FBD, Equilirbium, Plane Trusses and Virtual work

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

BD is a zero force member
F.B.D of joint C is shown below,


For equilibrium of joint C
\begin{array}{l} \sum \mathrm{F}_{\mathrm{y}}=0, \quad \mathrm{F}_{\mathrm{CD}} \cos 45=5 \\ \quad \therefore \mathrm{F}_{\mathrm{CD}}=\frac{5}{\cos 45} \\ \sum \mathrm{F}_{\mathrm{x}=0}, \quad \mathrm{F}_{\mathrm{BC}}=\mathrm{F}_{\mathrm{CD}} \cos 45=\frac{5}{\cos 45} \times \cos 45=5 \mathrm{kN} \end{array}

We analyse this problem in the frame of reference of car.
F.B.D of car is shown below,


As our frame of reference is accelerated hence, we have to apply a pseudo force 'ma' as shown
above, where, f_{1} and f_{2} are friction forces on rear and front wheels respectively.
For vertical equilibrium,
\begin{array}{l} \mathrm{R}_{\mathrm{r}}+\mathrm{R}_{\mathrm{f}}=\mathrm{W} \ldots(i)\\ \Sigma \mathrm{M}_{0}=0\\ W \times \frac{\ell}{2}-\frac{W a}{g} \times h-R_{f} \times \ell=0 \ldots(ii)\\ From (i) \& (ii)\\ R_{f}=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{\ell}\right) a \\ R_{r}=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{\ell}\right) a \end{array}

\begin{aligned} \frac{B C}{A B} &=\sin 45^{\circ} \\ A B &=\frac{B C}{\sin 45^{\circ}}=\frac{1}{\sin 45^{\circ}}=1.4142 \mathrm{m} \\ \text { Total travel, } O A+A B &=1+1.4142 \mathrm{m}=2.414 \mathrm{m} \end{aligned}

m = 2 kg,\; g = 10m/s^{2}
FBD of mass P at point Q


N-m g=m a_{r} \quad at no slip condition
N=m g+\frac{m v^{2}}{R}=2 \times 10+\frac{2 \times 20 \times 20}{2}=420 \mathrm{N}

\begin{aligned} T_{1}+T_{2} &=100 \mathrm{N} \qquad \ldots(i)\\ \Sigma M_{A} &=0 \\ T_{2} \cdot \frac{L}{2} &=100 \times \frac{L}{2} \\ \therefore \qquad T_{2} &=100 \mathrm{N} \\ T_{1} &=0 \mathrm{N} \end{aligned}

When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.
Option (B) is wrong because the ground should not be shown in FBD as the forces due to ground are already depicted.
Option (C) is wrong because x-component is not shown.
Option (D) is wrong because the rotation due to the force acting eccentrically causes moment which is not shown.