Roll No.
1
L
B.E / B.Tech ( Full Time ) ARREAR END SEMESTER EXAMINATIONS, APRIL / MAY 2014
COMMON TO ALL BRANCHES
Semester II
GE 9151 & ENGINEERING MECHANICS
(Regulation 2008)
Time: 3 Hours
Answer ALL Questions
Max. Marks 100
PART-A (10 x 2 = 20 Marks)
1.
The figure represents the principle of transmissibility.
The principle of transmissibility states that the conditions of equilibrium or motion of a rigid
body will remain unchanged if a force F acting at a given point of the rigid body is replaced
by a force F' of the same magnitude and same direction, but acting at a different point,
provided
that the two forces have the same line of action
K
fig 1
2.
R
R
Moment of the 250-N force on the handle of the monkey wrench about the center of the
bolt = - 250 cos 15° * 200 + 250 sin 15° * 30 = - 46.355 N.m * ANS
250 N
200
30 mm
1
3.
15^\
mm
a /
fig 2
The balanced teeter-totter (seesaw) signifies that a body is in equilibrium when the
moment of all the forces acting on the body is zero. The moment arm of the heavier
person is shorter than the moment arm of the lighter person. However the moment of
these forces about the point of support is equal and opposite thus resulting in a
balanced teeter-totter.
B
x
/ \
\
t.
A
V;
n
fig 3
/
(not a irec body diagram)
4
5.
a
fin
fig 4
4
The equivalent force system of these forces at point B as shown in the figure.
Answers:
(a) Force P (towards right),
Moment V3/2 Pa (clockwise)
(b) Force 4V3 P (upward),
Moment 3^3 Pa (anticlockwise)
(c) Force P (towards right),
ANS
Moment 3a/3 Pa (anticlockwise)
(d) Force P (towards left),
Moment 3V3/2 Pa (clockwise)
Parallel axis theorem used for finding the moment of inertia of an area.
1=1
+
Ad
z
This formula expresses that the moment of inertia I of an area
with respect to any given axis AA' is equal to the moment of inertia J
of the area with respect to a centroidal axis BB' parallel to AA'
plus the product of the area A and the square of the distance d
between the two axes. This theorem is known as the parallel-axis
6.
Volume V of the solid generated by revolving the 60-mm right triangular area through
180° about the z-axis.
V. 7
ynfao + §{60)][|C60){60)] = 2.83(10 ) mm
5
3
*
A
N
S
f
i
60
f
mm
30
rain
60
mm
,30*1
60
mm
1
fig 5
mm
30u m
»
If the car decelerates uniformly along the curved road from 25 m/s at A to 15 m/s at C
determine the acceleration of the car at B.
Tangential acceleration, a = dv/dt = (dv/ds) (ds/dt) = v (dv/ds)
t
Since car decelerates uniformly, a is constant.
t
a ds = v dv
t
Integrating between the points A and C, a ( s - s ) = /4 ( v
1
t
(15 m / s )
2
= (25 m / s )
a, = - 0 . 6 6 6 7 m / s
2
v
tt
+ 2^(300 m -
2
c
- v )
2
A
0)
$a)
+ 2 ( - 0 . 6 6 6 7 m / s ) ( 2 5 0 m - 0)
7
= 17.08 m/s
VJ5
(17.08 m/s)
300 m
= V(-0.6667 m/s )
2
-
A
2
u£ = v j j + 2<7,(^« vl = (25 m/s?
c
1.18 m / s
2
0.9722 m/s
+ (0.9722 m / s )
2
2
2
ANS
FI\press >our.anwvr w ith the nppropviM? \m\U<
S.-A
250 n
X
300 m
t
(V
HI •
v
2-3O0
3OT
8.
Conditions under which the motion of a projectile is parabolic:
(i) The air resistance is neglected
(ii) The acceleration due to gravity is constant and does not vary with altitude.
9.
The cylinder shown is of weight W and radius r, and the coefficient of static friction u is
the same at A and B. If the cylinder is in equilibrium, draw the free body diagram of the
cylinder.
s
fig 7
r
f
w
f
B
10.
Angle of repose
The value of the angle of inclination corresponding to impending motion is called the angle
of repose. Clearly, the angle of repose is equal to the angle of static friction q>s.
P a r t - B ( 5 x 1 6 = 80 m a r k s )
11.
Two soccer players approach a stationary ball 10 m away from the goal. Simultaneously, a player on
team O (offense) kicks the ball with force 100 N for a split second while a player on team D
(defense) kicks with force 70 N during the same time interval Does the offense score (assuming the
goalie is outside the range of the goal).
Goal
7 m
4
fig 8
5
A / * 70 N
D
10 m
- •
Tut*
10 m
& -
-2g8*
^e^uJtto^t
3f-an-7"G«
exX*> *o£ •o-~**. a~-gtt= 0-3.11° ,£o."£&t
of
ko>t£s»v«jfcaj£•
J tot*— *>-«•<*"
-
12.
a)
ANS
The special purpose milling cutter is subjected to a force of 1200 N and a couple of 240 Nm as
shown in fig. 9. Determine the moment of this force couple system about point O.
(10)
2tM
200 ram
1200 N 4 o < ^
240 N-ro
R = 1200 sin 30° j - 1200 cos 30° k = 600 j - 1039.23 k
Mo = - 600 * 0.25 i + 600 * 0.2 k + 1039.23 * 0.2 j + M
Ans
mm
150 i + 120 k + 207.846 j + 240 sin 30"0 i - 2 4 0 cos 300 k
u
150 i + 327.846 j - 87.846 k -» Ans
OR
B
<5
B
/
B,
n
6 m
2
/I
k N
D
j
2.5 m
n
1 mJ
/
I i x I - II
i
fig 10
s
Solution.
Tlu» sy-iv-n'i- cli-arlv thrw-diTmmswrui! with ;i" hue? or pUmei of
sjuimwry. and therefore the prubkm must'bo anaiy zod as a general -<pace *>•«•
loin offerees. The free-body diagram is drawn, where iho ring reaction is shown
in lorms of i u two components. All unknowns except T may he eliminated by a
moment" sum about- the line AB. The direction of All is- specified by the unit
vector 11
'4 "rj •; 6k:--
1
4(31 + 4k). The moment of T about- All
is the component in the direction of AC of the vet-tor moment about the point A
and equals r, x T- n. Similarly the moment of the applied load F about AB is
v., x F'-n. With CD - % 4(3.2 m, the vector expressions for T, F . r and i'i are
1 :
T
- J = (2i + 2.5j - 6 k )
. Ho;
r , = - t •+• 2.5j m
(J
F =• 2\ kN
r = 2.5i + 6k m
2
The moment equation now becomes
IXM,,, = 01 ( - i + 2.5j] x -JL=<2i • 2.5j - 6k)4i'3j + 4k)
v'46.2
s
+ (2.5i + 6k) x (2j)4i.3j + 4k) = 0
Completion of the vector operations gives
4 I L - 20 = 0 .
,46.2
T=2.83kN
.4ns.
and the components of T become
T
x
= 0.833 kN
T = 1.042 k N
T, = --2.50 kN
v
We may find the remaining unknowns by moment and force summations as
follows:
XiM, = 0]
11M
2(2.5) - 4.5B - 1.042(3) = 0
T
= OJ
4.5J3, - 2(6) - 1.042(6) = 0
\ZF, = 0]
4 . + 0.4.17 - 0.833 = 0
gj [XF = 01
A., -t 2 - 1.042 = 0
IE/-'. = 0J
.4. + 4.06 - 2.50 = 0
X
V
B , = 0.417 k N
Am.
B = 4.06 k N
Am.
T
A = -1.250 kN
Ans.
A = -3.04 k N
Ans.
x
y
A = -1.556 kN
z
Ans.
The plaic
<>--lx.
i> made of sice! having a densin
of
7SM) k a m ' . If the ihk&ncv <-'! the plate is Id mm. determine
f
the horizontal and \criicai components of reaction «t the pin
,-S and the tension in eahk- />'C".
Differential Element: The clement parallel to the3' aus shown shaded in Fig. <i will be considered. The
area of this dernenl is given by
dA -y<fc - 1.259ft*
10
A
Ontroid; Theccntroid of the dementis located at x =*;tand v - >-/2.
,\rca: Integrating,
(4m
.4 = J <M = J ° l . 2 5 W t
l / 5
iii=0.9«9j |^"
w
=6m
2
-4m-
Ttius, the mass of -the. plate can be obtained from
m = pAI= 7850<6KO.O!) = 47I I g
i«M
Jl.2599-t ^<-J
,C,
j
i.2S99x ctx
4n
'a.S39S* *
7/
>>t
Since the plate has a unifomt thickne.ES, its center of gravity coincides with its ceatroid.
Eqnafiom af EqtiBibrinm: By referring to the free body diagram shown in Fig. b.
ffIM
A
- 0:
^E/> =
+ T H J . = 0:
Pic (4) - 471(9.81X 22857) = 0
ft
Fgr; = 264037 N = 2.64 kK
Ans.
A -0
Ans.
x
A + 2640.27- 471(981 )= 0
v
A =1980.24 N = 1.9SkN
Ant
s
<— fig 11
1
4 m
r
b)
[OR]
Determine the moment of inertia of the beam's cross-sectional area about the x and y axes.
Composite Parts: The composite cross - sectional area of the beam can be subdi t ided into segments as shown in Fig. u. The
perpendicular distance measured from the ceniroid of each segment to lltejr axis is also indicated.
Moment of Inertia: The moment of inertia of each segment about the .* axis can be determined using the parallel - axis theorem.
Thus,
2| ^ < J 5 X 3 0 0 ) |+ 2(15X 300)(0)
3
^ Yj(120)(15 ) ]+ 2(120X15)(50)
3
2
= 61.% 10 ) +- 9.0675( 10 ) = 76.6(! 0 ) mm
6
6
6
2
4
CoOerm
15/iwri
C,
s
VP
150 mm
i
^ 5 0 mm
&
T
I5tm
l5rm
Campo-rit* Putts: llie ccHtipasifc cross-sectional area of the beam can be subdivided i t i t t ^ p ^ l t s as «fco*ti b R g . i t
perpendicular distance measured from ike oeotroid of each segment to (he}' axis is also indicatfed,
Momwt of Ificrifct: The moment of inertia of each segment about the x axis cari be d^tenraned using the paraJlel - axis IS
Thus.
- 41.17410 )+ 4 j a C 1 # ) ^ 4 5 J ( l 0 jruffl
6
14.
a)
6
4
(i) The two systems shown start from rest. On the left, two 200 N weights are connected by an
inextensible cord, and on the right, a constant 200 N force pulls on the cord. Neglecting all
frictional forces which of the following statements is true?
a) Blocks A and C will have the same acceleration.
b) Block C will have a larger acceleration than block A.
c)
Block A will have a larger acceleration than block C.
ANS
d) Block A and C will not have any acceleration.
e) None of the above
(8)
Justify your answer.
fig 13
c
200 N
200
N
,200 N
200 N
Let a be the acceleration of block A.
Let a be the acceleration of block B.
When block A moves to the right by x , block B moves down by yeX.A - y
Differentiating w.r.t. time
v = v
Differentiating w.r.t. time
a = a ->.(1)
Let T be the tension in the cable
A
B
A
B
A
A
B
B
i
1
2Fx= m.a
T = (200/9.8 1).8a"> (2)
ZFy = m.a
W - T = (200/9.81) a
Substituting from eqn. (1), 200 - T - (200/9.81) a ^ (3)
Substituting from eqn. (2), 200 - (200/9.8l).a = (200/9.81) a
9.81 = 2 a
a = 4.905 m / s
A
W-200N
B
B
A
A
A
2
A
A
T = 200N
Let ac be the acceleration of block C.
SFx= m.ac
T = (200/9.8 l).ac
200 = (200/9.8 l).ac
a = 9.81 m / s
2
c
T h u s Block C will have a larger acceleration than block A.
ANS
Solution,
The force registered by the scale and the velocity, both depend on
the acceleration of the elevator, which is constant during the interval for which
the forces are constant, f r o m the free-body diagram of the elevator, .scale, and
man taken together, the acceleration is found to be
iF..
8300 - 7360 = 750a.
ma.
1.257 m/s
2
The scale reads the downward'force exerted on i t by the man's feet . The equal
and opposite reaction J? to this action is shown on the free-body diagram of the
man alone together w i t h his weight;, and the equation of motion for him gives
R - ' 7 3 6 « ' 75(1.257)
JXF = maj
V
R *• 830 N
A}
IS:
The velocity reached at the end of the 3 seconds is
»J
> - 0 = I
adt)
1.257dt
v « 3.77 m/s
(8)
y
T * 8300 N
a
y
fig 14
750(9.81) = 7360 N
[OR]
Ans,
fig 15
Given:
= 10
r
0
ci
m
-
= 5m
g = 9.81
5
\0 .
t
f
n
it
iL
m
Solution:
Guesses
6
Given
d =r
C
= 70 dea
0
6
D
= 15 deg
0=
COS
d= r cos{d )(t0
D
At)
—f
4r =. 2 s
/ = Is
+' v sin( 6 ) t
0
C
0 = ~Y (t - AtY
v siii( <9/))(r - At)
0
f %
f 7.5.313'
r
= Fiud(c? . 0
c
t
\At)
D
J , At)
t = 1.972 s
At .-= 1.455 s
=
•
v
14.687,
dea
0
fig 16
"fit.
-tk
wtugW:
f^
/
cv~A.
-tie
;rw f>
hzcu-ko^.
^
f/tov—
^
tL.
^* V
^'
3
5
•
r
ANS
-//=0.25
p = 0.15
[OR]
Determine the range of cylinder mass m for which the system is in equilibrium. The coefficient of
friction between the 50-kg block and the incline is 0.15 and that between the cord and cylindrical
support is 0.25.
Wsin 6
y W cos 6
W = 490.5 N
Let T be the tension in the cable.
SFy= 0
N = W cosO = 490.5 * cos 20° = 460.919 N -> (1)
CASE I
When the block is moving up, friction force acts downward.
2Fx= 0
Ti - W sin0 - f = 0
T, - 490.5 * sin 20° - uN = 0
T, - 167.76 - 0.15 * 460.919 = 0 -> T j = 236.9 N
CASE II
When the block is moving down, friction force acts upward.
2Fx= 0
Ti - W sinO + f = 0
T , = 98.622 N
t
110°
p=
C A S E I : When The block moves up the inclined plane, T is the slack side, T2 is the tight side.
f
7;
T = Ti e
2
Mp
= 236.9 *
0 2 5
e
*
( 1 , 0 / 1 8 0
*
, t )
= 382.83 N
C A S E I I : When The block moves down the inclined plane, T is the slack side, Ti is the tight
side
2
T,
*
T = T, / e ^ = 98.622 / ^(nonm^
2
= 61.027 N
T
1
EFy= 0
W = T = 2 T * 9.81m = 2T
3
2
2
^
^
T
1
:
2
C A S E I:
9.81 m = 2 *382.83
T = 2T
m = 78.05 kg
3
2T,
CASE II:
9.81 m = 2 *61.027
m = 12.44 kg
12.44 kg < m < 78.05 kg
ANS
W= 9.81 m
2