NLM

Let $A$ be a point on cable 1
$L _ {1} = S _ {A} + 2 S _ {B}$
$0 = v _ {A} + 2 v _ {B}\tag{... (1}$
$L _ {2} = S _ {C} + \left(S _ {C} - S _ {B}\right)$
$0 = 2 v _ {C} - v _ {B}\tag{... (2}$

Combine (1) and (2) to obtain
$v _ {C} = - \frac {1}{4} v _ {A} = - \frac {1}{4} (3 2 0) = - 8 0 \mathrm{mms} ^ {- 1}$
So, W rises by $h = 80(5) = 400 \text{ mm}$

Let the datum be passing through the fixed pulley as shown

The length of the two ropes in terms of the position coordinates $x_{A}$ , $x_{B}$ and $x_{C}$ are
$x _ {A} + 2 x _ {C} = l _ {1}\tag{... (1}$
and $x_{B} + (x_{B} - x_{C}) = l_{2}$
$\ldots(2)$
Eliminating $s_{C}$ from equations (1) and (2),
$4 x _ {B} + x _ {A} = 2 l _ {2} + l _ {1}\tag{... (3}$
Taking the time derivative of equation (3), we get
$4 v _ {B} + v _ {A} = 0$
Since $v_{A}=4\ ms^{-1}$ , so we get
$\begin{array}{l l} & 4 v _ {B} + 4 = 0 \\ \Rightarrow & v _ {B} = - 1 \mathrm{ms} ^ {- 1} = 1 \mathrm{ms} ^ {- 1}, \text { upwards } \\ \Rightarrow & v _ {B} = 1 0 0 \mathrm{cms} ^ {- 1}, \text { upwards } \end{array}$

Let the datum be passing through both the fixed pulleys as shown.

The length of the rope in terms of the position coordinates $x_{A}$ and $x_{M}$ is
$3 x _ {A} + x _ {M} = l\tag{... (1}$
Taking the time derivative of equation (1), we get
$3 v _ {A} + v _ {M} = 0$
Taking downward direction as positive, so we have
$v _ {M} = 1 0 \mathrm{ms} ^ {- 1}$
$\Rightarrow 3 v _ {A} + 9 = 0$
$\Rightarrow v_{A} = -3\mathrm{ms}^{-1} = 3\mathrm{ms}^{-1}$ , upwards.

(a) Let $a$ be the acceleration of each block and $T_{1}$ and $T_{2}$ be the tensions, in the two strings as shown in figure.

Taking the three blocks and the two strings as the system.

Using $\Sigma F_{x} = ma_{x}$

b) Free body diagram (showing the forces in x-direction only) of 3 kg block and 1 kg block are shown in figure.
Using $\Sigma F_{x} = ma_{x}$
For 3 kg block,
$\begin{array}{r l} & T _ {2} = (3) (a) \\ \Rightarrow & T _ {2} = (3) (2) = 6 \mathrm{N} \end{array}$

Let a be the acceleration with which the masses move and $T_{1}$ and $T_{2}$ be the tensions in left and right strings. Friction on mass A is $\mu mg = 8 N$ . Then equations of motion of masses A, B and C are
For mass $A$
$T _ {1} - 8 = 4 a\tag{... (1}$
For mass $B$
$T _ {2} = 8 a$......(2)
For mass $C$
$2 0 0 - T _ {1} - T _ {2} = 2 0 a$......(3)
Adding the above three equations, we get
3 2 a = 1 9 2
$\Rightarrow a = 6 \mathrm{ms} ^ {- 2}$
From equations (1) and (2), we have
$T _ {2} = 4 8 \mathrm{N} \text {and} T _ {1} = 3 2 \mathrm{N}$

When force F is applied at the end of the string, the tension in the lower part of the string is also F as shown. If T is the tension in string connecting the pulley and the block, then,

$\begin{array}{r l} & {\mathrm{Also,} T = m a = (2 0 0) (1) = 2 0 0 \mathrm{N}} \\ & {\Rightarrow \quad 2 F = 2 0 0 \mathrm{N}} \\ & {\Rightarrow \quad F = 1 0 0 \mathrm{N}} \end{array}$

Free body diagram of crate A w.r.t. ground is shown in figure.
Equation of motion is given by
$1 0 0 - N = 1 0 a _ {A}\tag{... (1}$
Further, we have $a_{A}$ cosec30° = acceleration of crate
$\Rightarrow 2 a _ {A} = 2$
$\Rightarrow a _ {A} = 1 \mathrm{ms} ^ {- 2}$

Substituting in equation (1) we get $N = 90\mathrm{N}$

Let $\theta$ be the required angle, then maximum retardation is
$\begin{array}{r l} & a = \frac {f _ {L} - m g \sin \theta}{m} \\ \Rightarrow & a = \frac {\mu m g \cos \theta - m g \sin \theta}{m} \\ \Rightarrow & a = \mu g \cos \theta - g \sin \theta = g (\mu \cos \theta - \sin \theta) \end{array}$

Now, using $v^{2} = u^{2} - 2as$ , we get
$0 = (1 0) ^ {2} - 2 g (\mu \cos \theta - \sin \theta) (5)$
$\Rightarrow \mu g \cos \theta - g \sin \theta = 1 0$
$\begin{array}{r l} & {\Rightarrow \Bigg (\frac {4}{3} \Bigg) (1 0) \cos \theta - (1 0) \sin \theta = 1 0} \\ & {\Rightarrow 4 \cos \theta - 3 \sin \theta = 3} \\ & {\Rightarrow 4 \cos \theta = 3 (1 + \sin \theta)} \\ & {\Rightarrow 1 6 \cos^ {2} \theta = 9 (1 + \sin^ {2} \theta + 2 \sin \theta)} \\ & {\Rightarrow 1 6 (1 - \sin^ {2} \theta) = 9 (1 + \sin^ {2} \theta + 2 \sin \theta)} \\ & {\Rightarrow 2 5 \sin^ {2} \theta + 1 8 \sin \theta - 7 = 0} \\ & {\Rightarrow \sin \theta = \frac {- 1 8 \pm \sqrt {3 2 4 + 7 0 0}}{5 0}} \\ & {\Rightarrow \sin \theta = \frac {\sqrt {1 0 2 4} - 1 8}{5 0} = \frac {3 2 - 1 8}{5 0} = \frac {1 4}{5 0}} \\ & {\{\because \mathrm{Rejectingnegativesign} \}} \end{array}$
$\Rightarrow \theta \approx 1 6 ^ {\circ}$

Since the man is standing stationary w.r.t. the belt, so acceleration of the man equals the acceleration of the belt i.e. $a = 1 \, ms^{-2}$
Net force on the man is $F_{net} = Ma = 65 \times 1 = 65$ N
Limiting friction is
$f _ {L} = \mu \mathrm{mg}$
Let the man remains stationary with respect to the belt for maximum acceleration $a_{0}$ , then
$\begin{array}{r l} & m a _ {0} = f _ {L} = \mu \mathrm{mg} \\ \Rightarrow & a _ {0} = \mu g = (0. 2) (1 0) = 2 \mathrm{ms} ^ {- 2} \\ \Rightarrow & a _ {0} = 2 0 0 \mathrm{cms} ^ {- 2} \end{array}$

Equations of motion in this case are,
$T \cos \theta = \mu N\tag{... (1}$
and $N = W - T\sin \theta$
... (2)

(a) From equations (1) and (2), we get
$W = \frac {T (\cos \theta + \mu \sin \theta)}{\mu}\tag{... (3}$
b) For W to be maximum
$\mathrm{Since} \sin (1 0 ^ {\circ}) \cos (1 0 ^ {\circ}) = \frac {1}{6}$

$\Rightarrow 2 \sin (1 0 ^ {\circ}) \cos (1 0 ^ {\circ}) = \frac {1}{3}$
$\Rightarrow \sin (2 0 ^ {\circ}) = \frac {1}{3}$
$\Rightarrow \tan (2 0 ^ {\circ}) = \frac {1}{2 \sqrt {2}}$
So, $\theta = 20^{\circ}$
From equation (3), we get
$\begin{array}{l l} & W = \frac {1 2 0 0 \left(\cos (2 0 ^ {\circ}) + \frac {1}{2 \sqrt {2}} \sin (2 0 ^ {\circ})\right)}{\frac {1}{2 \sqrt {2}}} \\ \Rightarrow & W = \frac {1 2 0 0 \left(\frac {2 \sqrt {2}}{3} + \frac {1}{2 \sqrt {2}} \frac {1}{3}\right)}{\frac {1}{2 \sqrt {2}}} \\ \Rightarrow & W = 4 0 0 (2 \sqrt {2}) \left(2 \sqrt {2} + \frac {1}{2 \sqrt {2}}\right) \\ \Rightarrow & W = 4 0 0 (8 + 1) \\ \Rightarrow & W = 3 6 0 0 \mathrm{N} \end{array}$

(a) When the truck accelerates eastwards, force of friction on mass is eastwards, because the block has a tendency to move westwards w.r.t. the truck. So,
$f _ {\text { required }} = \text { mass } \times \text { acceleration } = 3 0 \times 1. 8 = 5 4 \mathrm{N}$
Since it is less than $\mu_{s}mg$ , hence
$f = 5 4 \mathrm{N} (\text { eastwards })$
b) When the truck accelerates westwards then similar to the argument given earlier, force of friction is westwards. So,
$f _ {\text { required }} = \text { Mass } \times \text { Acceleration } = 3 0 \times 3. 8 = 1 1 4 \mathrm{N}$
Since it is greater than $\mu_{s}mg$ , hence
$f = f _ {k} = \mu_ {k} m g \approx 5 9 \mathrm{N} (\text { westwards })$

$100 = \frac{mv^2}{r} = \frac{(4)(v^2)}{(1)}$
$\Rightarrow v = 5 \mathrm{ms} ^ {- 1}$
Now, $5 = (\mu_k g)t$
$\left\{\because v = a t \right\}$
$\Rightarrow t = \frac {5}{(0 . 1) (1 0)} = 5 \mathrm{s}$

For $10\mathrm{kg}$ , $10g - T_{2} = 10a$
... (1)
For $3\mathrm{kg}$ , $T_{2} - (T_{1} + 0.6g) = 3a$
$\ldots(2)$
For $2\mathrm{kg}$ , $T_{1} - 0.6g = 2a$
... (3)
Solving these equations, we get
$a = 6 \mathrm{ms} ^ {- 2}$

Since, $v_{C}^{2}=v_{A}^{2}+2a_{T}s$
Tangential acceleration in interval AC is
$a _ {T} = \frac {v _ {C} ^ {2} - v _ {A} ^ {2}}{2 s} = 1. 4 4 7 \mathrm{ms} ^ {- 2}$
Centripetal i.e. Normal acceleration $a_{N}$ at C is
$a _ {N} = \frac {v ^ {2}}{r _ {C}} = 2. 4 1 \mathrm{ms} ^ {- 2}$
Tangential force $F_{T}$ at C is
$F _ {T} = m a _ {T} = 1 5 0 0 \times 1. 4 4 7 = 2 1 7 0 \mathrm{N}$
Normal force $F_{N}$ at C is
$F _ {N} = m a _ {N} = 1 5 0 0 \times 2. 4 1 = 3 6 2 0 \mathrm{N}$
Total force at $C$ is
$F = \sqrt {F _ {N} ^ {2} + F _ {T} ^ {2}}$
$\Rightarrow F = \sqrt {(2 1 7 0) ^ {2} + (3 6 2 0) ^ {2}}$
$\Rightarrow F = 4 2 2 0 \mathrm{N}$

$R=\frac{v^{2}}{\mu g}=\frac{5\times5}{0.5\times10}=5\ m$

$m_{A}=m_{B}=m_{C}=m$
Since, $F = \left( m_{A} + m_{B} + m_{C} \right) g \sin \theta$ and $N = m_{C} g \sin \theta$
$\Rightarrow N = \frac {m _ {C} F}{m _ {A} + m _ {B} + m _ {C}} = \left(\frac {m}{3 m}\right) 1 5 = 5 \mathrm{N}$

There is no tendency of relative slipping between blocks.

For equilibrium,
$1 0 = 8 + T\tag{... (1}$
$T + f _ {2} = 2 0$
$\Rightarrow f _ {2} = 1 8 \mathrm{N}\tag{... (2}$

Let $T$ be the tension in the string. The upward force exerted on the clamp is
$T \sin (3 0 ^ {\circ}) = \frac {T}{2}$

$\begin{array}{r l} & {\mathrm{Now} \frac {T}{2} = 4 0 \mathrm{N}} \\ & {\Rightarrow T = 8 0 \mathrm{N}} \\ & {\Rightarrow a = \frac {T - m g}{m} = \frac {8 0 - 5 0}{5} = 6 \mathrm{ms} ^ {- 2}} \end{array}$

$v = 2 t ^ {2}$
$ \Rightarrow a = 4 t$

At t = 1 s, slipping occurs, so we have
$\Rightarrow m a = \mu_ {s} m g$
$\Rightarrow 4 t = \mu_ {s} g$
$\Rightarrow \mu_ {s} = 0. 4$
So in the interval from t = 1 to t = 3 s, we observe that the slipping occurs between t = 1 and t = 2 s.
$\Rightarrow 4 t - \mu_ {k} g = \frac {d v}{d t}$
Integrating, we get
$v = \left(2 t ^ {2} - \mu_ {k} g t\right) \Big | _ {1} ^ {2}$
$\Rightarrow \mu_ {k} = 0. 3$
$\Rightarrow \frac {3 \mu_ {s}}{\mu_ {k}} = 4$

$a = \frac{160 - 120}{40} = 1\mathrm{ms}^{-2}$

Since wedge is stationary and block is slipping down with a constant velocity, so normal force on the wedge due to the table is

$N = (m + M) g = 8 \mathrm{N}$

Let tension the thread connecting $m_{2}$ to the pulley $P_{1}$ be T and acceleration of $m_{2}$ be a. Then tension in the thread connecting lowest pulley $P_{3}$ to $m_{1}$ is 4 T and acceleration of $m_{1}$ is $\frac{a}{4}$ .
For $m_{1}$ : 40 - 4T = 4 $\left(\frac{a}{4}\right)$
For $m_{2}$ : $T = 1a$
$\Rightarrow 4 0 - 4 a = a$
$\Rightarrow 5 a = 4 0$
$\Rightarrow a = 8 \mathrm{ms} ^ {- 2}$

Pseudo force is zero w.r.t. an observer in an Inertial or ground frame.

The system can be redrawn as

$\Rightarrow a = \frac {4 m g}{8 m} = \frac {g}{2} = 5 \mathrm{ms} ^ {- 2}$
Since, $v = u + at$
$\Rightarrow v = 0 + (2) (5) = 1 0 \mathrm{ms} ^ {- 1}$
$\Rightarrow v = 1$ decametre per second