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Theory of machines _ Khurmi
Solution of Exercises .
Ch.7(Relative velocity method
Solved & prepared By :
Bakhtiar Essa , Ac.year : 2016 – 2017
Edition one
UNIVERSITY OF SALAHADDIN
COLLEGE OF ENGINEERING
MECHANICAL DEPARTMENT

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EXERCISES Chapter 7 : Velocity in Mechanisms _Page : 169
Given : N = 600 rpm (c.w) , Va =? ,
Vg = ? , Wab = ?
Wob = = 62.83 rad/s
Vob = Wob * OB = 62.83 * 0.125
= 7.85 rad /s
v. d.scale [ 7.85 m /s = 45 mm ]
VA = 37 * = 6.45 m / s
VG = 38.7 * = 6.75 m/s
VAB = ab * = 5.4 m/s
WAB = = 10.8 rad /s
1. In a slider crank mechanism, the length of crank OB and connecting rod AB
are 125 mm and 500 mmrespectively. The centre of gravity E of the
connecting rod is 275 mm from the slider A. The crankspeed is 600 r.p.m.
clockwise. When the crank has turned 45° from the inner dead centre
position,determine: 1. velocity of the slider A, 2. velocity of the point E, and
3. angular velocity of theconnecting rod AB[Ans. 6.45 m/s ; 6.75 m/s ; 10.8 rad/s]

4. 3
Solution :
WOA = WOB =
= 4.188 rad / s .
VA= WOA * OA ,
VA = 4.18 * 50 = 209.4 mm /s
b
[209.4 mm/s =60mm] a
velocity diagrams scale .
Vg = fg * c g o d
Vg = 19.5 * = 68mm/s f e
2. In the mechanism, as shown in Fig. 7.32, OA and OB are two equal cranks
at right angles rotating about O at a speed of 40 r.p.m. anticlockwise. The
dimensions of the various links are as follows :OA = OB = 50 mm ; AC = BD
= 175 mm ; DE = CE = 75 mm ; FG = 115 mm and EF = FC.Draw velocity
diagram for the given configuration of the mechanism and find velocity of the
slider G. [Ans. 68 mm/s]

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We can find ( dc ) on the S.DIA. by this relationship .
= . b
de = dc * = 36.8 mm f e
VF = af *
VF = 46.45 * = 250mm/s o , d
c
3. The dimensions of various links in a mechanism, as shown in Fig. 7.33,
are as follows :AB = 60 mm ; BC = 400 mm ; CD = 150 mm ; DE = 115
mm ; and EF = 225 mm. Find the velocity of the slider F when the crank
AB rotates uniformly in clockwise direction at a speedof 60 r.p.m.
[Ans. 250 mm/s]

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Given :
N = 150 rpm ,
d = 50 mm .
R = 0.25 cm
Space diagram scale
1 : 10 it means
150 mm ( given at the statement ) 15 mm (on the space diagram)
WBA = = 15.7 rad/s
VBA = WBA* AB
VBA= 15.7 * 150 mm = 2355 mm/s Chose this it as the velocity diagram scale .
So 2355m m/ sec = 45 mm ( in v.daigram )
4. In a link work, as shown , the crank AB rotates about A at a uniform speed of
150 r.p.m.The lever DC oscillates about the fixed point D, being connected to
AB by the connecting link BC. The block F moves, in horizontal guides being
driven by the link EF, when the crank AB is at 30°. The dimensions of the
various links are : AB = 150 mm ; BC = 450 mm ; CE = 300 mm ; DE = 150
mm ; and EF = 350 mm. Find, for the given configuration, 1. velocity of slider
F, 2. angular velocity of DC, and 3. Rubbing speed at pin C which is 50 mm in
diameter. [Ans. 500 mm/s ; 3.5 rad/s ; 2.4 cm/s]

7. 6
Since the point E lies on DC, therefore divide vector dc in e in the same ratio
as E dividesCD In other words Ce = cd * = 30 * = 20 mm
VF = fd * = 9.56 mm * VF = 500 mm/s
VDC = dc * = 1575 mm/s when [ dc = 30.1mm ]
WDC = = = 3.5 rad/s
VCB = cb * = 2669 mm /s when [ c b = 51mm ]
WCB = = = 6 rad /s
V.R© = ( WCB + WDC ) * R
V.R © = ( 6 + 3.5 ) * 0.25 = 2.38 cm

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WOB = = 9.42 rad /s .
VOB = WOB * OB
VOB = 9.42 * 0. 225 = 2.12 m /s .
v.d Scale : [VOB = 9.42m/s = 50mm ]
= oa = ob * =
33. 3 mm
VC = oc = 30mm * = 1.2 m/s
a b
VAC = ac = 10 mm * = 0.424 m/s
WAC = = = 1.6 rad / s . o c
b
b
b
b
b
b
b
b
b
b
b
5. The oscillating link OAB of a mechanism,as shown , is pivoted at O and is
moving at 90 r.p.m. anticlockwise. If OA= 150 mm ; AB = 75 mm, and AC =
250 mm ,calculate 1. the velocity of the block C; 2. the angular velocity of
the link AC; and 3. the rubbing velocities of the pins at O, A and C,
assuming that these pins are of equal diameters of 20 mm
[Ans. 1.2 m/s ; 1.6 rad/s clockwise ; 200 mm/s, 782 mm/s, 160 mm/s]

9. 8
WOB = = 9.42 rad /s . = WOA { because on the one link}
WAC = = = 1.6 rad / s .
R = = 10 mm .
 Ru.O = ( WOA ) * R
Ru.O = ( 9.42 ) * 10 = 94 .2 mm /s
 Ru.A = ( WOB + WAC ) * R
Ru.A = ( 9 . 42 – 1.6 ) * 10 = 78 . 2 mm /s
 Ru.C = ( WAC ) * R
Ru.C = ( 1.6 ) * 10 mm = 16 mm /s

10. 9
WBA = = 4 rad /s
VBA = 4 *30 = 377 mm /s
[ V.d scale 377mm /s = 50 mm ]
VEA = ae =16 *
VBC = bc = 44 *
WBC = = = 4 .14 rad /s .
VCD = dc = 24 * = 180 . 96 mm /s WCD =
= = 2. 2 rad /s
VCE = ce = 20 . 5 * = 154 . 5 mm /s
WCE = = = 1.28 rad /s
6. The dimensions of the various links of a mechanism, as shown , : AB = 30
mm ; BC = 80 mm ; CD = 45 mm ; and CE = 120 mm. The crank AB rotates
uniformly in the clockwise direction at 120 r.p.m. Draw the velocity diagram
for the given configuration of the mechanism and determine the velocity of
the slider E and angular velocities of the links BC, CD and CE.
[Ans. 120 mm/s ; 2.8 rad/s ; 5.8 rad/s ; 2 rad/s ]

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N = 120 rpm
WOA = = 4
VOA = WOA * OA
VOA =
VOA = 2.51 mm /s = 70 mm]V.D.S : [
Vp = op = 17 .6 * = 0.63 m /s
VQB = qb = 26 .4 * = 0 . 946 m /s
o,q p
WQB = = = 6 .31 rad /s .
b
a
9. In a Whitworth quick return motion mechanism, as shown , the dimensions
of various links are as follows : OQ = 100 mm ; OA = 200 mm ; BQ = 150
mm and BP = 500 mm. If the crank OA turns at 120 r.p.m. in clockwise
direction and makes an angle of 120° with OQ, Find : 1. velocity of the block
P, and 2. angular velocity of the slotted link BQ. [Ans. 0.63 m/s ; 6.3 rad/s ].

12. 11
Given :
N = 60 r.p.m
Tp = 115 N.m
 = 60 % = 0.6
Ts = ? , Fs = ?
Solution :
WP =
WP = = rad /s
VP = WP * OP s q , o
VP = * 50 m = 0 .3141 m /s
V.D Scale : [ 0.3141 m /s = 60 mm ] p
r
VS = os * , when{ os = 6mm}
VS = 0 .03141 m /sec
10. A toggle press mechanism, as shown, has the dimensions of various
links as follows :OP = 50 mm ; RQ = RS = 200 mm ; PR = 300 mm. Find
the velocity of S when the crank OP rotates at 60 r.p.m. inthe anticlockwise
direction. If the torque on P is 115 N-m, what pressure will be exerted at S
when the overall efficiency is 60 percent. [Ans. 400 m/s ; 3.9 kN]

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TP * WP = in put
Fs * VS = out put
 =
 = , FS =
FS = = 13802 .6 N = 13 .8 KN .

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WOA = = 2 rad /s
VOA = WOA * OA
VOA = 2 * 150 * = 0.942 m /s
Velocity diagram Scale
[ 0.942m/s = 70 mm ]
VCB = ob = 67 * = 0.9 m /s
VD = od = 39* = 0.52 m /s 0.5 m /s . a b
VAB = ab = 67 * = 0.9 m /s
VBD = bd = 66 * = 0.88 m /s
WBD = = 1.76 rad /s d o,c
11. in the figure shows a toggle mechanism in which link D is constained to
move inzhorizontal direction.For the given configuration, find out : 1.velocities
of points band D ; and 2. angular velocities of links AB, BC, and BD.The rank
OA rotates at 60 r.p.m. in anticlockwise direction. [Ans. 0.9 m/s; 0.5 m/s;
0.0016 rad/s (anticlockwise) 0.0075 rad/s (anti-clockwise),0.0044 rad/s
(anticlockwise)] .
35 degree

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Given :
FE = 2.2 KN ,  = 72 %
= ? , FD = ?
Solution :
To drow space diadram we know that
BC = 2AB , if we assume that AB = 40
mm , so BC = 80 mm .
aslo we assume the scale of velocity
digram is 60 mm = ab .
VE = ae = 58 mm and VD = ad= ac = 18 mm . d,c a
= = 3 .222
 = FD =  *
FD = 0 .72 * = 5 KN b e,f
12. A riveter, as shown below , is operated by a piston F acting through the
links EB, AB and BC. The ram D carries the tool. The piston moves in a line
perpendicular to the line of motion of D. The length of link BC is twice the
length of link AB. In the position shown, AB makes an angle of 12° with AC and
BE is at right angle to AC. Find the velocity ratio of E to D. If, in the same
position, the total load on the piston is 2.2 kN, find the thrust exerted by D when
the efficiency of the mechanism is 72 per cent, Ans. [3.2 ; 5 kN]

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Q1: Describe the method to find the velocity of a point on a link whose direction (or
path) is known and the velocity of some other point on the same link in magnitude
and direction is given.
ANS 1: by Relative Velocity Method The relative velocity method is based upon
the relative velocity of the various points of the link ,Consider two points A and B
on a link . Let the absolute velocity of the point A i.e. vA is known in magnitude and
direction and the absolute velocity of the point B i.e. vB is known in direction only.
Then the velocity of B may be determined by drawing the velocity diagramas .
Q2: Explain how the velocities of a slider and the connecting rod are obtained in a
slider crankmechanism.
ANS,2 : A slider crank mechanism is shown below (a). The slider A is attached to
the connecting rod AB. Let the radius of crank OB be r and let it rotates in a
clockwise direction, about the point O with uniform angular velocity rad/s.
Therefore, the velocity of B i.e. vB is known in magnitude and direction. The slider
reciprocates along the line of stroke AO. The velocity of the slider A (i.e. vA) may
be determined by relative velocity method as discussed below :

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Q3 : Define rubbing velocity at a pin joint. What will be the rubbing velocity at pin
joint when the two links move in the same and opposite directions ?
ANS 3 : The rubbing velocity is defined as
the algebraic sum between the angular velocities of
the two links which are connected by pin joints,
multiplied by the radius of the pin.Consider two links
OA and OB connected by a pin joint at O as shown
Let w1 = Angular velocity of the link OA or
the angular velocity of the point A with respect to O.w2 = Angular velocity of the
link OB or the angular velocity of the point B with respect to O, and r = Radius of the
pin. According to the definition, Rubbing velocity at the pin joint O = (w1 – w2) r, if
the links move in the same direction = (w1 + w2) r, if the links move in the opposite
direction .
Q4 :What is the difference betweenideal mechanical advantage and actual
mechanical advantage ?
ANS 4 : 1- Ideal mechanical advantage : FA × vA = FB × vB , = 100 %
( neglecting effect of feiction )
If we consider the effect of friction, less resistance will be overcome with the given
effort.Therefore the actual mechanical advantage will be less.
Let : = Efficiency of the mechanism.
2- Actual mechanical advantage,
≠ 100 %

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Solved & prepared By : Bakhtiar Essa Omar , 2016 – 2017
with Good luck for all

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