FBD, Equilirbium, Plane Trusses and Virtual work

If at a point 3 member are meeting and two are colinear then in 3rd member force will be zero.
F_{GH}=0

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

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)

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).

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

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}

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.)

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

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.

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