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)?

Which one or more of the following members do not carry any load (force)?

AB | |

CD | |

EF | |

GH |

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

0.953 Ns | |

1.403 Ns | |

0.814 Ns | |

1.172 ns |

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

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).

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).

1 | |

2 | |

3 | |

4 |

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)

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?

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

**MSQ**Configuration (P) | |

Configuration (Q) | |

Configuration (R) | |

Configuration (S) |

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).

0 | |

0.25 | |

0.5 | |

0.75 |

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).

5.76 | |

9.32 | |

4.25 | |

8.65 |

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).

12 | |

15 | |

18 | |

22 |

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).

10 | |

20 | |

25 | |

40 |

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}

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).

The coefficient of static friction between the block and surface is _____ (round off to two decimal places).

0.28 | |

0.36 | |

0.17 | |

0.05 |

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

The forces in the members PR andRS, respectively, are

F\sqrt{2} (tensile) and F (tensile) | |

F\sqrt{2} (tensile) and F (compressive) | |

F (compressive) and F\sqrt{2} (compressive) | |

F (tensile) and F\sqrt{2} (tensile) |

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.

Excuse me sir where is Engineering economic 2021 set1 and set 2 questions?

Sorry engineering that is mechanics!