Kom Tutorials & Assignments Unit 1 & 2

8/9/2019 Kom Tutorials & Assignments Unit 1 & 2

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UNIT 1

TUTORIAL

1. For the Kinematic linkages shown in fig, Calculate the number of binary links,

ternary links, links and degree of freedom.

Solution :

(a) For fig (i):

binary links = 4 ; higher pairs (h) = 0 fig.

ternary links = 4

total links = 8

Degree of freedom,


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n = 3(l 1) 2 j h

=3(8 1) 2 x 11 0

n = 1.

The link is a unconstrained chain.

(b) For fig(ii)

binary links = 4

Ternary links = 4

Total links = 8

Degree of freedom , n = 3(l 1)2j h

n = 3(8 1) 2 x 10

n = 1

The linkage is a constrained chain.

(c) For fig (iii)

binary links = 7

Ternary links = 2

Total links = 11

Degrees of freedom, n = 3(l 1)2j h


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=3(11 1) 2 x 15

n = 0

The linkage is a locked chain.

2. Determine the mobility of all the linkages shown in fig.

Solution :


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For Fig(i)

Mobility (or) No

of degree of freedom

n = 3(l

1)2j h

From the Fig (a),= 5; h

= 0; It has two binary jointsand two ternary joints,

so j = 2 + 2 x 2 = 6

n = 3(l 1)2j h

n = 3(5 1)2 x 6 0

n = 0

Ans : Therefore linkage forms structure.

(2) For Fig.(b): From the Fig.(b), l = 6; h = 0;

It has seven binary joints, so j = 7

n = 3(l 1) 2j h

n = 3(6 1) 2 x 7 0 = 1

Ans: Therefore it has single degree of freedom.


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(3) For Fig.1(C):From the Fig.(C), l = 7;h = 0; It has eight binary joints, so j = 8.

n = 3(l 1) 2j h

n = 3 (7 1) 2 x 8 0 = 2

Ans: Therefore, the linkage has two degrees of freedom.

(4) For Fig.(d): l = 4; h = 1 (i.e., roller constitutes a higher pair); and

j = 3 (i.e., it has three binary joints)

n = 3(l 1) 2j h

=3(4 1) 2 x 3 1 = 2

Ans: Therefore the linkage has two degrees of freedom.

(5) For Fig.(e): From the Fig(e), l = 3; j = 2 and h = 1(i.e., roller constitutes a higher

pair)

n = 3(l 1) 2j h

=3(3 1) 2 x 2 1

Ans: Therefore the linkage has single degree of freedom.

3. Determine the degree of freedom of linkages shown in Fig.


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Solution :

(i)

For Fig.

(i): l = 6; j = 7; h

= 0

Degree of

freedom,

n = 3(l

1) 2j h

n = 3(6

1) 2 x 7 0 = 1.

(ii) For Fig (ii): 1 = 8; j = 10; h = 0

n = 3(l 1) 2 j h

=3( 8 1 ) - 2 x 10 10 = 1 Ans


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(iii) For Fig.(iii): l = 4; h = 0

n = 3(l 1) 2j h

= 3(4 1) 2 x 4 0 = 0

(iv) For Fig.(iv): l = 4; j = 4 (three turning pair and one sliding pair); h = 0

n = 3(l 1) 2j h

= 3 (4 1) 2 x 4 0 = 1

(v) For Fig.(v): l = 3; j = 3; h = 0

n = 3(l 1) 2j h

=3(3 1) 2 x 3 0 = 0

(vi) For Fig.(vi): l = 4; j = 4; h 0

n = 3(l 1) 2j h

=3 (4 1) 2 x 4 0 = 1

(vii) For Fig (vii): l = 6; j = 7; h 0

n = 3(l 1) 2j h


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=3(6 1) 2 x 7 0 = 1

4. For the kinematic linkages shown in Fig. Find the degrees of freedom.

Solution :

(1) 1 = 6; j = 7; h = 0

Number of degree of freedom,

n = 3(l 1) 2j h

n = 3(6 1) 2 x 7 0 = 1

Thus the motion is constrained.


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(2) For Fig.(ii): l = 7; j = 8; h = 0

n = 3(l 1) 2j h

= 3(7 1) 2 x 8 = 2

Thus the motion is constrained.

(3) For Fig.(iii): l = 5; j = 6; h = 0

n = 3 (l 1) 2j h

= 3 (5 10 2 x 6 0 = 0

Thus it is a structure. Ans.

(4) For Fig.(iv): In this Fig (iv), the link 3 can slide without causing any movement in

the rest of the mechanism. Thus this link is said to have redundant degree of freedom.

L = 4; j = 4; h = 0; r = 1

n = 3 (l 1) 2j h r

=3 (4 1) 2 x 4 0 1 = 0

Thus it is a structure.

5. Find the degrees of freedom for the linkages shown in Fig below.


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Solution:

(1) For Fig.(i): l = 14; j = 18; h = 1

n = 3(l 1) 2j h

= 3(14 1) 2 x 18 1 = 2 Ans

(2) For Fig.(ii): l = 7; j = 8; h = 0

n = 3(l 1) 2j h

= 3(7 1) 2 x 8 0 = 2 Ans

(3) For Fig.(iii): l = 8;

j = 10 [ 2 binary joints and 2 ternary joints, so j = 2 + 4 x 2 = 10]; h = 0

n = 3(l 1) 2j h

= 3(8 1) 2 x 10 0 = 1


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(4) For Fig.(iv): l = 7; j = 7; h = 1

n = 3(l 1) 2j h

= 3(7 1) 2 x 7 1 = 3 Ans.

6. Examine the mechanism shown in fig and indicate the cases where unique

relation between the motions of the input and output links exists.

Solution:

(i) For Fig.(i): l = 6; j = 7; h = 0

n = 3(l 1) 2j h

=3(6 1) 2 x 7 0 = 1

Thus it is a constrained motion. This mechanism is driven by single input motion.


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(2) For Fig.(ii): In the Fig.(ii), the link 3 can slide without causing any

movement in the rest of the mechanism. Thus this link is said to have redundant degree

of freedom.

L = 4; j = 4; h = 0; r = 1

n = 3(l 1) 2j h r

=3(4 1) 2 x 4 0 1 = 0

Thus it is a structure.

(3) For Fig.(iii): l = 10; h = 0; h = 0; j = 12

[because ten binary joints and one ternary joint, so 10 + (1 x 2) = 12]

n = 3(l 1) 2j h

=3(10 1) 2 x 12 0 = 3

So it does not have unique relation between input and output link's motions.

7. Prove that the mechanism shown in Fig. Below is a constrained kinematic chain.


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Solution:

From the Fig., l = 6; j = 7 and h = 0; p = 5

Therefore Degree of freedom,

n = 3(l 1) 2j h

=3(6 1) 2 x 7 0 = 1

Thus it is a single degree of freedom system.

Two conditions to be satisfied are:

l = 2p 4 .... (i)

and j =2

3

l 2 ....(ii)

where l = Number of links = 6

p = Number of pairs = 5

j = Number of joints = 7

From equation (i), 6 = 2 x 5 4 = 6

So equation (i) is satisfied.


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From equation (ii) 7 =2

3

x 6 2 = 7

L.H.S. = R.H.S.

So equation (ii) is also satisfied.

Hence, the given mechanism is a constrained kinematic chain. Ans.

8. In a crank and slotted lever quick return mechanism, the distance between the fixed

centres is 150mm and the driving crank is 75mm long. Determine the ratio of the time

taken on the cutting and return strokes.

Solution :

=360

storkereturnofTime

strokecuttingofTime

linkfixedoflength

crankoflength

290sin =

= 75/150 =

90 -2

= sin1(0.5) = 30


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2

= 90 30 = 60

= 120

Stroke ratio =120

120360

= 2.

9. In a crank and slotted level quick return motion mechanism, the length of thefixed link is 300m and that of the driving crank is 150 mm. Determine the maximum

angle the slotted lever will make with the fixed link. Also determine the ration of time

of cutting and return strokes.

If the length of the slotted lever is 700mm, what would be the length of the stroke

assuming that the line of the stroke passes through the extreme positions of the free end

of the slotted lever?

Solution :

Given data :

AB = 300mm = 0.3

AE = 150mm = 1.15m

BP1= 7==mm = 0.7m


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Solution: (i)Inclination of the slotted bar with the fixed link:

LetABE = inclination of the slotted bar with the vertical

Fig.1.67 shows the extreme positions of the crank.

We know that, sinABE = sin

290

=ABAE

=

5030

150=

ABE = 90

2

= sin1(0.5)

= 30Ans.

(ii) Time ratio of cutting stroke to the return stroke:

We know that 902

= 30

or= 120


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strokereturnofTime

strokecuttingofTime

=

=

360

=2

(iii) Length of the stroke:

Length of the stroke = P1P2= 2(P1P)

= 2(BP) sin

290

= 2 x 0.7 sin (90

60

)

L = 450 mm

:

strokereturnofTime

strokecuttingofTime

=

=

360

=

!352"96

"96360=

10. The distance between two parallel shafts is 18mm and they are connected by an

oldham's coupling. The driving shaft revolves at 160 rpm. What will be the maximum

speed of the tongue of the intermediate piece along its groove?


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Solution :

r = Distance between the area of shafts = 18 mm = 0.018m

N = 160 rpm

= 16.75 rad/rad/s

Maximum sliding velocity V = r.

= (0.018)(16.75)

= 0.302 m/s


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ASSIGNMENT UNIT 1

1. Explain the four bar chain and its inversions stating its applications with neat

sketches.

Four bar chain:

A four link mechanism or linkage is the most fundamental and the simplest kinematic

chain. It consists of four turning pairs.

Four links of a four bar chain are:


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1.Frame: The fixed link is known as frame.

2.Crank (or) driver: A link that makes complete revolutions is called as crank.

3.Coupler: The link opposite to the fixed link is known as coupler.

4.Lever or Rocker or follower: The link which makes a partial rotation or

oscillation is known as lever.

Inversions of four bar chain:

I. First Inversion (Crank and lever Mechanism):

As shown in Fig (a), link 1 is the crank, link 4 is fixed and link 3 oscillates

whereas in Fig. (b),lnk 2 is fixed and link 3 oscillates. The mechanism is also known as

crank-rocker mechanism or a crank-lever mechanism or a rotary-oscillating converter.

Application: Beam engine

1. Beam Engine:


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This is an example of crank-lever mechanism, where one link oscillates, while the

other rotates about the fixed link, as shown in Fig:

This mechanism is used to convert rotary motion into reciprocating motion.

In this mechanism, when the crank rotates about a fixed centre A, the lever

oscillates about a fixed centre D. The end E of the lever CDE is connected to a piston

rod which reciprocates due to rotation of the crank.

II. Second inversion: (Double crank mechanism)

If the shortest link, i.e., link 1 (crank) is fixed, the adjacent links 2 and 4 would

make complete revolutions, as shown in Fig. The mechanism thus obtained is known

as crank-crank or double crank mechanism or rotary-rotary converter.

Application: Coupling rod of a locomotive.

2. Coupling Rod of a Locomotive:


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This is an example of a double crank mechanism where both cranks rotate about

the points in the fixed link. It consists of four links. The opposite links are equal in

length, since links 1 and 3 work as two cranks, the mechanism is also known as rotary-rotary converter. Refer Fig.

III. Third inversion:

If the link opposite to shortest link is fixed. i.e., link 3 is fixed, then the shortest

link (link 1) is made coupler and the other two links 2 and 4 would oscillate as shown inFig. The mechanism thus obtained is known as rocker:- rocker or double-rocker or

double-lever mechanism or an oscillating-oscillating converter.

Applications: 1.Watt's indicator mechanism

2.Pantograph

3.Ackermann steering

3. Watt's Indicator Mechanism:


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This mechanism was invented by Watt for his steam engine to guide the piston

rod. It is also known as simplex indicator. It consists of four links: fixed link at A, link

BC and link DEF, connected to the piston of the indicator cylinder. Links BC and DEFwork as levers and due to this, the mechanism is also known as double lever

mechanism. The displacement of the lever DEF is directly proportional to the steam or

gas pressure in the indicator cylinder.

In Fig. full lines depict the initial position of the mechanism, whereas the dotted

line shows the position of the mechanism when gas or steam pressure acts on the

indicator plunger.

2. Sketch the single slider crank chain and its inversions. Explain any one of its

applications.

Single slider crank chain is a modification of four bar chain. It consists of one

sliding pair and three turning pair. This is used to convert reciprocating motion to

rotary motion and vice versa.

Inversions of single slider crank chain


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(i) First inversion:

First inversion is obtained when link 1 (i.e. Frank is frame is fixed and link 2 and

4 are made the crank and slider respectively.

(ii) Second inversion:

Second inversion is obtained by fixing the link 2 (i.e. crank) of a slider crank

chain. As shown in Fig., when the link 2 is fixed then the link 3 along with the slider at

its end E becomes a crank. This makes link 1 to rotate about O along with the slider

which also reciprocates on it.

(iii)Third inversion:

This is obtained by fixing the link 3 of the slider crank mechanism. Link2 acts as a crank

and link 4 oscillates.


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(iv) Fourth inversion:

Fourth Inversion:

Fourth inversion is obtained by fixing the link 4 of the slider crank mechanism.As shown in Fig. In the inversion, link 3 can oscillate about the fixed pivot B on link 4.

This makes end A of link 2 to oscillate about B and end O to reciprocate along the axis

of the fixed link 4.

Application: Quick Return Mechanism:

The two inversions of single slider crank chain which are used for the sameapplication (i.e. Quick return motion) are:

(a) Whitworth quick return mechanism

(b) Crank and slotted lever mechanism


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Kinematic difference:

(i) In crank and slotted lever mechanism the output line oscillates.

(ii) In whit worth mechanism the output link rotates.

Description:

(i) Whitworth quick return mechanism:

It is a mechanism used in workshop i.e. In shaping and slotting machines. In this

mechanism, link 2 (i.e. Crank) is fixed, link 3 rotates, link 4 reciprocates and link 1

oscillates.


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Initially, let the slider 4 be at B1point C2, then the tool will be in its extreme left position.

When the crank further rotates and the slider 4 reaches to point C1, the tool will be in its

extreme right position. The distance between extreme left and right positions is thestroke length.

When the link AE rotates from the position AB2to AB2, then the ram moves from

left to right and corresponding movement of the crank link will be from BC2to B C1.

When the line AE further rotates, link AE move from AB2to AB1, then the ram moves

from right to left and the crank BC correspondingly moves from BC2to BC1.

When crank moves from BC2to BC1, it is the backward stroke. The time taken

during the left to right motion of the ram will be equal to the time taken by the crank

from BC1portion to BC2. The time taken during the right to left motion of the ram will

be equal to the time taken by the crank from BC2position to BC1. When the tool is in its

left extreme position, it is the return or backward stroke.

Let = obtuse angle C1BC1at B

= acute angle at C1B2at B.

Then,

=

=360

#or$360strokereturnofTime

strokecuttingofTime

(b) Crank and slotted Lever Quick return mechanism:-

This mechanism is mostly used in shaping and slotting machines. This

mechanism is also obtained by fixing the link 3, as shown in Fig. From the figure, OA is

the driving crank, BP is the slotted lever, PT is the connecting rod and T is the cutting

tool.


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As the link OA revolves clockwise, the link BP oscillates between the two extreme

positions BP1and BP2. In the extreme pistons, BP1and BP2are tangential to the circle

with A as center and AO radius. Hence the two strokes of the tool are made while the

crank makes over the arcs FGE and EHF respectively. The cutting stroke occurs when

the crank moves over the arc FGE.

Let, = Obtuse angle FAE at A

= Acute angle FAE atA


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Then,

=

=360

#or$360strokereturnofTime

strokecuttingofTime

The travel of the tool or length of stroke P1P2:

P1P2= 2 BP sin P1BA

= 2BP sin (90 -2

)

= 2BP cos2

P1P2= 2BP AB

A%

The ratio of the times given by the above formula can be varied by varying the distance

AB.

3. Sketch the double slider crank chain and its inversions. Explain the working of old

ham's coupling.

DOUBLE SLIDER CRANK CHAIN

As its name implies, in double slider crank chain, there are two sliders and one

crank. It is also a four bar pneumatic chain consists of two sliding pairs and two turning

pairs as shown in Fig. The link 1 and link 2 form one sliding pair and link 4 and link 1

from the second sliding pair. The link 2 and link 3 form one turning pair and link 3 and

link 4 form second turning pair.


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Inversions:

(i) First inversion:

First inversion is obtained by fixing the link 1. In this, the two adjacent pairs 2 3

and 3 4 are turning pairs and the other two pairs 1 2 and 1 4 are sliding pairs.

Refer Fig (a) Application: Elliptical trammel


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(ii) Second inversion:

Second inversion is obtained by fixing any one of the slider blocks of the first

inversion. When link 4 is fixed end B of crank 3 rotates about A and link 1 reciprocates

in the horizontal direction.

Application: Scotch yoke mechanism

(iii) Third inversion;

Third inversion is obtained by fixing the link 3 of the first inversion. In this inversion

link 1 is free to move.

Application: Oldham's coupling.

Oldham's Coupling:

This mechanism is used for transmitting motion between two shafts which are

parallel but not coaxial. This inversion is obtained by fixing the link 3 as shown in Fig.

It consists of a driving shaft, fitted with a flange(link 2) having a diametrical slot on its

face; a driven shaft fitted with flange C (link 4) also has diametrical slot on its face. This


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whole makes link 4. The slots on the two flanges are at right to each other. An

intermediate piece circular shape and having tongues X and Y at right angles on

opposite sides, is fitted in between the flanges of the two shafts in such a way that the

tongues of the intermediate piece get fitted closely in the slots of flanges. The

intermediate circular piece E forms link 1 which slides between the flanges C and D.

When driving shaft rotates through certain angle the driven shaft also rotates through

the same angle. Motion is transmitted through intermediate link1. If the distance

between the axis of the shafts is x, it will be the diameter of a circle traced by the center

of intermediate piece.

4.Differentiate Kinematic pair and Kinematic chain.

Kinematic pair Kinematic chain

1. A pair is a joint of two links or

elements of a machine.

Ex: 1.Crank and connecting road.

1. When the Kinematic pairs are coupled in such a

way that the last link is joined to the first link to

transmit definite motion, it is called a Kinematic

chain.


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Kinematic pair Kinematic chain

2.Piston and cylinder. Ex: 1. Four bar chain.

2. Double slider crank chain.

5. Structure

An assembly of resistant bodies, known as members having no relative motion

between them and meant for carrying loads is called structure.

6. Compare Machine and Structure:

Machine Structure

1. The different parts of a machine has

relative motion.

2. It transforms energy into useful work.

3. Links are used to transmit motion.

4. Example: Shaping m/c

1. There is no relative motion between the

members.

2. It does not convert the energy into work.

3. Members are used for carrying loads.

4. Example: bridges, roof trusses m/c frames.

7. Mechanical advantage of a mechanism.


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It is defined as the ratio of output torque to the input torque. It is also defined as

the ration of load to the effort.

8. Differents in Kinematic chain and Mechanism.

Kinematic chain Mechanism

1. It is a combination of Kinematic

pairs.

2. Types:

a) Four bar chain,

b) Slider crank chain

1. If any one of the link in the Kinematic chain is

fixed then the Kinematic chain is called amechanism.

2. Types:

a) Simple mechanism

b)Compound mechanism.

9. Transmission angle?

In a four bar chain mechanism, the angle between the coupler and the follower

link is called as the transmission angle.

10. Diffrents in mechanism and machine:


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Mechanism Machine

1. Mechanism transmits and modifies

motion.

2. It is the outline of the m/c to

produce definite motion between

various links.

1.M/C transmits forces and couples.

2. Machine may have many mechanisms for

transmitting power or work.

11. The Grashof's law for a four bar mechanism.

Grashof's law states that the sum of the shortest and longest links cannot be

greater than the sum of the remaining two link lengths, if there is to be continuous

relative motion between the two members.

The significance of Grashof's law:

(i) Grashof's law specifies the order in which the links are connected in a Kinematic

chain.

(ii) It specifies, which link of the four bar chain is fixed.

(iii) A four bar chain should satisfy Grashaf's law, otherwise no link will make a

complete revolution relative to one another.

i.e. ( S + l) * (P + Q)

where

S = length of the shortest link


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L = Length of the longest link

P & Q = Length of the other two links.

12. Differentiate between Movability and Mobility

(i) Movability includes six degrees of freedom of the devices as a whole, as if the

ground link were not fixed and thus applies to a Kinematic chain.

(ii) Mobility neglects these and considers only the internal relative motions, thus

applying to a mechanism.

13. Classify the 'Constrained motion', and example for each.

1.Completely constrained motion [Ex: Square bar moving in square

hole]

2.Incompletely constrained motion [Ex:circular shaft in a hole]

3.Successfully constrained motion [Ex: Motion of IC engine value]

14. Explain the The Grubler's criterion for plane mechanisms.

n = 3(l 1) 2j

for single degree of freedom joints => 3l 2j = 4


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where:

l = No of links

j = No of binary joints.

15. In a four bar chain, which link is called crank.

In a four bar chain, the smallest link, which makes complete revolution is called

crank.

2 Marks Questions: (UNIT 1)

1. What is Kinematics?Kinematics is the study of motion (position, velocity, acceleration). A major goal ofUnderstanding inematics is to develop the a!ility to design a system that "ill satisfy#pecified motion re$uirements. %his "ill !e the emphasis of this class.

&. What is Kinetics?Kinetics is the study of effect of forces on moving !odies. 'ood inematic design should

produce good inetics.. efine *in.

A lin is defined as a mem!er or a com!ination of mem!ers of a mechanism connecting

other mem!ers and having relative motion !et"een them. %he lin may consist of one or

more resistant !odies. A lin may !e called as inematic lin or element. +g

-eciprocating steam engine.. efine Kinematic /air.

Kinematic pair is a joint of t"o lins having relative motion !et"een them. %he types of

inematic pair are classified according to 0ature of contact (lo"er pair, higher pair)

0ature of mechanical contact (losed pair, unclosed pair)

0ature of relative motion (#liding pair, turning pair, rolling pair, scre" pair,

spherical pair)2. efine Kinematic hain

When the inematic pairs are coupled in such a "ay that the last lin is joined to the first

lin to transmit definite motion it is called a inematic chain.%he cran shaft of an engine forms a inematic pair "ith the !earings "hich are fi3ed in

a pair, the connecting rod "ith the cran forms a second inematic pair, the piston "ith

the connecting rod forms a third pair and the piston "ith the cylinder forms the fourth

pair. %he total com!ination of these lins is a inematic chain. +g *a"n mo"er.


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4. efine egrees of 5reedom.6t is defined as the num!er of input parameters "hich must !e independently controlled

in order to !ring the mechanism in to useful engineering purposes.6t is also defined as the num!er of independent relative motions, !oth translational and

rotational, a pair can have.

7. efine /antograph./antograph is used to copy the curves in reduced or enlarged scales. 8ence this

mechanism finds its use in copying devices such as engraving or profiling machines.9. What is meant !y spatial mechanism?

#patial mechanism have a geometric characteristics in that all revolute a3es are parallel

and perpendicular to the plane of motion and all prism lie in the plane of motion.:. lassify the onstrained motion?

onstrained motions are classified into three types

1. ompletely constrained motion.&. 6ncompletely constrained motion.

. #uccessfully constrained motion.1;. What is %oggle position?

6t is the position of a mechanism at "hich the mechanical advantage is infinite and the

sine of angle !et"een the coupler and driving lin is


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UNIT 2

TUTORIAL

1. In a revolving stage with a speed of 3 rpm, a person is walking with a speed of 0.5

m/s along a radial path. Determine the magnitude of the coriolis component of

acceleration in this motion. NOV 2003

Solution :

Coriolis acceleration = 2 V

= 2 x (0.5)

60

32

= 0.3141 rad/sec.

2. In a four bar chain mechanism crank AB = 0.4m rotates with 150 rpm and lever CD

oscillates with a linear velocity of 7 m/s and what is the mechanical advantage of thesystem.

Solution :

Mechanical advantage =&'

AB

(

(

VAB= ABAB

=60

)2 AB

AB


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=

( )

60

1502

0.4

=6.2 m/s

VDC= 7 m/s

Mech. Advantage =!

26

= 0.898.

3. In a four bar !ain A"#$% A$ is &'e an is 12 ** +on, an rotates at

1 r-*. #+okise !i+e t!e +ink #$ / 0**% si++ates about $% "# an

A$ are of eua+ +en,t!. in t!e an,u+ar 4e+oit5 of +ink #$ !en an,+e

"A$ / 0.

Solution:

(Relative velocity method)

Given Data: AD = 120mm; AB = 30mm;

D = !0mm;"BA = 100 #.$.m.;

% BA =

60

1002

60

)2 AB

= 10.&' #adec

(a) on*+u#ation Dia+#am: (,) -elocity Dia+#am

Scale : 1 cm = 20mm Scale: 0.0!2 m = 1 cm

-,a = /BA AB = 1.0&' 0.03

-,a = 0.31&1


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#ocedu#e:

(a) on*+u#ation dia+#am: Ree# i+(a)

1. /ith uita,le cale d#a4 AD and AB 4ith an an+le !00.

2. #om $oint B d#a4 an a#c o 120 mm #adiu.

3. #om $oint D d#a4 anothe# a#c o #adiu !0mm.

&. 5a6e the inte#ectin+ $oint a 7 and 8oinB'

and'&

7 a ho4n in i+

(,) -elocity dia+#am: Ree# i+ (,)

/ith uita,le cale d#a4 -BA (i.e.7a*

) $e#$endicula# to AB at any $oint.

Since the lin6A&

i *ed 4hoe #elative velocity i 9e#o hence a and d a#e lie on

the ame $oint.

D#a4 line $e#$endicula# to'&

andB'

#om $oint d and , #e$ectively o a

to +et the inte#ection $oint.

ocate the inte#ectin+ $oint a 7 a ho4n in i+.(,)

#om i+.(,)7

-BA = a, on vecto# velocity dia+#am7

-B = ,c on vecto# velocity dia+#am and

-D = cd (o#) ca on vecto# velocity dia+#am

Reult: By meau#ement7 4e *nd that

-D = 0.23 m

/D =060

23+!0

'&

23+!0=


42/75

= & #adec (cloc64ie a,out D) An.

6. In a four% +ink *e!anis*% t!e rank A" rotates at 30 ro7se. T!e

+en,t! of t!e +inks are A" / 2 **% "# / 6**% #$ / 68 ** an A$ /

0 **. A$ is t!e &'e +ink. At t!e instant !en A" is at ri,!t an,+e to

A$. $eter*ine t!e 4e+oit5 of :

(i) T!e *i9-oint of +ink "#.

(ii) A -oint on +ink #$% 1** fro* t!e -oints onnetin, t!e +inks

#$ an A$.

Solution :

Given Data:

/BA = 3! #adec and 0mm AD = !00mm

AD = ied.

(a) on*+u#ation dia+#am(,) -elocity dia+#am

Scale: 100 mm = 1 cm Scale: 1.&&m = 1 cm

-BA = /BA AB

= 3! 0.2 = '.2m

#ocedu#e:

(a) on*+u#ation dia+#am: Re i+.(a). D#a4 the $ace dia+#am a to the

dimenion.

(,) -elocity dia+#am: Ree# i+(,). D#a4 the velocity dia+#am a dicued in the

$#eviou


43/75

?am$le.

1. ocate the $oint e at the cent#e o vecto# ,c to *nd velocity o mid $oint o

lin6 B.

2. @oin e and a. "o4 the vecto# ea 4ill +ive the velocity o the mid $oint ?.

i.e.7 -?A =ea

= (&.>> 1.&&) = !.>>2 m

5o *nd out the velocity o $oint on'&

cd

df

(

(

&'

&,=

&'&,(( cddf =

By meau#ement7

-D = &.! 1.&& = !.!2& mec #om i+.(,)

%-d = !.!2&

5

= 1.&'2 mec An.

Reult:

1. -elocity o mid $oint o lin6 B 4ith #e$ect to A i

-?A = !.>>2 m An.

2. -elocity o a $oint on a lin6 D 100 mm #om D i

-A = -D = 1.&'2 m An.

8. In a s+ier rank *e!anis*% t!e +en,t! of rank " an onnetin,

ro A" are 128** an 18** res-eti4e+5. T!e entre of ,ra4it5 G of t!e

onnetin, ro is 28** fro* t!e s+ier A. T!e rank s-ee is 0 r-*

+okise. ;!en t!e rank !as turne 68 fro* t!e inner ea entre

-osition% eter*ine:

1.


44/75

3. An,u+ar 4e+oit5 of t!e onnetin, ro A".

Solution :

Given Data:

CB = 12>mm

AB = >00mm

AG = 2'>mm.

"BC = !00 #$m( % )

%BC =60

6002

= !2.3 #

= &> #om inne# dead cent#e

-BC = % BC CB = !2.3 0.12> = '.>& m

(a) on*+u#ation dia+#am: Scale : !2.> mm = 1 cm

(,) -elocity dia+#am: Scale: 1.! m = 1 cm

AB

A-

(

(

a*

a- =

500

2!59631+52(a- =

-aG = 3.0' m


45/75

#ocedu#e:

(a) on*+u#ation dia+#am: Re i+ (a).

i#t d#a4 line o t#o6e.

At any $oint on the line o t#o6e d#a4 a c#an6 CB o len+th 12> mm and = &>

#om $oint B d#a4 an a#c o #adiu >00 to cut the t#o6e7 then the cuttin+ $oint ,e A.

ocate G on A, at a ditance AG = 2'> mm.

(,) -elocity dia+#am: Re i+.(,).

D#a4o*

vecto# $e#$endicula# to CB 4ith uita,le cale

#om C d#a4 an ho#i9ontal vecto# $a#allel to the line o t#o6e.

#om $oint , d#a4 a line $e#$endicula# to AB7 it 4ill inte#ect the line o t#o6e CA

at a.

#om velocity dia+#am it i unde#tood that

-CA =oa

= 3.3 1.! = !.&!m

-BA =a*

= 2.> 1.! = >.>! m

5o locate the $oint + on vecto# velocity dia+#am7 ta6e -a+ = 3.0' m

(al#eady calculated) and 8oin the $oint o and + 4hich 4ill +ive the velocity 4ith

#e$ect to o.

-GC = 3.&> 1.! = !.'' m

Reult: -elocity o the lide# A = -CA = !.&! m An.

-elocity o the +#avity G = -CG = -GC = !.'' m An.

An+ula# velocity o the connectin+ #od AB i +iven ,y7

%AB =50

5+65

AB

(AB =

%AB = 11.1 #adec


46/75

0. In t!e *e!anis* as s!on in i,.% t!e rank rotates at 2 r-*

anti+okise an ,i4es *otion to t!e s+iin, b+oks " an $. T!e

i*ensions of 4arious +inks are oA / 3 **= A" / 12 **= "# / 68

** an #$ / 68 **. or t!e ,i4en on&,uration% eter*ine.

(i) 0 mm; D = &>0 mm;

% CA =60

202

= 2.0& #adec;

-CA = 0.!23 m.

#ocedu#e:

(a) on*+u#ation dia+#am: Ree# i+.(a). D#a4 the con*+u#ation dia+#am a $e#

the +iven data.

(,) -elocity dia+#am: Ree# i+:(,);

1. D#a4oa

vecto# #om $oint o in the di#ection $e#$endicula# to the c#an6 CA.

2. #om o d#a4 a line $a#allel to the line o action o the lide# B


47/75

3. #om a d#a4 a line $e#$endicula# to the lin6AB

to com$lete vecto# t#ian+le

oa,.

&. #om $oint o d#a4 a line $a#allel to line o action o lide# D.

>. 5o locate $oint on a,

AB

A'

AB

AB =

1200

!505530

AB

A'ABA' ==

%A = a, cale = &.& 0.12>! = 0.>>3 m

%A = 0.3&> m

!. #om the $oint c d#a4 a line to4a#d d $e#$endicula# to the lin6'&

.

Reult:

(i) -elocity i lide# at B:

-B = -ecto# o, = 3.2 0.12>! = 0.&01 m An.

And -elocity o lide# at D:

-D = 1. 0.12>! = 0.23 m An.

(ii) An+ula# velocity o D:"50

1256092

'&

'&'&

=

=


48/75

%D =

sec.rad+090"50

36"0=

. In a *e!anis* s!on in i,. T!e rank A is 1 ** +on, an

rotates +okise about at 12 r-*. T!e onnetin, ro A" is 6 **

+on,. At a -oint # on A"% 18 ** fro* A% t!e ro #E 38 ** +on, is

atta!e. T!is ro #E s+ies in a s+ot in a trunion at $. T!e en E is

onnete b5 a +ink E% 3 ** +on, to t!e !ori>onta++5 *o4in, s+ier .

or t!e *e!anis* in t!e -osition s!on% in

1.


49/75

%CA =60

)2 %A

%CA =60

1202

= 12.>!!& #adec

%CA = %CA F CA = 12.>!! F 0.1 = 1.2>!! m

#ocedu#e:

(a) on*+u#ation dia+#am: Ree# i+ a.

(,) -elocity dia+#am: Ree# i+ ,.

D#a4 -CA = 1.2>! m7 #om 0 to a in the di#ection $e#$endicula# to the c#an6 CA.

#om $oint o d#a4 a line $a#allel to line o t#o6e o the lide# B.

#om $oint a d#a4 a line $e#$endicula# to lin6 AB. "ame the $oint o inte#ection

a ,.

5o locate on a,.

BA

B'

*a

*c

BA

B' ==

-B = ,c = 1.!'> F 0.2>2

-B = 0.&2> m.

#om c d#a4 a line $e#$endicula# to the lin6 ?.

#om $oint d#a4 a line $a#allel to the lin6 ?.

5o locate $oint e:


50/75

G

120

'&

'E

cd

ce

'&

'E ==

-? = ce = cd F250

350

= 0.31> F250

350

-? = 0.&&1 m.

= cd F cale

#om $oint e7 d#a4 a line $e#$endicula# to the lin6 ? and d#a4 a line $a#allel

to the line o t#o6e at C and locate the cuttin+ $oint a .

Reult:

1. -elocity o :

- = -ecto# o = 2.1 0.2>2

= 0.>2 m An.

o = 2.1 cm;Scale = 0.2>2

2. -elocity o lidin+ o ? in the t#unnion:

-? = -ecto# ec = 0.&&1 m An

3. An+ula# velocity o ?:

%? =350

""10

'E

'E =

= 1.2! #adec

?. An en,ine *e!anis* s!on in i,. T!e rank #" /1 ** an t!e

onnetin, ro "A/3** it! entre of ,ra4it5 G% 1** fro* ". In t!e

-osition s!on t!e ranks!aft !as a s-ee of 8 ra7s an an an,u+ar

ae+eration of 12 ra7s2. &n: 1: 4e+oit5 of G an an,u+ar 4e+oit5 of

A"% an 2 ae+eration of G an an,u+ar ae+eration of A".


51/75


52/75

1. B ( ) Ans.

;.

ABAB

Vrad s Clockwise

BA = = =

(a) S$ace dia+#am. (,) -elocity

dia+#am.

2. Accele#ation o G and an+ula# accele#ation o AB

/e 6no4 that #adial com$onent o the accele#ation o B 4ith #e$ect to 7

& &&(7.2) 24&.2 B

;.1

r BC

BC

va m s

CB= = =

and #adial com$onent o the accele#ation o A 4ith #e$ect to B7

& &&(7.2) 2. B

;.

r BC

BC

va m s

BA= = =

"o4 the accele#ation dia+#am7 a ho4n in i+ H i d#a4n.


53/75

D#a4 vecto# cI ,J $a#allel to B7 to ome uita,le cale7 to #e$#eent the #adial

com$onent o the accele#ation o B 4ith #e$ect to . i.e aIB7 uch that -ecto# cI

,J = a #B=>!2.>m2

2. #om $oint ,J 7d#a4 vecto# ,J,I $e#$endicula# to vecto# cI ,J o# B to #e$#eent

the tan+ential com$onent o the accele#ation o B 4ith #e$ect to i.e a1B7 uchthat

-ecto# ,J ,I = aIB=120m2 K(Given)

3. @oin cI,I. the vecto# cI,I #e$#eent the total accele#ation o B 4ith #e$ect to i.e

aB accele#ation o A 4ith #e$ect to i.e. a1B.

&. #om $oint ,I7 d#a4 vecto# L $a#allel to BA to #e$#eent #adial com$onent o the

accele#ation o A 4ith #e$ect to B i.e a#AB uch that

-ecto# ,J I = a #B=>3.3m2

>. #om $oint 7 d#a4 vecto# I $e#$endicula# to vecto# ,I o# BA to #e$#eent

tan+ential com$onent o the accele#ation o A 4ith #e$ect to B i.e a#AB. /hoe

ma+nitude i yet 6no4n.

!. "o4 d#a4 vecto# cI aI $a#allel to the $ath o motion o A (4hich i alon+ A) to

#e$#eent the accele#ation o A i.e aA. 5he vecto# aI and cIaI inte#ect at aI @oin ,I

aI. 5he vecto# ,I aI #e$#eent the accele#ation o A 4ith #e$ect to B i.e aAB.

'. n o#de# to *nd the accele#ation o G7 divide vecto# aI ,I in +I in the ame #atio a

G divide BA in i+ . (a) . @oin cI+I. 5he vecto# cI+I #e$#eent the accele#ation o G.

By meau#ement7 4e *nd that accele#ation o G7

&c@ g@1 mBs Ans.Ga vector =

#om accele#ation dia+#am7 4e *nd that tan+ential com$onent o the

accele#ation o A 4ith #e$ect toB.

AIAB=vecto# aI =>&! m

.. (By meau#ement )

An+ula# accele#ation o AB7

&@ 2419&; B ( ) .

;.

ABAB

arad s Clockwise Ans

BA = = =


54/75

1. In t!e *e!anis* s!on in &,% t!e s+ier # is *o4in, to t!e ri,!t it!

a 4e+oit5 o 1 *7s an ae+eration of 2.8 *7s2. T!e i*ensions of 4arious

+inks are A"/ 3* in+ine at 68 it! t!e !ori>onta+ $eter*ine: 1. t!e

*a,nitue of 4ertia+ an !ori>onta+ o*-onent of t!e ae+eration of t!e

-oint "% an 2. t!e an,u+ar ae+eration of t!e +inks A" an "#.

Solution: Given: v=1m. ac=2.>m2; AB=3m; B=1.>m

i#t o all7 d#a4 the $ace dia+#am7 a ho4n in *+. (a)7to ome uita,le

cale. "o4 the velocity dia+#am7 a ho4n in *+. (,) 7 i d#a4n a dicued ,elo4:

Since the $oint A and D a#e *ed $oint7 the#eo#e they lie at one $lace in the

velocity dia+#am. D#a4 vecto# dc $a#allel to Dc7 to ome uita,le cale7 4hich

#e$#eent the velocity o lide# 4ith #e$ect to D o# im$ly velocity o 7 uchthat

-ecto# dc=vD=v=1 m.

Since $oint B ha t4o motion7 one 4ith #e$ect to A and the othe# 4ith #e$ect to 7

the#eo#e #om $oint a7 d#a4 vecto# a, $e#$endicula# to AB to #e$#eent the velocity

o B 4ith #e$ect to A7 i.e.vBA and #om $oint c d#a4 vecto# c, $e#$endicula# to B

to #e$#eent the velocity o B 4ith #e$ect to i.evB. 5he vecto# a, and c,

inte#ect at ,.


55/75

By meau#ement 4e *nd that velocity o B 4ith #e$ect to A.

-BA= vecto# a, =0.'2 m and velocity o B 4ith #e$ect to 7

c!;.7&mBsBCv vector =

/e 6no4 that #adial com$onent o accele#ation o B 4ith #e$ect to 7

& &r &(;.7&)a ;.4 B

1.2

BC

BC

vm s

CB= = =

and #adial com$onent o accele#ation o B 4ith #e$ect to A.

& &&(;.7&) ;.17 B

r BA

BA

va m s

AB

= = =

"o4 the accele#ation dia+#am7 a ho4n in i+ H7 i d#a4n a dicued ,elo4:

1.M Since the $oint A and D a#e *ed $oint7 the#eo#e they lie at one $lace in the

accele#ation dia+#am. D#a4 vecto# dIcI $a#allel to D7 to ome uita,le cale to

#e$#eent the accele#ation o 4ith #e$ect to D o# im$ly accele#ation o i.e aD

o# a uch that

vecto# dIcI=aD=ac=2.> m2

2. 5he accele#ation o B 4ith #e$ect to 4ill have com$onent7 i.e. one #adialcom$onent o B 4ith #e$ect to (a#B) and the othe# tan+ential com$onent o B

4ith #e$ect to (aIB). 5he#eo#e #om $oint LcI7 d#a4 vecto# cI $a#allel to B to

#e$#eent a#B= 0.3&!m2

3. "o4 #om $oint 7 d#a4 vecto# ,I $e#$endicula# to vecto# cI o# B to #e$#eent

atBA). /hoe ma+nitude i yet un6no4n.


56/75

&.5he accele#ation o B 4ith #e$ect to A 4ill alo have t4o com$onent o B 4ith

#e$ect to A (a#BA) and othe# tan+ential com$onent o B 4ith #e$ect to A (atBA).

5he#eo#e #om $oint aI d#a4 vecto# aIy $a#allel to AB to #e$#eent atBA7 uch that

vecto# aIy =a#BA=0.1'3 m2

#om $oint y7 d#a4 vecto# y,I $e#$endicula# to vecto# aIy o# AB to #e$#eent atBA.

5he vecto# y,I inte#et the vecto# ,Iat ,I.@oin aI ,I and cI ,I. 5he vecto# aI,I

#e$#eent the accele#ation o $oint B (aB) and the vecto# cI ,I #e$#eent the

accele#ation o B 4ith #e$ect to .

1. Ea+nitude o ve#tical and ho#i9ontal com$onent o the accele#ation o the $oint B

D#a4 ,I ,J i the ho#i9ontal com$onent o the accele#ation o the $oint B. By

meau#ement7

-ecto# ,I ,J = 1.13 m2 and vecto# aI ,J = 0. m2 An.

2. An+ula# accele#ation o AB and B

By meau#ement #om accele#ation dia+#am7 4e *nd that tan+ential

com$onent o accele#ation o the $oint B. 4ith #e$ect to A.

atBA = vecto# y,I = 1.&1 m2

and tan+ential com$onent o accele#ation o the $oint B 4ith #e$ect to 7

atB= vecto# ,I = 1.& m2

/e 6no4 that an+ula# accele#ation o AB7

&1.1 ;.7 B Ans.

t

A

AB

arad s

AB = = =B

and an+ula# accele#ation o B7

&1.: 1. B Ans.1.2

t

C

BC

arad s

BC = = =B

11. @QS is a four bar !ain it! +ink @S &'e. T!e +en,t!s of t!e +ink are

@Q /02.8 **= Q/18**= S/112.8 **= an @S /2**. T!e rank @Q

rotates at 1 ra7s +okise. $ra t!e 4e+oit5 an ae+eration ia,ra*

!en an,+e Q@S/0 ab Q ab +ie on t!e sa*e sie of @S. &n t!e

an,u+ar 4e+oit5 an an,u+ar ae+eration of +inks Q an S.

Solution: Given N= 10 #ad; N= !2> mm = 0.0!2>m; NR=1'>mm=0.1'>m;

RS=112.>mm= 0.112>m; S=200=0.2m


57/75

/e 6no4 that velocity o N 4ith #e$ect to o# velocity o N.

-N=vN=N N = 10 0.0!2> = 0.!2> m.

(e#$endicula# to N)

An+ula# velocity o lin6 (NR and R)

i#t o all7 d#a4 the $ace dia+#am o a ou# ,a# chain7 to ome uita,le

cale7 a ho4n in i+ (a)."o4 the velocity dia+#am a ho4n in *+ (,)7 i d#a4n a

dicued ,elo4:

1. Since and S a#e *ed $oint7 the#eo#e thee $oint lie at one $lace in velocity

dia+#am. D#a4 vecto# $O $e#$endicula# to N7 to ome uita,le cale7 to #e$#eentthe velocity o N 4ith #e$ect to o# velocity o N i.e.vN o# -N uch that

vecto# $O= vN =vN =0.!2> m.

2.#om $oint O7 d#a4 vecto# O# $e#$endicula# to NR to #e$#eent the velocity o R

4ith #e$ect to N (i.e.vRN) and #om $oint 7 d#a4 vecto# # $e#$endicula# to SR to

#e$#eent that velocity o R 4ith #e$ect to S o# velocity o R (i.e vRS o# vR). 5he

vecto# O# and # inte#ect at #. By meau#ement 74e *nd that.

-RN=vecto# O# = 0.333m7 and vRS=vR = vecto# #=0.&2! m.

/e 6no4 that an+ula# velocity o lin6 NR.

NR=

;.1.: B (Anticloc"ise) Ans.

;.172

RQv

rad sQR

= =


58/75

and an+ula# velocity o lin6 RS7

RS=

;.&4.79 B (cloc"ise) Ans.

;.11&2

RSv rad sSR

= =

An+ula# accele#ation o lin6 NR and R.

Since the an+ula# accele#ation o the c#an6 N i not +iven7 the#eo#e the#e

4ill ,e no tan+ential com$onent o the accele#ation o N 4ith #e$ect to .

/e 6no4 that #adial com$onent o the accele#ation o N 4ith #e$ect to (o#

the accele#ation o

& &&(;.4&2) 4.&2 B

;.;4&2

QPr

QP QP Q

va a a m s

PQ= = = = =

Radial com$onent o the accele#ation o R 4ith #e$ect to N.

& &&(;.) ;.4 B

;.172

RQr

RQ

va m s

QR= = =

And #adial com$onent o the accele#ation o R 4ith #e$ect to S (o# the accele#ation

o R).

& &&(;.&4) 1.41 B

;.11&2

r RS

RS RS R

va a a m s

SR= = = = =

5he accele#ation dia+#am 7 a ho4n in i+ H i d#a4n a ollo4:

Since and S a#e *ed $oint7 the#eo#e thee $oint lie at one $lace in the

accele#ation dia+#am. D#a4 vecto# $IOI $a#allel to N7 to ome uita,le cale7 to

#e$#eent the #adial com$onent o accele#ation o N 4ith #e$ect to o#

accele#ation o N i.e a#N o# aN uch that

-ecto# $IOI=a#N = aN =!.2> m2

2. #om $oint OI d#a4 vecto# OI $a#allel to NR to #e$#eent the #adial com$onent o

accele#ation o R 4ith #e$ect to N i.e. a#RN uch that

-ecto# OI = a#RN = 0.!3& m2

3. #om $oint 7 d#a4 vecto# #I $e#$endicula# to NR to #e$#eent the tan+ential

com$onent o accele#ation o R 4ith #e$ect to N i.e. atR 4hoe ma+nitude i not

yet 6no4n.


59/75

&. "o4 #om $ointI d#a4 vecto# Iy $a#allel to SR to #e$#eent the #adial com$onent

o the accele#ation o R 4ith #e$ect to S i.e. a#RS. Such that

-ecto# Iy =a#RS = 1.!13 m2

>. #om $oint y7 d#a4 vecto# y#I $e#$endicula# to SR to #e$#eent the tan+ential

com$onent o accele#ation o R 4ith #e$ect to S i.e atRS.

!. 5he vecto# #I and y#I inte#ect at #I.@oin $I# and OI #I. By meau#ement 4e *nd

that

& t &

-#@ .1 B a @ 2. Bt

RQa vectorxr m s and vectoryr m s= = = =

/e 6no4 that an+ula# accele#ation o lin6 NR7

&.1 &. radBs ( ) .;.172

t

RQ

QR

aAnticlockwise Ans

QR = = =

and an+ula# accele#ation o lin6 RS.

&2. 7.1 radBs ( )Ans.;.11&2

t

RS

RS

aAnticlockwise

SR = = =

12. T!e i*ension an on&,uration of t!e four bar *e!anis*% s!on in

i,% are as fo++os:

@1A / 3**= @2"/30 **= A" / 30 **% an @1@1/0 **.


60/75

5he an+le A12= !0. 5he c#an6 1A ha an an+ula# velocity o10 #ad andan an+ula# accele#ation o 30 #ad27 ,oth cloc64ie. Dete#mine the an+ula#

velocitie and an+ula# accele#ation o 2B7 and AB and the velocity and

accele#ation o the 2,7 and the velocity and accele#ation o the 8oint B.

Solution: Given:A1=10 #ad; A1=30 #ad7 1A= 30mm= 0.3 m; 2B=AB=3!0

mm= 0.3!m

/e 6no4 that the velocity o A 4ith #e$ect to 1 o# velocity o A7

-A1=vA=A1 $1A= 10 0.3 = 3 m.

-elocity o B and an+ula# velocitie o 2B and AB

i#t o all7 d#a4 the $ace dia+#am7 to ome uita,le cale7 a ho4n in *+.

(a)."o4 the velocity dia+#am7 a ho4n in *+ (,)7 i d#a4n a dicued ,elo4.

Since 1 and 2 a#e *ed $oint7 the#eo#e thee $oint lie at one $lace in velocity

dia+#am. D#a4 vecto# $1 $e#$endicula# to 1A7 to ome uita,le cale to

#e$#eent the velocity o A 4ith #e$ect to 1 o# velocity o Ai.e. vA1=vA=3m

#om $oint a7 d#a4 vecto# a, $e#$endicula# to AB to #e$#eent velocity o B 4ith

#e$ect to A (i.e.vBA) and #om $oint $2 d#a4 vecto# $2, $e#$endicula# to 2B to

#e$#eent the velocity o B 4ith #e$ect to 2o# velocity o B i.e vB2 o# vB. 5he

vecto# a, and $2, inte#ect at ,.

By meau#ement 74e *nd that

-B2=vB= vecto# $2,=2.2 m An.

And vBA=vecto#a,= 2.0>m

/e 6no4 that an+ula# velocity o 2B7

&&

&

&.&4.1 radBs (cloc"ise) Ans.;.4

BPp B

v

P B = = =

and an+ula# velocity o AB7

&.;22.7radBs (cloc"ise) Ans.

;.4

BAAB

v

BA = = =


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Accele#ation o B and an+ula# accele#ation o 2B and AB

/e 6no4 that tan+ential com$onent o the accele#ation o A 4ith #e$ect to 17

&

1 1 1; ;. : Bt

AP AP Aa P m s= = =

Radial com$onent o the accele#ation o A 4ith #e$ect to 17

&& & &1

1 1 1

1

1; ;. ; Br APAP AP

va P A x m s

P A= = = =

Radial com$onent o the accele#ation o B 4ith #e$ect to A

& &&(&.;2) 11.47 B

;.4

r BA

BA

va m s

AB= = =

and #adial com$onent o the accele#ation o B 4ith #e$ect to 2I

& &&&

&

&

(&.&)1. B

;.4

r BPBP

va m s

P B= = =

5he accele#ation dia+#am a ho4n in *+ H7 i d#a4n a ollo4.

1. Since 1 and 2 a#e *ed $oint7 the#eo#e thee $oint 4ill lie at one $lace in the

accele#ation dia+#am. D#a4 vecto# $1I $a#allel to 1A7 to ome uita,le cale7 to#e$#eent the #adial com$onent o the accele#ation o A 4ith #e$ect to 17 uch

that.

-ecto# $I1=a#A1 = 30 m2


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2. #om $oint 7 d#a4 vecto# aI $e#$endicula# to $1Ato #e$#eent the tan+ential

com$onent o the accele#ation o A 4ith #e$ect to 17 uch that vecto#

aI=atA1= m2

3. @oin 1I aI. the vecto# $1 LaI #e$#eent the accele#ation o A. By meau#ement7

4e *nd that the accele#ation o A.

aA=aA1=31.! m2

&. #om $oint aI7 d#a4 vecto# aI y $a#allel to AB to #e$#eent the #adial com$onent o

the accele#ation o B 4ith #e$ect to A7 uch that

vecto# aIy=aIBA=11.!' m2

#om oint y7 d#a4 vecto# y,I $e#$endicula# to AB to #e$#eent the tan+ential

com$onent o the accele#ation o B 4ith #e$ect to A(i.e. atBA) 4hoe ma+nitude iyet un6no4n.

"o4 #om $oint 2I d#a4 vecto# $29 $a#allel to $2B to #e$#eent the #adial

com$onent o the accele#ation B 4ith #e$ect to 27 uch that

-ecto# $2I9 =a#B2 =13.&&m2

#om $oint 97 d#a4 vecto# 9,I $e#$endicula# to 2B to #e$#eent the tan+ential

com$onent o the accele#ation o B 4ith #e$ect to 2i.e atB2.

5he vecto# y,I and 9,I inte#ect at ,I. "o4 the vecto# $I2,I #e$#eent the

accele#ation o B 4ith #e$ect to 2 o# the accele#ation o B i.e aB2 o# aB. Bymeau#ement7 4e *nd that

AB2=aB=vecto# $2I,I= 2.! m2 An.

Alo vecto# y,I =atBA=13.!m27 and vecto# 9,I=atB2= 2!.!m2

/e 6no4 that an+ula# accele#ation o 2B7

&&&

&

&4.47.9 B ( ) .

;.4

t

BPP B

arad s Anticlockwise Ans

P B = = =

and an+ula# accele#ation o AB=

&1.4 7.9 B ( ) .;.4

t

BA

AB rad s Anticlockwise Ans

AB

= = =


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13. In t!e *e!anis* as s!on in &, % t!e rank A rotates at 2 r.-.*.

anti+okise an ,i4es *otion to t!e s+iin, b+oks " an $. T!e

i*ensions of t!e 4arious +inks are oA/3**= A"/ 12**= "#/ 68

** an #$ / 68*.

o# the +iven con*+u#ation 7 dete#mine :1. velocitie o lidin+ at B and D7 2 an+ula#

velocity o D7 3.linea# accele#ation o D7 and &7 an+ula# accele#ation o D.

Solution: Given "AC=20 #.$.m. o# AC=2 20!0=2.1 #ad; CA=300 mm = 0.3m;

AB=1200 mm = 1.2 m; B=D=&>0 mm =0.&>m

/e 6no4 that linea# velocity o a 4ith #e$ect to C o# velocity o A7

-AC =vA=AC CA = 2.1 0.3 = 0.!3 m (e#$endicula# to CA)

1.-elocite o lidin+ at B and D.

i#t o all7 d#a4 the $ace dia+#am 7to ome uita,le cale7 a ho4in+ *+ (a)

ne4 the velocity dia+#am7 a o4n in i+ (,)7 i d#a4n a dicued ,elo4.


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1. D#a4 vecto# on $e#$endicula#

to CA7 to ome uita,le cale7 to

#e$#eent the velocity o A 4ith

#e$ect to C (o# im$ly velocity o

A)7 uch that. -ecto#

oa =vAC=vA= 0.!3

m

2. #om $oint a7 d#a4 vecto# a,

$e#$endicula# to AB to #e$#eent

the velocity o B 4ith #e$ect to A

(i.e vBA) and #om $oint 0

d#a4 vecto# a, $a#allel to $ath

o motion , (4hich i alon+ BC) to #e$#eent the velocity o B 4ith #e$ect to C (o#

im$ly velocity o B). 5he vecto# a, and a, inte#ect at ,.

3. Divide vecto# a, at c in the ame #atio a divide AB in the $ace dia+#am .n

othe# 4o#d.

BA = ,cca

&. "o4 #om $oint c7 d#a4 vecto# D to #e$#eent the velocity o D 4ith #e$ect to

(i.e. vD) and #om $oint a d#a4 vecto# ad $a#allel to the $ath o motion o d (4hich

alon+ the ve#tical di#ection ) #o #e$#eent the velocity o lidin+ at B7

By meau#ement7 4e *nd that velocity o lidin+ at B7

-B=vecto# o, = 0.&m An.

And velocity o lidin+ at D. vD= vecto# od = 0.2& m An.

2. an+ula# velocity o D

By meau#ement #om velocity dia+#am7 4e *nd that velocity o D 4ith

#e$ect to 7

-D= vecto# cd= 0.3' m

An+ula# velocity o D.

;.7;.9& B ( ) Ans

;.2

DCCD

vrad s Anticlockwise

CD = = =


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3. inea# accele#ation o D

/e 6no4 that the #adial com$onent o the accele#ation o A 4ith #e$ect to 0 o#

accele#ation o A.

&& & &

(&.1) ;. 1.& Br

AO A AO

v AO

a a OA m sOA = = = = =

Radial com$onent o the accele#ation o B 4ith #e$ect to A.

& &&(;.2) ;.& B

1.&

r BABA

va m s

AB= = =

K (BP meau#ement 7 vBA= 0.>& m.

Radial com$onent o the accele#ation o D 4ith #e$ect to 7

& &(;.7);.; B

;.2

r DCDC

va m s

CD= = =

"o4 the accele#ation dia+#am7 a ho4 in *+ (c)7 i d#a4n a dicued ,elo4 :

D#a4 vecto# oI aI $a#allel to CA to ome uita,le cale 7to #e$#eent the #adial

com$onent o the accele#ation o a 4ith #e$ect to o o# im$ly the accele#ation o

7uch that

-ecto# oIaI =aIAC=aA = 1.323 m2

2. #om $oint aI d#a4 vecto# a L $a#allel to AB to #e$#eent the #adial com$onent

oten accele#ation o B 4ith #e$ect to A7 uch that

-ecto# aI=a#BA=0.2&3 m2

3. #om $oint 7 d#a4 vecto# ,I $e#$endicula# to AB to #e$#eent the tan+ential

com$onent o the accele#ation o B 4ith #e$ect to A (i.e atBA) 4hoe ma+nitude i

not yet 6no4n.

&. #om $oint d#a4 vecto#22

oI,I $a#allel to the $ath o motion o B (4hich i alon+ BC) to #e$#eent the

accele#ation o B(aB). 5he vecto# ,I and oI,I inte#ect at ,I. @oin aI,I. the vecto#

aI,I #e$#eent the accele#ation o , 4ith #e$ect to A.

>. Divide vecto# aI ,I at cI in the ame #atio a divide AB in the $ace dia+#am.

n othe# 4o#d7


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BBA=,I cI,IaI

!. #om $oint I7 d#a4 vecto# cI y $a#allel to D to #e$#eent the #adial com$onent o

the accele#ation o D 4ith #e$ect to 7 uch that

vecto# cIy =a#D= 0.30& m2

'. #om $oint y7 d#a4 ydI $e#$endicula# to D to #e$#eent the tan+ential

com$onent o accele#ation o d 4ith #e$ect to (i.e.atD) 4hoe ma+nitude i not

yet 6no4n.

. #om $oint 0I d#a4 vecto# o oI dI $a#allel to the $ath o motion o D (4hich i

alon+ the ve#tical di#ection) to #e$#eent the accele#ation o D (aD). 5he vecto# ydI

and oIdI inte#ect at dI.

By meau#ement7 4e *nd that linea# accele#ation o D.

AD=vecto# 0IdI=0.1!m2 An.

&. An+ula# accele#ation o D

#om the accele#ation dia+#am 74e *nd that the tan+ential com$onent o the

accele#ation o D 4ith #e$ect to 7

AtD=vecto# ydI =1.2m2 ..(By meau#ement)

An+ula# accele#ation o D7

D=

&@ 1.&9 &.9 B ( ) Ans.

;.2

DCa

rad s clockwise

Cd

= =

UNIT 2 (BINEMATI#S)

ASSIGNMENT

2 Marks Questions:

1. What are the important concepts in velocity analysis?i. %he a!solute velocity of any point on a mechanism is the velocity of that point "ith

reference to ground.ii. -elative velocity descri!es ho" one point on a mechanism moves relative to another

point on the mechanism.&. efine 6nstantaneous centre and 6nstantaneous a3is.


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6nstantaneous centre of a moving !ody may !e defined as that centre "hich goes on

changing from one instant to another.6nstantaneous a3is is a line dra"n through an instantaneous centre and perpendicular to

the plane of motion.. 8o" to represent the direction of linear velocity of any point on a lin "ith respect to

another point on the same lin?%he direction of linear velocity of any point on a lin "ith respect to another point on the

same lin is perpendicular to the line joining the points.. efine Kennedys theorem.

%he Kennedys theorem states that if three !odies move relatively to each other, they

have three instantaneous centers and lie on a straight line.2. efine displacement.

6t may !e defined as the distance moved !y a !ody "ith respect to a fi3ed certain fi3ed

point. When there is no displacement in a !ody it is said to !e at rest and "hen it is

!eing displaced, it is said to !e in motion.4. What are the types of motions?1. -ectilinear motion.&. urvilinear motion.. ircular motion.

7. What are the methods for determining the velocity of a !ody?6mportant methods for determining the velocity of a !ody are1. 'raphical method i) -elative velocity method

ii) 6nstantaneous centre method&. Analytical method.

9. efine velocity and #peed

Celocity may !e defined as the rate of change of displacement of a !ody "ith respect to

the time. #ince the velocity has !oth magnitude and direction, therefore it is a vector

$uantity.

#peed may !e defined as the rate of change of linear displacement of a !ody "ith

respect to the time. #ince the speed is irrespective of its direction, therefore it is a scalar

$uantity.:. efine Acceleration and eceleration

%he rate of change of velocity "ith respect to time is no"n as acceleration.%he negative acceleration is also no"n as deceleration or retardation.

1;. efine coincident points.When a point on one lin is sliding along another rotating lin, then the point is no"n as

coincident point.11. efine centrode, A3ode and >ody centrode.

%he locus of all instantaneous centres (i.e., 61, 6&,D) is no"n as centrode.%he locus of all instantaneous a3is is no"n as a3ode.


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%he locus of all instantaneous centre relative to the !ody itself is called the !ody

centrode.

UNIT 4 - CAMS ASSIGNMENT

2 Marks Questions:

1. What is cam?A cam is a rotating machine element "hich gives reciprocating (or) oscillating motion to

another element no"n as follo"er&. efine tangent cam?

When the flans of the cam are straight and tangential to the !ase circle and nose circle,

the cam is no"n as tangent cam.. istinguish radial and cylindrical cams.

-adial cam ylindrical cam

6n this cam, the follo"er reciprocates (or) oscillates in a

direction perpendicular to the a3is.

6n this the follo"er reciprocates (or) oscillates in a

direction parallel to the cam a3is.

. What are the different motions of the follo"er? (i) Uniform motion, (ii) #imple harmonic motion, (iii) Uniform acceleration and retardation, and (iv) ycloidal motion.

2. ompare -oller and mushroom follo"er of a cam.


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#.0o -oller 5ollo"er Eushroom 5ollo"er

1. -oller follo"ers are e3tensively used "here more

space is availa!le.

%he mushroom follo"ers are generally used "here

space is limited.

&. 6t is used in stationary gas engines, oil engines and

aircraft valves in engines.

6t is used in cams "hich operate the valves in

automo!ile engines.

4. +3plain offset follo"er.When the motion of the follo"er is along an a3is a"ay from the a3is of the cam centre, it

is called offset follo"er.7. efine trace point in the study of cams.

6t is a reference point on the follo"er and is used to generate the pitch curve. 6n case of

nife edge follo"er the nife edge represents the trace point and the pitch curve

corresponds to the cam profile. 6n a roller follo"er the centre of the roller represents the

trace point.9. efine pressure angle "ith respect to cams.

6t is the angle !et"een the direction of the follo"er motion and a normal to the pitch

curve. %his angle is very important in designing a cam profile. 6f the pressure angle is too

large, a reciprocating follo"er "ill jam in its !earings.:. efine *ift (or) #troe in cam.

6t is the ma3imum travel of the follo"er from its lo"est position to the topmost position.1;. efine undercutting in cam. 8o" is occurs?

%he cam profile must !e continuous curve "ithout any loop. 6f the curvature of the pitch

curve is too sharp, then the part of the cam shape "ould !e lost and thereafter the

intended cam motion "ould not !e achieved. #uch a cam is said to !e undercut.Undercutting occurs in the cam !ecause of attempting to achieve too great a follo"er lift

"ith very small cam rotation "ith a smaller cam.11. What do you no" a!out 0omogram?

6n 0omogram, !y no"ing the values of total lift of the follo"er (*) and the cam rotation

angle (F) for each segment of the displacement diagram, "e can read directly the

ma3imum pressure angle occurring n the segment for a particular choice of prime circle

radius (-;).

1&.What are the classifications of cam !ased on the follo"er movement?1) -iseG-eturnG-ise (-G-G-) cams,

&) "ellG-iseG-eturnG"ell (G-G-G) cams,) "ellG-iseG"ellG-eturnG"ell (G-GG-G) cams,) "ellG-iseG"ell (G-G) cams.

1. What are the different types of cams?1. Wedge (or) flat cams&. -adial (or) isc cams. #piral cams. ylindrical (or) >arrel (or) rum ams


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2. onjugate cams4. 'lo!oidal cams7. #pherical cams

1.What do you no" a!out gravity cam?6n this type, the rise of the cam is achieved !y the rising surface of the cam and the

return !y the force of gravity of die to the "eight of the cam.12. efine %race point.

6t is a reference point on the follo"er to trace the cam profile. 6n case of a nife edge

follo"er, the nife edge itself is a tracing point and in roller follo"er, the centre of the

roller is the tracing point.14. efine pressure angle.

6t is the angle !et"een the direction of the follo"er motion and a normal to the pitch

curve. %his is very important in cam design as it represents steepness of the cam profile.

6f the pressure angle is too large, a reciprocating follo"er "ill jam in its !earings.17. efine /rime circle.

%he smallest circle dra"n tangent to the pitch curve is no"n as the prime circle.19. efine Angle of Ascent.

%he angle of rotation of cam from the position "hen the follo"er !egins to rise till it

reaches its highest position is no"n as angle of ascent. 6t is also no"n as out stroe

and is denoted !y H;.1:. What is meant !y #imple 8ormonic Eotion?

When a !ody rotates on a circular path "ith uniform angular velocity, its projection on the

diameter "ill have simple harmonic motion. %he velocity of the projection "ill !e

ma3imum at the centre of and


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UNIT 5 HIGHER AIRS! GEAR TRAINSASSIGNMENT

2 Marks Questions:

1. #tate la" of 'earing.

%he la" of gearing states that for o!taining a constant velocity ratio, at any instant of

teeth the common normal at each point of contact should al"ays pass through a pitch

point, situated on the line joining the centre of rotation of the pair of mating gears.


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&. efine normal and a3ial pitch in helical gears.

0ormal pitch is the distance !et"een similar face of adjacent teeth, along a heli3 on the

pitch cylinder normal to the teeth.

A3ial pitch is the distance measured parallel to the a3is !et"een similar faces of a

adjacent teeth.

. What is the ma3imum efficiency in "orm and "orm gear?

Ima3 1GsinJB1sinJ

. What are the advantages and limitations of gear drive? Write any t"o.

A"#anta$es:

1. #ince there is no slip, so e3act velocity ratio is o!tained.

&. 6t is more efficient and effective means of po"er transmission.

%i&itations:

1. Eanufacture of gear is complicated.

&. %he error in cutting teeth may cause vi!ration and noise during operation.

2. efine interference.

%he phenomenon "hen the tip of tooth undercuts the roots on its mating gear is no"n

as interference.

4. efine cycloidal tooth profile and involute tooth profile.

A cycloid is the curve traced !y a point on the circumference of a circle "hich rolls

"ithout slipping on a fi3ed straight line.

6nvolute profile is defined as the locus of a point on a straight line "hich rolls "ithout

slipping on the circumference of a circle.

7. efine circular pitch and diametral pitch in spur gears.


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ircular pitch (pc) 6t is the distance measured along the circumference of the pitch circle

from a point on one teeth to the corresponding point on the adjacent tooth.

pcLB%

iametral pitch (p) 6t is the ratio of num!er of teeth to the pitch circle diameter.

'(T)'(*)+,

9. efine >acslash.

6t is the difference !et"een the tooth space and the tooth thicness along the pitch

circle.

Ba,ksas. ( Toot. s+a,e Toot. t.i,kness

:. What is gear train of train of "heels?

%"o or more gears re made to mesh "ith each other to transmit po"er from one shaft to

another. #uch a com!ination is called a gear train or train of "heels.

1;. Write velocity ratio in compound train of "heels?

#peed of last follo"er G /roduct of teeth on drivers

#peed of first driver M /roduct of teeth on follo"ers.

11. efine simple gear train.

When there is only one gear on each shaft, it is no"n as simple gear train.

1&. What is reverted gear train?

When the a3es of the first and last "heels are coGa3ial, the train is no"n as reverted

gear train.

1. Where the epicyclic gear trains are used?

%he epicyclic gear trains are used in the !ac gear of lathe, differential gears of the

automo!iles, pulley !locs, "rist "atches, etc.

1. Write do"n the difference !et"een involute and cycloidal tooth profile.


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#.0o 6nvolute %ooth /rofile ycloidal %ooth /rofile

1. Cariation in centre distance does not affect the

velocity ratio.

%he centre distance should not vary.

&. /ressure angle remains constant throughout

the teeth.

/ressure angle varies. 6t is


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&) Alge!raic method.&2. What is the e3ternally applied tor$ues used to eep the gear train in e$uili!rium?

1) 6mpart tor$ue on the driving mem!er.&) -esisting or holding tor$ue on the driven mem!er.) 8olding or !raing tor$ue on the fi3ed mem!er.

Kom Tutorials & Assignments Unit 1 & 2

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