 = 10000 mm2

Y2 = 100 + = 100 + = 133.33 mm

③ for circle

A3 = 2 = 2

A3 = 7853.98 mm2

Y3 = 100 mm

Y =

Y = 69.89 m

1. A thin rod is bent into a shape ABCD as shown in figure.

 Determine the centroid of bent rod w.r.t. origin ‘O’.

① line segment OA

Length = L1 = 100 mm

X1 = 50 mm                   co –ordinate of point C

Y1 = 0 mm

② Line segment AB

Length = L2 = 50 mm

X2 = 100 mm

Y2 = -25 mm    co-ordinate of C2

③ Line segment BC

L3 = 10 mm

X3 = 150 mm

Y3 = -50 mm    co-ordinate of C3

④ Line segment CD

L4 = 50 mm

X4 = 200 mm

Y4 = -25 mm    co-ordinate of C4

2.     A slender rod is welded into the shape as shown in figure

Locate the position of centroid of the rod w.r.t. origin ‘O’ if

AO = BO = CO = 50 mm.

25 cos 45 = 17.68 mm

25 sin 45 = 17.68 mm

Thus,

X =

X =

X = 00 mm    from origin ‘O’

Y =

Y =

Y = 3.45 mm    from origin ‘O’

2.3  Moment of Inertia :

Moment of force about any point is the product of magnitude of force and perpendicular distance of line of action of force the point

Mo F d

              O

      d

                  F

This is known as first moment of force.

• When this product Mo i.e. (F d) is again multiplied with distance ‘d’ then

2

2 is known as second moment of force or moment of inertia of force.

• When force is replaced by area of figure or mass of the body, then second moment is known as area moment of inertia or mass moment of inertia.

A)    Area moment of inertia (I)

The second moment of area about a particular axis is known as area moment of inertia about axis

Unit – mm4, Cm4, m4……

First moment of area = area distance

     = mm2 mm

      = mm3

Second moment of area = moment (1st) of area distance

      = area distance distance

     = mm3 mm

     = mm4

B)    Mass Moment of Inertia :

The second moment od mass of the body about a particular axis is known as ‘’mass moment of inertia’’ about thet axis

Unit – kg.mm2, kg.m2

=

= kgm2

• Radius of Gyration :

If entire mass of body be assumed to be concentrated at a certain point which is located at a distance K from given axis such that

2

=

Then

Distance K is known as Radius of Gyration

Thus Radius of Gyration is defined as the Distance from the axis of reference where whole mass of the body is assumed to be concentrated

• For plane figure having negligible mass, we can consider area for finding radius of gyration

Thus

of an area (plane figure)

2

• Perpendicular Axis Theorem :

It states that , if be the moments of inertia of figure or plane section about two axis x and y which are perpendicular to each other at point O, then moment of Inertia of that plane section about Z axis which is perpendicular to the plane [&( is given by :

of plane area about

Z axis passing through C.G.

of plane area about

X axis passing through C.G

of plane area about

Y axis passing through C.G

is also called as polar moment of Inertia (J)

• Parallel axis Theorem:

It states that moment of Inertia of plane area about an axis passing through its centre of gravity is denoted by G, then moment of Inertia of that area about any other axis AB which is parallel to the first is given by:

AB = G + Ah2

AB = .M. of plane area about axis AB.

G = of plane area

it C.G.

A = Area of plane figure

h = distance between centre of gravity of

  area & axis AB.

• Determine moment of Inertia for section about x & y axis as shown in figure.

Q: Determine the moment of inertia of shaded area as shown in figure about its centroid as axis.

As this figure is symmetrical about

Y axis

X = 100 mm

Y =

There ,

area ① = rectangle

area ② = triangle

area ③ = circle

Y = 79.95 mm

to find M.I. of shaded portion , let G is the centroid of shaded area which is at y = 79.95 mm from base.

of shaded portion @ x-x axis passing through its centroid G will be,

x-x) +

x-x) –

x-x axis)

1 + 23

= (G1 + A1h12) + G2 + A2h22) - G3 + A3h32)

= + -60] +

+ –

84329013.21 mm4

of shaded portion about y-y axis passing through its centroid G will be,

y-y) +

y-y axis) –

y-yaxis)

1 + 23

=

= 80000000 + 16666666.67 – 3220623.34

93446043.33 mm4

2.4  Friction

It is a tangential force developed between the bodies (or two surfaces) which are in contact when one body moves or tends to move over another body (or surface).

Frictional force always acts in the opposite direction of motion or in the opposite direction of impending motion.

• Impending Motion :

It is the state of body in which the body is on the verge of motion i.e. about to move.

Frictional force is denoted by Fs or FK

• Characteristics Frictional Force

1. Its reactive and self-adjusting force
2. Frictional force increases when external force is increased upto limiting condition.

At the limiting particle try to impend the motion.

When particle is in motion, frictional force decreases with increases in external force.

3.     Frictional force always acts in the opposite direction of motion or in the opposite direction of impending motion

4.     Upto limiting condition, frictional force is equal in magnitude of external force\.

• Friction

Friction is defined as the resistive force which acts in the opposite direction of motion of body.

• Classification of Friction:

1. Static Friction
2. Kinetic Friction  A) sliding Friction , B) Rolling Friction

• Static Friction :

Friction experienced by a body when it is at rest condition

• Kinetic Friction :
  • Friction experienced by a body when it moves
  • When one body slides over another, then friction experienced by a body is          sliding friction
  • When one body rolls over another, then friction experienced by body is called as Rolling Friction.
• Laws of Friction (Coulombs Law)

1. For Static Friction :

1) frictional force always acts in opposite direction in which body tends to move

2) frictional force is directly proportional to normal reaction

3) Frictional force is equal to net force acting in the direction in which object

tends to move.

4)Frictional force does not depends an the surface area contanct.

5) Frictional force depends upon roughness of surface.

2.5 Application of friction on inclined planes Wedges and ladders friction Application to flat belt

• For kinetic friction :

1)     Frictional force always acts in opposite direction in which body moves

2)     Frictional force is directly proportional to normal reaction.

3)     For low speeds, frictional force is constant.

4)     For higer speeds, frictional force decreases with increasing speed and vice-versa.

• Coefficient of friction :

1. Coefficient of static friction () : from Culombs law

Magnitude of frictional force is directly proportional to normal reaction

FS RN

FS = S RN

S =

B.     Coefficient of kinetic friction (K) :

C.    FK RN

FK = K RN

K =

• Angle of Friction () :

Angle of static Friction (S) :

For the body at rest, the angle between

Normal reaction and resultant reaction

Is called as angle of static friction (S)

tan (S) =

tan S = S

S =tan Sand

S = tan-1S

• Angle of kinetic friction (K) :

For the body in motion, angle made by normal reaction with resultant is called as angle of kinetic friction (K)

tan (K) =

tan K = K

K =tan K and

K = tan-1K

• Limiting Friction  / Limiting equilibrium :

It is the max value of frictional force when body just begins to slide over a surface.

Or

It is the max value of frictional force when body is just on the verge of sliding.

•   Angle of Repose () :

It is the angle made by inclined surface with the horizontal, when the block kept on inclined plane just begins to slide down the plane due to its self-weight.

Angle of friction static and angle of repose are equal.

S =

tan S = S

Cone of Friction :

Here

= semi vertex angle

= angle of friction

In the limiting state of equilibrium, of block , if line of action applied force P is rotated in a horizontal plane, then line of action of resultant reaction RR will also rotate. Due to rotation line of action of RR will generate a cone which is known as cone of friction.

Altitude of Height of Cone = Normal Reaction = RN

Base Radius = Limiting Frictional Force = FS = SRN

Generator of Cone = Slant Height = RR = Resultant Reaction.

•    Belt friction :

1. Flat Belt :

Let,

T1 Tension in Tight side of belt

T2 Tension in Slack side of belt

= Lap angle or angle of contact in radians

S = coeff. Of static friction

K = coeff. Of kinetic friction

when belt is about to slip :

= ……… (T1 > T2)

(in motion) :

= ……… (T1 > T2)

2.     V-Belt

=

Where = abgle of V groove or belt

3.     Bond – Brake :

Q: Determine the horizontal force P needed to just start moving the 300 N crate up the plane as shown in figure. Take = 0%

For the equilibrium

= 0

P – RNcos 70 – FScos 20 = 0

P – RNcos 70 – s FScos 20 = 0

P – 0.342 RN  0.094 RN = 0

P = 0.436 RN

– 300 + RN Sin 70 – FS Sin 20 = 0

– 300 + RN Sin 70 – s FS Sin 20 = 0

– 300 + RN Sin 70 – 0.1 RN Sin 20 = 0

– 300 + 0.94 RN – 0.342 RN = 0

0.9058 RN = 300

RN = 331.2 N

n 1

P = 0.436 331.2

P = 144.4 N ------------------- force required

3.     Placed

Q: A body of weight 50 N is along rough horizontal plane A pull of 18 N acting at an angle of 140 with the horizontal find coefficient of static friction.

Considering the equilibrium of block

18 Cos 14 – Fs = 0

Fs = 17.46 N

RN + 18 Sin 14 – 50 = 0

RN = 45.64 N

As Fs = S RN

17.46 = S 45.64

S = 0.386

Q: A block of mass m rests on frictional plane which makes an angle with horizontal as shown if the coeff. Of friction between the block & frictional plane is 0.2, determine angle for limiting friction.

For limiting friction

Tan tan S = S

tan-1S

tan-1

11.310

Q: Determine whether the block shown is in equilibrium and find the magnitude and direction of friction force when = 350 and P = 100 N

Let us select inclined plane as x-x axis & perpendicular to plane as y-y axis

Let Fs = static frictional force which is required to maintain equilibrium.

--------  resolving force along x-axis

Fs + 100 Cos 35 – 250 Sin 35 = 0

Fs – 61.48 = 0

Fs = 61.48 N ----- frictional force

--------  resolving force along y-axis

-(250 Cos 35 + P Sin 35) + RN = 0

-250 Cos 35 – 100 Sin 35 + RN = 0

-262.15 + RN = 0

RN = 262.15 N -------- normal reaction

Maximum frictional force can be developed is given by

Fs max = S RN = 0.3 262.15

Fs max = 78.65 N

As frictional force required to maintain equilibrium (61.48 N) is less than max. Frictional force,

Equilibrium will be maintained

     Q:

At what position from the base, will be induce the slipping of ladder.

Take S = 0.34 at all contact surface.

Weight of man = 609.81 =588.6 N

Weight of ladder

AB = 8 m

Impending motion

Ac = 8 Cos 54

BC = 8 Sin 54

ɑ = 2.35 m

b = x Cos 54

FSA = S RNA 

FSB = S RNB

Q: A 137 kg block is supported by a rope which

Is wrapped 1.5 times around a horizontal

Rod knowing that coeff. Of static friction

Between rope and rod is 0.15, determine

The range of values of P for which

Equilibrium is maintained.

Weight of block =

Angle of contact = LAP Angle

A)    Case I : when P > 1343.97 N

Force P tends to move the rope down ward and block moves upward. Thus the part rope on which force P is acting is height side and rope to which block is attached is slack side.

= =

P = 5525.34 N

B)    Case II : when P < 1343.97 N

force P is less than 1343.97 N, so block will move down ward & the side rope on which P acts will move upward.

Rope side to which block is attached is tight side and another side is slack side.

= =

P = 326.92 N

range of P for equilibrium is 326.92 N

Q: Two cylinders are connected by a rope that passes

Over two fixed rods as shown. Knowing that the

Coefficient of static friction between the rope and

The rods is 0.40, determine the range of the mass m

Of cylinder D for which equilibrium is maintained.

Weight of cylinder = 50 9.81 = 490.5 N

Lap angle =

S = 0.40

A)    Case I :

Weight of cylinder A > weight of cylinder B.

In this case cylinder A will tends to move downward while B will tends to move upward

Consider part AB of the Rope as AB rope is moving down,

tension in hight side T1 =

tension in slack side = T2 = unknown.

= = = 1.874

= T2

T2 = 261.74 N

Consider part CD, of the rope. As rope CD is moving upward,

Tension in tight side T1 = T2 of rope AB

Tension in tight side, T1 = 261.74 N

Tension in slack side = T2 = (m 9.81)

B)    Case II :

Weight of cylinder A < weight of cylinder B

In this case cylinder a will tends to move upward while cylinder B will tends to move downward

Consider part AB of the rope moving upwards

Tension in slack side = side A =

Consider part CD of rope, Rope CD is moving downwards

Tension in tight side = T1 = (m 9.81) N

Tension in slack side = T2 = T1 of rope AB = 919.197 N

Range of mass m for equilibrium to be maintained is

14.23 kg m 175.59 kg

Q: A Force P = mg/6 is required to lower the cylinder with card making 1.25 turns around the fixed shaft. Determine the coeff. Of friction “S” between the card and the shaft

Refer the figure

Lap angle

= 2.5 radians

For flat belt,

=

=

6

1.79 = S 2.5

S = 0.23

Q: Determine the range of P for the equilibrium of block of weight W as shown in fig. the coeff. Of friction between rope and pulley is 0.2

Case I : when P is max.

=

=

Pmax = 1.369 w

Case II : when P is min.

=

Pmin = 0.73 w

For equilibrium P must be between 0.73 P 1.369 w

Reference Books:

1. Engineering Mechanics by S. P. Timoshenko and D. H. Young, McGraw- Hill publication

2. Engineering Mechanics by J. L. Meriam and Craige, John Willey

3. Engineering Mechanics by F L Singer, Harper and Rowe publication

4. Engineering Mechanics by A. P. Boresi and R. J. Schmidt, Brooks/Cole Publication