NLM

$f_{\max} = \mu mg \cos \theta = 0.6 \times mg \times \frac{\sqrt{3}}{2} = 0.52 mg$

$m g \sin \theta = \frac {m g}{2} = 0.5 m g$
Since, $f_{\mathrm{max}} > mg\sin \theta$
Block will decelerate and come to rest.

$N = w + F \sin \theta$
:
$f _ {\mathrm{max}} = \mu N = (\tan \phi) (w + F \sin \theta)$

To move the body, $F \cos \theta = f_{\max} = (\tan \phi) (w + F \sin \theta)$
Solving this equation, we get
$F = \frac {w \sin \phi}{\cos (\theta + \phi)}$

Maximum acceleration due to friction of mass m over mass 2m can be $\mu g$ . Now, for the whole system,
$a = \frac {\text { Net pulling force }}{\text { Total mass }}$
:
$\mu g = \frac {m g}{4 m}$
or
$\mu = \frac {1}{4}$

$2T_{1}\cos45^{\circ}=mg$
:
... (i)




:

From the free body diagram,

So, as the man moves up, $\theta$ increases, $\cos\theta$ decreases.
:

Now, when the man is at depth h, then the force is
$F = \frac {m g}{2 \cos \theta} = \frac {m g \sqrt {h ^ {2} + \frac {d ^ {2}}{4}}}{2 h} = \frac {m g}{4 h} \sqrt {d ^ {2} + 4 h ^ {2}}$

Writing equation of motion for two weights

Acceleration relative to pulley Acceleration relative to ground
$w _ {1} - T = \frac {w _ {1}}{g} (a _ {r} - a)\tag{... (i}$
$T - w _ {2} = \frac {w _ {2}}{g} (a _ {r} + a)\tag{... (ii}$
Solving Eqs. (i) and (ii), we get
$T = \frac {4 w _ {1} w _ {2}}{w _ {1} + w _ {2}} \mathrm{with} a = g$

Acceleration of system,
$a = \frac {F - f}{M + m} = \frac {2 0 - 1 0}{6 + 4} = 1 \mathrm{ms} ^ {- 2} (\text { towards left })$
Let $F_{0}$ be the reading of dynamometer, then equation of motion of mass m would be
⇒
$\begin{array}{r} F _ {0} - f = m a \\ F _ {0} = f + m a = 1 0 + (4) (1) = 14 \mathrm{N} \end{array}$

Net pulling force on the system should be zero, as velocity is constant. Hence,
$\begin{array}{r l} & m _ {A} g \sin 3 0 ^ {\circ} = \mu m _ {A} g \cos 3 0 ^ {\circ} + m _ {B} g \\ & m _ {B} = \left(\frac {m _ {A}}{2}\right) - \left(\frac {\mu m _ {A} \sqrt {3}}{2}\right) \\ & = 1 0 \left[ \frac {1}{2} - 0.2 \times \frac {\sqrt {3}}{2} \right] = 3.3 \mathrm{kg} \end{array}$

$5 g - T = 5 a\tag{... (ii}$
From Eqs. (i) and (ii), we get
$\begin{array}{r} 5 \times 9.8 - 0.5 \times 5 \times 9.8 = 1 0 a \\ 0.5 \times 5 \times 9.8 = 1 0 a \\ a = 2.4 5 \mathrm {m/ s^ {2}} \end{array}$
From Eq (ii), we get $5 \times 9.8 - 5 \times 2.4 5 = T$
$T = 36.75 \mathrm{N}$

Since, the block is at rest under two forces
(i) weight of block.
(ii) contact force from the plane (resultant of force of friction and normal reaction).
Contact force should be equal to weight (or 30 N) in upward direction. Because under the action of two forces, a body remains in equilibrium when both the forces are equal in magnitude, but opposite in direction, i.e. 30 N.

When the system of two blocks of masses $M$ and $m$ is either at rest or moving with the same speed, no net force acts on them.

In this equilibrium state,
$N = m g - Q \cos \theta$

$P + Q \sin \theta = \mu N = \mu (m g - Q \cos \theta)$
$\mu = \frac {(P + Q \sin \theta)}{(m g - Q \cos \theta)}$

$m_{1}g > m_{2}g \sin\theta + \mu m_{2}g \cos\theta$
⇒
$\frac {m _ {1}}{m _ {2}} > \sin \theta + \mu \cos \theta$

$N = Mg - F\cos \theta = Mg - Mg\cos \theta = 2Mg\sin^2\frac{\theta}{2}$
Further, block can be pulled, if $F \sin\theta \geq \mu N$
$\begin{array}{l l} \Rightarrow & 2 M g \sin \frac {\theta}{2} \cdot \cos \frac {\theta}{2} \geq 2 \mu M g \sin^ {2} \frac {\theta}{2} \\ \Rightarrow & \cot \left(\frac {\theta}{2}\right) \geq \mu \end{array}$

$a = \frac{2F - F}{2m} = \frac{F}{2m}$
(towards left)
Writing equation of right hand side block,
$\begin{array}{r l} 2 F - N \sin 3 0 ^ {\circ} & = m a = \frac {F}{2} \\ \frac {N}{2} & = 2 F - \frac {F}{2} = \frac {3 F}{2} \Rightarrow N = 3 F \end{array}$

In critical case, weight of hanging part = force of friction of the part of rope lying on table.
$\frac {m}{l} \cdot l _ {1} g = \mu \frac {m}{l} (l - l _ {1}) g$
Solving above equation, we get
$l _ {1} = \left(\frac {\mu}{1 + \mu}\right) l$

Block $A$ moves due to friction. Maximum value of friction can be $\mu m_A g$ . Therefore, maximum acceleration of $A$ can be $\frac{\mu m_A g}{m_A}$ or $\mu g = 0.2g = \frac{g}{5}$ . When force $2mg$ is applied on lower block common acceleration (if both move together) will be
block, common acceleration (if both move together) will be
$\begin{array}{c} m \xrightarrow {0.2 \text { mg }} \\ 3 m \xrightarrow {2 \text { mg }} \end{array}$
$a = \frac {\mathrm{Netforce}}{\mathrm{Totalmass}} = \frac {2 m g}{4 m} = \frac {g}{2}$
Since, a = g/2 is greater than maximum acceleration of A which can be given to it by friction. Therefore, slipping will take place.
$\begin{array}{r l} & {a _ {A} = 0.2 g = \frac {g}{5}} \\ & {a _ {B} = \frac {2 m g - 0.2 m g}{3 m} = 0.6 g = \frac {3 g}{5}} \\ & {a _ {A B} = a _ {A} - a _ {B} = \left(\frac {g}{5} - \frac {3 g}{5}\right)} \\ & {\qquad = \frac {- 2 g}{5} (\mathrm{inbackwarddirection})} \end{array}$

$f_{1}$ = maximum value of friction between m and 2m = $\mu mg$ $f_{2}$ = maximum value of friction between 2m and ground $=\frac{\mu}{3}(3m)g=\mu mg$
Now, let us draw free body diagram of m and 2m in all four cases.
(i)
$\begin{array}{c} \frac {\mu}{3} m g \quad \boxed {m} \longrightarrow F = \frac {\mu}{3} m g \\ T = 0 \quad \boxed {2 m} \longrightarrow \frac {\mu}{3} m g \\ \frac {\mu}{3} m g \\ \mu m g \quad \boxed {m} \longrightarrow F = \mu m g \end{array}\tag{ii}$
$\begin{array}{c} T = 0 \\ \leftarrow \quad \boxed {2 m} \quad \mu m g \\ \mu m g \end{array}\tag{iii}$
$\mu m g \boxed {m} \longrightarrow F = 2 \mu m g$
$\begin{array}{c} T = 0 \\ \leftarrow \quad \boxed {2 m} \quad \mu m g \\ \mu m g \end{array}$
$\mu m g \boxed {m} \longrightarrow F = 3 \mu m g$
$\begin{array}{c} T = 0 \\ \leftarrow \quad \boxed {2 m} \quad \mu m g \\ \mu m g \end{array}$
As tension is zero in all four cases. So, option (c) is incorrect.

For block,
$a = g \sin \theta = \frac {g}{\sqrt {2}}$

$\begin{array}{r} a _ {H} = a _ {V} = a \cos 4 5 ^ {\circ} \\ = \frac {g}{\sqrt {2}} \times \frac {1}{\sqrt {2}} = \frac {g}{2} \\ \frac {N}{\sqrt {2}} = \frac {m g}{2} \\ m g - \frac {N}{\sqrt {2}} = \frac {m g}{2} \end{array}\tag{... (i}$
... (ii)
For wedge,

$\begin{array}{c} {N ^ {\prime} = \frac {N}{\sqrt {2}} + 2 m g} \\ {\mu N ^ {\prime} = \frac {N}{\sqrt {2}}} \end{array}\tag{... (iii}$
... (iv)
Solving above four equations, we get
$\mu = 0.2$

2T = 250

Also, $T + 0.4F = 150$
$F = 6 2.5 \mathrm{N}$

$N_{1}\sin37^{\circ}=N_{2}\sin74^{\circ}$
$\Rightarrow N_{1} = 2N_{2}\cos 37^{\circ}$

Now, $N_{1}\cos 37^{\circ} - N_{2}\cos 74^{\circ} = mg$
$\Rightarrow 2N_{2}\cos^{2}37^{\circ} - N_{2}(2\cos^{2}37^{\circ} - 1) = mg$

$T \sin \theta - mg \sin \alpha = ma$

and $T\cos \theta = mg\cos \alpha$
From these two equations, we get
$\tan \theta = \frac {a + g \sin \alpha}{g \cos \alpha} \Rightarrow \theta = \tan^ {- 1} \left(\frac {a + g \sin \alpha}{g \cos \alpha}\right)$

A

$mg \sin \theta = 10 \times 10 \times \frac{1}{2} = 50 \, N$
$f_{\mathrm{friction}} = \mu mg\cos \theta = 0.7\times 10\times 10\times \frac{\sqrt{3}}{2} = 60.62\mathrm{N}$
Since, $\mu mg \cos \theta$ is more, block will remain stationary.

A