# FBD, Equilirbium, Plane Trusses and Virtual work

 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
GATE ME 2022 SET-2   Engineering Mechanics
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
GATE ME 2022 SET-2   Engineering Mechanics
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
GATE ME 2022 SET-2   Engineering Mechanics
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)
GATE ME 2022 SET-2   Engineering Mechanics
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).

 A 12 B 15 C 18 D 22
GATE ME 2022 SET-1   Engineering Mechanics
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).

 A 10 B 20 C 25 D 40
GATE ME 2022 SET-1   Engineering Mechanics
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}
 Question 7
A plane truss $PQRS(PQ=RS, \text{and }\angle PQR=90^{\circ})$ is shown in the figure

The forces in the members $PR$ and$RS$, 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)
GATE ME 2021 SET-2   Engineering Mechanics
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.
 A A B B C C D D
GATE ME 2020 SET-2   Engineering Mechanics
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
 A BF and DH only B BF, DH and GC only C BF, DH, GC, CD and DE only D BF, DH, GC, FG and GH only
GATE ME 2020 SET-1   Engineering Mechanics
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.
 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.
 A 0 B 10 C 100 D 200
GATE ME 2019 SET-2   Engineering Mechanics
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.

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

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