Engineering Mechanics

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.)

For pure rolling


\begin{array}{l} v_{P}=v=(P R) \omega \quad \ldots(i)\\ v_{Q}=2 v=(Q R) \omega \quad \ldots(ii) \end{array}
Divide by (ii) to (i),
2=\frac{Q R}{P R} \Rightarrow Q R=2(P R)
\begin{aligned} P R+Q R &=2 r \\ P R+2(P R) &=2 r \\ P R &=\frac{2}{3} r \\ \text{From equation(i) }\quad v &=\left(\frac{2}{3} r\right) \omega \Rightarrow \omega=\frac{3 v}{2 r} \end{aligned}

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.

Let V_{1} is the speed of 3 kg mass after collision
\mathrm{V}_{2} is the speed of m kg mass after collision
\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}
By linear momentum conservation
\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}
By using (1), (2) \& (3) we get
m = 1 kg (or) 9 kg

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}

Maximum friction force, f_{\max }=\mu \mathrm{N}=0.2 \times 10 \times 9.81=19.62 \mathrm{N}
Applied force, P=10 \mathrm{N} \lt \mathrm{f}_{\max }
\therefore Friction force = Applied force =10 \mathrm{N}