Friction

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}

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}

We can solve this question by moment or by force
triangle as this is condition of equilibrium (just to topple)
\begin{aligned} \sum M_{Q} &=0 \\ P \times 2-m g \times 1 &=0 \\ P &=\frac{m g}{2}=\frac{981}{2} \\ &=490.5 \mathrm{N} \end{aligned}
as P is horizontal force and it will be over come
by friction force
P=\left(f_{s}\right)_{\max } [ as this is at verge of motion ]
\begin{aligned} 490.5 &=\mu_{s} m g \\ \mu_{s} &=0.5 \end{aligned} OR

\begin{aligned} \tan \theta &=\frac{2}{1} \\ \theta &=63.4349^{\circ} \end{aligned}
Wardrobe is about to topple about pt Q so there
will be contact force from point Q on the wardrobe


\begin{aligned} \frac{W}{P} &=\tan \theta=2 \\ P &=\frac{W}{2}=\frac{981}{2}=490.5 \mathrm{N} \end{aligned}
Two components of R will balance W and P and
horizontal component (parallel to contacting
surface is called friction and perpendicular to
contacting surface is called normal reaction)
\begin{aligned} R \cos \theta &=\text { friction force } \\ &=P(\text { in magnitude }) \\ \left(f_{s}\right)_{\max } &=P=490.5 \\ \text{and}\qquad \left(f_{s}\right)_{\max } &=\mu_{s} m g\\ \mu_{s} m g &=490.5 \\ \mu_{s} &=0.5 \end{aligned}

\begin{aligned} & v^{2}=u^{2}+2 a s \\ & u=0 \quad(\text { given }) \\ \therefore \quad & \quad s=\frac{v^{2}}{2 a}=\frac{v^{2}}{2 \mu g \cos 45^{\circ}} \\ \therefore \quad & \quad s=\frac{400}{2 \times 0.5 \times 9.81 \times \cos 45^{\circ}} \\ &=57.664 \mathrm{m} \end{aligned}

Crate will acquire max acceration when \left(f_{s}\right)_{\max } will
be acting on crate by truck and at that condition
acceleration of truck and crate wil be same.
Increasing acceleration of truck after that will cause
slipping of crate on truck.
\left[\left(f_{s}\right)_{\max }\right]_{C T}=\mu N_{C T}

\begin{aligned}\left[\left(f_{s}\right)_{\max }\right]_{C r}-m g \sin 10^{\circ} &=m a \\ \mu m g \cos 10^{\circ}-m g \sin 10^{\circ} &=m a \\ a &=1.1948 \mathrm{m} / \mathrm{s}^{2} \end{aligned}

\begin{aligned} N &=m g=200 \mathrm{N} \\(f_{s})_{max} &=\mu_{s} N \\ &=0.4 \times 200=80 \end{aligned}
upto 8 \sec [P=10 t] \rightarrow no motion
\Rightarrow from 8 sec to 10 sec
\begin{aligned} f_{k} &=\mu_{k} N \\ &=0.2 \times 200=40 \mathrm{N} \end{aligned}
Average applied force
=\frac{P_{\max }+P_{\min }}{2}
=\frac{100+80}{2}=90 \mathrm{N}
Average resultant force = Avg. applied force kinetic friction
=90-40=50 N
Average acceration
\begin{aligned} &=\frac{50 \times 9.8}{200} \\ \Rightarrow \quad V &=u+a t\\ u&=0\\ a&=\frac{50 \times 9.8}{200} \\ t&=2 \text{second}\\ \therefore V&=\frac{50 \times 9.8}{200} \times 2=4.9 \mathrm{m} / \mathrm{s} \end{aligned}

\begin{aligned} F &=\mu\left(W_{S}+W_{R}\right)+\mu W_{R} \\ F &=\mu\left(W_{S}+2 W_{R}\right) \\ &=0.4(150+2 \times 100) \times 9.81 \\ &=1373.4 \mathrm{N}=1.37 \mathrm{kN} \end{aligned}

\begin{aligned} \text { Friction force } &=\mu \times N \\ \Sigma F_{V} &=0\\ \therefore \qquad N-m g &=0 \\ \text{Friction force} &=m g=1 \times 9.8 \\ &=0.1 \times 9.8 \\ &=0.98 \mathrm{N} \end{aligned}
since applied force is less than friction force, hence friction force acting on body at equlibrium is 0.8 \mathrm{N}

Free body Diagram

Assume normal reaction =R (Newton)
Balancing force in horizontal direction
(\Sigma F x=0)
\begin{aligned} \mu R &=100 \\ (0.2) R &=100 \\ R &=500 \mathrm{N} \end{aligned}
Balancing the force in vertical direction
\begin{aligned} (\Sigma F y&=0) \\ T+R-W &=0 \\ \Rightarrow T+500-981 &=0 \\ T &=481 \mathrm{N} \end{aligned}

FBD of block:

N=m g \cos \theta
From Newton's second law
mg \sin \theta-\mu m g \cos \theta=m a
a=g \sin \theta-\mu g \cos \theta
S=u t+\frac{1}{2} a t^{2}
S=0+\frac{1}{2}(g \sin \theta-\mu g \cos \theta) t^{2}
t=\sqrt{\frac{2 s}{g \cos \theta[\tan \theta-\mu]}}