Engineering Mechanics

Question 1
The lengths of members BC and CE in the frame shown in the figure are equal. All the members are rigid and lightweight, and the friction at the joints is negligible. Two forces of magnitude Q \gt 0 are applied as shown, each at the mid-length of the respective member on which it acts.

Which one or more of the following members do not carry any load (force)?
A
AB
B
CD
C
EF
D
GH
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Question 1 Explanation: 


If at a point 3 member are meeting and two are colinear then in 3rd member force will be zero.
F_{GH}=0
Question 2
A rigid homogeneous uniform block of mass 1 kg, height h= 0.4 m and width b= 0.3 m is pinned at one corner and placed upright in a uniform gravitational field (g = 9.81 m/s^2), supported by a roller in the configuration shown in the figure. A short duration (impulsive) force F, producing an impulse I_F, is applied at a height of d = 0.3 m from the bottom as shown. Assume all joints to be frictionless. The minimum value of I_F required to topple the block is

A
0.953 Ns
B
1.403 Ns
C
0.814 Ns
D
1.172 ns
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Question 2 Explanation: 
Mass moment of inertia of block about hinge 'O' =\frac{M}{12}(h^2+b^2)+Mr^2
where, h=0.4m, b=0.3m
r=\sqrt{0.15^2+0.2^2}=0.25m


I_o=\frac{1}{2} \times (0.4^2+0.3^2)+1 \times 0.25^2=0.083 kg-m^2
For block to topple about 'O' it should just reach the position such that centre of mass reaches from point (G) to point (G')

[Beyond this the torque of gravity will rotate the block itself]
Now, y(increase in height of centre of mass)
= OG' - OP
= OG - OP
= 0.25 - 0.20 = 0.05

So the block should have initial angular velocity to just shift centre of mass at this position (i.e., G')
F.B.D of block is shown below.

\frac{d\vec{L}}{dt}=\vec{\tau _{ext}}\;(about\;\;O)
[ \vec{L} is angular momentum, \vec{\tau _{ext}} is external torque]
\tau _{ext} dt=d \vec{L}
F\Delta t \times d=L_f-L_i
(Because, Gravity force is small compared to impulsive force hence we neglect its torque)
I_f \times d =I_o \times \omega . . . (i)
By using conservation of mechanical energy
\frac{1}{2}I_o\omega ^2=mg\;y
\frac{1}{2}0.083 \times \omega ^2=1 \times 9.81 \times 0.05
\Rightarrow \omega =3.44 \; rad/s
Putting in eq.(i)
I_F \times 0.3 =0.083 \times 3.44
\Rightarrow I_F =0.951 N-s
Question 3
A rope with two mass-less platforms at its two ends passes over a fixed pulley as shown in the figure. Discs with narrow slots and having equal weight of 20 N each can be placed on the platforms. The number of discs placed on the left side platform is n and that on the right side platform is m.
It is found that for n = 5 and m = 0, a force F = 200 N (refer to part (i) of the figure) is just sufficient to initiate upward motion of the left side platform. If the force F is removed then the minimum value of m (refer to part (ii) of the figure) required to prevent downward motion of the left side platform is______ (in integer).

A
1
B
2
C
3
D
4
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Question 3 Explanation: 
F.B.D of fig.(i) :

For upward impending motion of left side platform
T_1=n \times 20=5 \times 20=100N\;\; . . . (i)
F=T_2=200N . . . (ii)
\therefore \; \frac{T_2}{T_1}=2
So, ratio of tensions on two sides remains [in figure (i) and (ii)] same as 2 because it depends on \mu \text{ and } \theta which is same in fig.(i) and fig.(ii).
F.B.D of fig.(ii) :

T_3=100
Now since impending motion of left side platform is downward therefore T_3=100 is tight side tension.
T_4=\frac{T_3}{2}=\frac{100}{2} =50 \; N
T_4=m \times 20
50=m \times 20
m=2.5
Hence, m = 3 (As no. of discs is integer)
Question 4
A square plate is supported in four different ways (configurations (P) to (S) as shown in the figure). A couple moment C is applied on the plate. Assume all the members to be rigid and mass-less, and all joints to be frictionless. All support links of the plate are identical.

The square plate can remain in equilibrium in its initial state for which one or more of the following support configurations?

MSQ
A
Configuration (P)
B
Configuration (Q)
C
Configuration (R)
D
Configuration (S)
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Question 4 Explanation: 
In case P it forms collinear force system. i.e., all the force passes through the intersection point. Hence, they can't balance couple (As they can't create couple).
Question 5
A cylindrical disc of mass m=1 \; kg and radius r=0.15\;m was spinning at \omega =5 \; rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity g acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive. Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).

A
0
B
0.25
C
0.5
D
0.75
GATE ME 2022 SET-1      Friction
Question 5 Explanation: 


About point P there is no external torque.
[Because Torques of N and 'mg' balances and friction force pass through point P, hence its torque is zero]
\therefore \left [ \frac{d\vec{L}}{dt} =\vec{\tau }_{ext}\right ]_{\text{about P}}=0
where \vec{\tau }= Torque and \vec{L }= Angular momentum
\begin{aligned} \vec{L_i}&=\vec{L_f}\\ I_0\omega &=I_0\omega '+mvr\\ \frac{mr^2}{2}\omega &=\frac{mr^2}{2}\omega'+m\omega 'r^2\\ \therefore \omega '&=\frac{\omega }{3}=\frac{5}{3}\\ \Rightarrow v&=\omega ' \times r\\&=\frac{5}{3} \times 0.15=0.25 m/s \end{aligned}
Question 6
Two rigid massless rods PR and RQ are joined at frictionless pin-joint R and are resting on ground at P and Q, respectively, as shown in the figure. A vertical force F acts on the pin R as shown. When the included angle \theta \lt 90^{\circ} , the rods remain in static equilibrium due to Coulomb friction between the rods and ground at locations P and Q. At \theta = 90^{\circ} , impending slip occurs simultaneously at points P and Q. Then the ratio of the coefficient of friction at Q to that at P P(\mu _Q/\mu _P) is _________ (round off to two decimal places).

A
5.76
B
9.32
C
4.25
D
8.65
GATE ME 2022 SET-1      Friction
Question 6 Explanation: 


By pythagoras theorem:
PQ = 13 m
By Geometry:
\begin{aligned} 5^2-PR^{'2}&=10^2-AR^{'2}\\ QR'&=13-PR'\\ \Rightarrow PR'&=\frac{25}{13}m\\ \Rightarrow QR'&=\frac{144}{13}m\\ \Sigma M_P&=0\\ \Rightarrow F(PR')&=N_Q(PQ)\\ \Rightarrow N_Q&=\frac{25}{169}F\\ \Sigma F&=0\\ \Rightarrow N_P&=F-\frac{25}{169}F\\ &=\frac{144}{169}F \end{aligned}
Now, f_P=\mu _PN_P, f_Q=\mu _QN_Q, \Sigma F_x=0
\begin{aligned} \mu _PN_P&=\mu _QN_Q\\ \frac{\mu _Q}{\mu _P}&=\frac{N_P}{N_Q}\\ &=\frac{144}{25}=5.76 \end{aligned}
Question 7
A structure, along with the loads applied on it, is shown in the figure. Self-weight of all the members is negligible and all the pin joints are frictionless. AE is a single member that contains pin C. Likewise, BE is a single member that contains pin D. Members GI and FH are overlapping rigid members. The magnitude of the force carried by member CI is ________ kN (in integer).

A
12
B
15
C
18
D
22
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Question 7 Explanation: 


By method of sections cut CI, EI and EF and prefer the right hand side
For equilibrium:
\Sigma M_E=0 Anticlockwise positive
T_{CI}(1.5)-2(1.5)-4(6)=0
\Rightarrow T_{CI}=\frac{3+24}{1.5}=18kN
Question 8
The plane of the figure represents a horizontal plane. A thin rigid rod at rest is pivoted without friction about a fixed vertical axis passing through O. Its mass moment of inertia is equal to 0.1 kg.cm^2 about O. A point mass of 0.001 kg hits it normally at 200 cm/s at the location shown, and sticks to it. Immediately after the impact, the angular velocity of the rod is ___________ rad/s (in integer).

A
10
B
20
C
25
D
40
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Question 8 Explanation: 
Given,
I_{rod} = 0.1 kgcm^2 = 1\times 10^{-5} kg m^2
m_{point} = 10^{-3} kg
V_{point} = 2 m/s
\omega _{o(rod)}=0 rad/s

To find: \omega _{final}
Since there is no external moment involved about O. Therefore the Angular momentum of the system about O in conserved.

\begin{aligned} \therefore \; I_{rod}\; \omega _{initial}+(mV_0r)_{point}&=(I_{rod}+I_{point})\omega _{final}\\ 10^{-3} \times 2 \times 0.1&=(10^{-5}+10^{-3}(10^{-1})^{2})\omega _{final}\\ 2 \times 10^{-4}&=2 \times 10^{-5}\omega _{final}\\ \omega _{final}&=10rad/s \end{aligned}
Question 9
A block of negligible mass rests on a surface that is inclined at 30^{\circ} to the horizontal plane as shown in the figure. When a vertical force of 900 N and a horizontal force of 750 N are applied, the block is just about to slide.

The coefficient of static friction between the block and surface is _____ (round off to two decimal places).
A
0.28
B
0.36
C
0.17
D
0.05
GATE ME 2021 SET-2      Friction
Question 9 Explanation: 

After forces are applied block is just about move (mass is negligible). Calculate coefficient of friction
\begin{aligned} F_{H} &=750 \mathrm{~N} \\ F_{V} &=900 \mathrm{~N} \\ \theta &=30^{\circ} \end{aligned}


\begin{aligned} N &=900 \cos \theta+750 \sin \theta \\ N &=900 \cos 30^{\circ}+750 \sin 30^{\circ} \\ &=1154.4228 \mathrm{~N} \\ F_{\max }+900 \sin 30^{\circ} &=750 \cos 30^{\circ} \\ \mu N &=199.519 \\ \mu &=\frac{199.519}{1154.4228} \\ \mu &=0.1728 \end{aligned}
Question 10
A plane truss PQRS(PQ=RS, \text{and }\angle PQR=90^{\circ}) is shown in the figure



The forces in the members PR andRS, respectively, are
A
F\sqrt{2} (tensile) and F (tensile)
B
F\sqrt{2} (tensile) and F (compressive)
C
F (compressive) and F\sqrt{2} (compressive)
D
F (tensile) and F\sqrt{2} (tensile)
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Question 10 Explanation: 




Joint C,
\Sigma F_{H}=0
\begin{aligned} \Rightarrow \qquad F_{P R} \sin 45^{\circ}&=F \\ F_{P R}&=\sqrt{2} F(\text { Tensile }) \\ \Rightarrow\qquad F_{P R} \cos 45^{\circ}&=F_{R S}\\ \end{aligned}
\Sigma F_{V}=0

\Rightarrow F_{RS}=F(Comp.)
There are 10 questions to complete.

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