Question 1 |

The options show frames consisting of rigid bars connected by pin joints. Which
one of the frames is non-rigid?

A | |

B | |

C | |

D |

Question 1 Explanation:

Nonrigid frames: Relative movement of joints is
large then a frame said to be nonrigid.

Here, P force cause large deformation therefore. It is non rigid.

Here, P force cause large deformation therefore. It is non rigid.

Question 2 |

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 2 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 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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).

There are 5 questions to complete.

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

Thank You Gliesha,

We have updated the solution.

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