# FBD, Equilirbium, Plane Trusses and Virtual work

 Question 1
The options show frames consisting of rigid bars connected by pin joints. Which one of the frames is non-rigid? A A B B C C D D
GATE ME 2023   Engineering Mechanics
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.
 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)?
 A AB B CD C EF D GH
GATE ME 2022 SET-2   Engineering Mechanics
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 A 0.953 Ns B 1.403 Ns C 0.814 Ns D 1.172 ns
GATE ME 2022 SET-2   Engineering Mechanics
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)
$I_F \times 0.3 =0.083 \times 3.44$
$\Rightarrow I_F =0.951 N-s$
 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). A 1 B 2 C 3 D 4
GATE ME 2022 SET-2   Engineering Mechanics
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)
 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?

MSQ
 A Configuration (P) B Configuration (Q) C Configuration (R) D Configuration (S)
GATE ME 2022 SET-2   Engineering Mechanics
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.

### 7 thoughts on “FBD, Equilirbium, Plane Trusses and Virtual work”

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

• Thank You Gliesha,
We have updated the solution.

2. Options for question no. 13 are not showing up…!

3. Q5 explanation says collinear but it is concurrent system of forces

4. all last 10 years pyqs are added in this website? or not?

• 5. 