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 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 5 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 6 |

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 6 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 7 |

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 7 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.)

Question 8 |

An attempt is made to pull a roller of weight W over a curb (step) by applying a horizontal
force F as shown in the figure.

The coefficient of static friction between the roller and the ground (including the edge of the step) is \mu. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.

The coefficient of static friction between the roller and the ground (including the edge of the step) is \mu. Identify the correct free body diagram (FBD) of the roller when the roller is just about to climb over the step.

A | |

B | |

C | |

D |

Question 8 Explanation:

Weigh = W

Note:

(i) When the cylinder is about to make out of the curb, it will loose its contact at point A, only contact will be at it B.

(ii) At verge of moving out of curb, Roller will be in equation under W, F and contact force from B and these three forces has to be concurrent so contact force from B will pass through C.

(iii) Even the surfaces are rough but there will be no friction at B for the said condition.

FBD

Question 9 |

The members carrying zero force (i.e. zero-force members) in the truss shown in the
figure, for any load P \gt 0 with no appreciable deformation of the truss (i.e. with no
appreciable change in angles between the members), are

BF and DH only | |

BF, DH and GC only | |

BF, DH, GC, CD and DE only | |

BF, DH, GC, FG and GH only |

Question 9 Explanation:

If at any joint three forces are acting out of which
two of them are collinear then force in third member
must be zero.

For member ED look at joint E.

Similarity look for other members.

For member ED look at joint E.

Similarity look for other members.

Question 10 |

The figure shows an idealized plane truss. If a horizontal force of 300 N is applied atpoint A, then the magnitude of the force produced in member CD is ______ N.

0 | |

10 | |

100 | |

200 |

Question 10 Explanation:

Adopting method of joints and taking FBD of joint B

F_{BC} = 0 (zero force member)

Further by taking FBD of joint C

F_{CD} = 0

There are 10 questions to complete.

Solution to question no. 10 is incorrect. Please check.

Thank You Gliesha,

We have updated the solution.

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