Engineering Mechanics

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).

About point P there is no external torque.
[Because Torques of N and 'mg' balances and friction force pass through point P, hence its torque is zero]
\therefore \left [ \frac{d\vec{L}}{dt} =\vec{\tau }_{ext}\right ]_{\text{about P}}=0
where \vec{\tau }= Torque and \vec{L }= Angular momentum
\begin{aligned} \vec{L_i}&=\vec{L_f}\\ I_0\omega &=I_0\omega '+mvr\\ \frac{mr^2}{2}\omega &=\frac{mr^2}{2}\omega'+m\omega 'r^2\\ \therefore \omega '&=\frac{\omega }{3}=\frac{5}{3}\\ \Rightarrow v&=\omega ' \times r\\&=\frac{5}{3} \times 0.15=0.25 m/s \end{aligned}

By pythagoras theorem:
PQ = 13 m
By Geometry:
\begin{aligned} 5^2-PR^{'2}&=10^2-AR^{'2}\\ QR'&=13-PR'\\ \Rightarrow PR'&=\frac{25}{13}m\\ \Rightarrow QR'&=\frac{144}{13}m\\ \Sigma M_P&=0\\ \Rightarrow F(PR')&=N_Q(PQ)\\ \Rightarrow N_Q&=\frac{25}{169}F\\ \Sigma F&=0\\ \Rightarrow N_P&=F-\frac{25}{169}F\\ &=\frac{144}{169}F \end{aligned}
Now, f_P=\mu _PN_P, f_Q=\mu _QN_Q, \Sigma F_x=0
\begin{aligned} \mu _PN_P&=\mu _QN_Q\\ \frac{\mu _Q}{\mu _P}&=\frac{N_P}{N_Q}\\ &=\frac{144}{25}=5.76 \end{aligned}

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}

After forces are applied block is just about move (mass is negligible). Calculate coefficient of friction
\begin{aligned} F_{H} &=750 \mathrm{~N} \\ F_{V} &=900 \mathrm{~N} \\ \theta &=30^{\circ} \end{aligned}


\begin{aligned} N &=900 \cos \theta+750 \sin \theta \\ N &=900 \cos 30^{\circ}+750 \sin 30^{\circ} \\ &=1154.4228 \mathrm{~N} \\ F_{\max }+900 \sin 30^{\circ} &=750 \cos 30^{\circ} \\ \mu N &=199.519 \\ \mu &=\frac{199.519}{1154.4228} \\ \mu &=0.1728 \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.)