Laws of Motion

Let v be speed of the bead when it makes an angle $\theta$ with the vertical direction.

The forces acting on the bead are its weight mg and normal reaction N. Since all the forces acting on the bead are conservative, the mechanical energy of the bead is conserved i.e.,

$mgr(1 - \cos \theta ) = {1 \over 2}m{v^2}$ ..... (1)

The radially inward components of force provide centripetal acceleration to the bead i.e.,

$mg\cos \theta - N = m{v^2}/r$ ...... (2)

Eliminate v² from equations (1) and (2) to get

$N = mg(3\cos \theta - 2)$ ...... (3)

When 3cos$\theta$ > 2, direction of N will be same as shown in figure, i.e., radially outward.

When 3cos$\theta$ < 2, direction of N will be radially inward. Now, normal on the bead is opposite to the force applied on the wire by the bead, following Newton's third law. Therefore, the force applied on the bead by the wire is radially inwards initially and radially outwards later, as it moves from A to B.

The forces on the block of mass m₁ are its weight m₁g, normal reaction from the inclined plane N₁, and the reaction from the second block R.

Similarly, forces on the block of mass m₂ are m₂g, N₂, R, and the frictional force f. If $\theta$ is slowly increased, f starts increasing and attains its maximum value f = $\mu$N₂ at $\theta$ = $\theta$_r (angle of repose). The blocks are stationary if $\theta$ $\le$ $\theta$_r otherwise they are moving. Consider the limiting case, $\theta$ = $\theta$_r, when the blocks are at rest and

$f = \mu {N_2}$ ...... (1)

Apply Newton's second law to m₁,

$R = {m_1}g\sin \theta $ ....... (2)

${N_1} = R = {m_2}g\cos \theta $ ....... (3)

and to m₂,

$f = R + {m_2}g\sin \theta $ ...... (4)

${N_2} = {m_1}g\cos \theta $ ...... (5)

Eliminate R, N₂ and f from equations (1)-(5) to get

${\theta _r} = {\tan ^{ - 1}}\left( {{{\mu {m_2}} \over {{m_1} + {m_2}}}} \right) = {\tan ^{ - 1}}\left( {{{0.3 \times 2} \over {1 + 2}}} \right)$

$ = {\tan ^{ - 1}}(0.2) = 11.5^\circ $

Thus, for $\theta$ = 5$^\circ$ and $\theta$ = 10$^\circ$, blocks are at rest with frictional force

$f = R + {m_2}g\sin \theta = ({m_1} + {m_2})g\sin \theta $

For $\theta$ = 15$^\circ$ and $\theta$ = 20$^\circ$, blocks are moving with frictional force

$f = \mu {N_2} = \mu {m_2}g\cos \theta $, (limiting value).

The forces acting on the block are F = 1 N towards the left, weight mg = 0.1 $\times$ 10 = 1 N downwards, normal force N, and the frictional force f. Resolve F and mg along and perpendicular to the inclined plane.

When $\theta$ = 45$^\circ$, the net force that brings the block down is

${F_d} = mg\sin \theta - F\cos \theta = {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 0$.

Thus, the block is stationary if $\theta = 45^\circ $. When $\theta > 45^\circ $, the force ${F_d} > 0$, and hence the block has a tendency to move down. Thus, the frictional force f acts on the block upwards i.e., towards Q.

Resolve T along the horizontal and the vertical directions. As the particle moves in a horizontal plane, net vertical fore on it is zero. Net horizontal force on it provides the necessary centripetal acceleration for circular motion. Apply Netwon's second law in the horizontal direction to get

$T\sin \theta = m{\omega ^2}r = m{\omega ^2}(l\sin \theta )$

$T = m{\omega ^2}l$

$\omega = \sqrt {{T \over {ml}}} $

Substituting the given values, we get

$\omega = \sqrt {{{324} \over {0.5 \times 0.5}}} = 36$ rad/s

The forces acting on the block are its weight mg, normal reaction N, applied force P and frictional force f.

Resolve mg along and normal to the inclined plane and apply Newton's second law to get

$0 = P + f - mg\sin \theta $,

which gives

$f = - P + mg\sin \theta $. ...... (1)

This is a straight line with slope $-$1. Substitute the values of P₁ and P₂ in equation (1) to get the frictional force at those points i.e.,

${f_1} = \mu mg\cos \theta $ and ${f_2} = - \mu mg\cos \theta $.

While the parabola is moving right side, the bead experiences a force with a magnitude $ma$ to the left. That is, at equilibrium,

$N\cos \theta = mg$

$N\sin \theta = ma$

Now, $\tan \theta = {a \over g}$

Also, $\tan \theta = {{dy} \over {dx}} = 2kx \Rightarrow x = {a \over {2gk}}$

The block slides on the incline when

$mg\sin \theta \ge \mu mg\cos \theta $

$ \Rightarrow \tan \theta \ge \mu $

$ \Rightarrow \tan \theta \ge \sqrt 3 = 1.732$

For the block to topple, we have

$f\left( {{{15} \over 2}} \right) \ge N(5)$ (about centre of mass)

$ \Rightarrow \mu N\left( {{{15} \over 2}} \right) \ge N(5)$

$ \Rightarrow \mu \ge {2 \over 3} = 0.67$

or $\tan \theta \ge 0.67$

That is, the block neither slides nor topples till $\theta \le {\tan ^{ - 1}}(0.67)$ (i.e. it remains at rest) and on exceeding this, it topples first before sliding.

Case I : Free body diagram.

Frictional force ($fr$) = $\mu$N

Normal force (N) = N$_1$ + $mg$

Frictional force ($fr$)$_\mathrm{I}$ = $\mu$(N$_1$ + $mg$) ..... (i)

Where N = Normal force

$\mu$ = Coefficient of friction

$m$ = mass of object

$g$ = acc. due to gravity

N$_1$ = Normal force i.e., vertical component of force due to push.

N$_2$ = Parallel component of force due to push.

Case II : Free body diagram (Pull)

For balancing the force,

N = N$_1$ $-$ $mg$

Frictional force $(fr)$$_\mathrm{II}$ = $\mu$N ..... (ii)

From eq. (i) and (ii)

$(fr)$$_\mathrm{I}$ > $(fr)$$_\mathrm{II}$

In case of pull, it is easy to move object.

Statement 1 is true. Statement 2 is true.

Force of friction depends on roughness of two surfaces in contact. If roughness will increase, the force of friction will also increase.

Velocity along the plane just before collision,

${v^2} - {u^2} = 2gs \Rightarrow v = \sqrt {2g \times 3} = \sqrt {60} \,m/s$

Velocity along the plane just after collision

${v_B} - v\cos 30 = \sqrt {45} \,m{s^{ - 1}}$

$v_C^2 - v_B^2 = 2as$

$v_C^2 - 45 = 2 \times 10 \times 3$

${v_c} = \sqrt {60 + 45} = \sqrt {105} \,m{s^{ - 1}}$ is the velocity of block just before leaving incline.

Given that,

$\overrightarrow{\mathrm{P}}(t)=\mathrm{A}[\hat{i} \cos (k t)-\hat{j} \sin (k t)]$

where, $\mathrm{A}$ and $k=$ constant

$\begin{aligned}\text { Force }\overrightarrow{\mathrm{F}}) & =\frac{d \overrightarrow{\mathrm{P}}}{d t}\left(\text { from Newton's } 2^{\text {nd }}\right. \text { law) } \\\\ \Rightarrow \quad \overrightarrow{\mathrm{F}} & =\frac{d}{d t}[\mathrm{~A}(\hat{i} \cos (k t)-\hat{j} \sin (k t))] \\\\& =\mathrm{A} k[\hat{i} \cos (k t)-\hat{j} \sin (k t)]\end{aligned}$

By applying dot product between $\vec{F} \& \vec{P}$ we have

$\begin{aligned}& \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}=0 \\\\ & \Rightarrow|\vec{F}||\vec{P}| \cos \theta=0 \\\\ & \Rightarrow \quad \cos \theta=0 \\\\ & \text {Hence,} \theta=90^{\circ}\end{aligned}$

Statement 1 : Cloth can be pulled out without dislodging the dishes from the table because of inertia. Therefore, statement 1 true.

Statement 2 : This is Newton's third law and hence true. But statement 2 is not a correct explanation of statement 1 .

The acceleration of mass 'm' is due to the force $T\cos \theta $, from 2^nd law of motion we have,

$T\cos \theta = ma$

$ \Rightarrow a = {{T\cos \theta } \over m}$ ..... (i)

and also, $F = 2T\sin \theta $

$ \Rightarrow T = {F \over {2\sin \theta }}$ ..... (ii)

From (i) & (ii), we have,

$a = \left( {{F \over {2\sin \theta }}} \right)\,.\,{{\cos \theta } \over m}$

$a = {F \over {2m\tan \theta }}$

$\therefore$ $a = {F \over {2m{{\sqrt {{a^2} - {x^2}} } \over x}}}$

$\therefore$ $a = {F \over {2m}}\,.\,{x \over {\sqrt {{a^2} - {x^2}} }}$

Two blocks A and B of mass 2 m and m respectively are connected by a massless and in extensible string.

The equilibrium condition FBD of 2 m

$ k x=\mathrm{T}+2 m g=m g+2 m g=3 m g $

After the string is cut.

$ \mathrm{T}=0 $

FBD of $2 m, a_2=\frac{3 m g-2 m g}{2 m}=\frac{m g}{2 m}=\frac{g}{2}$

FBD of $m, m a_1=m g=\mathrm{T}$

$ \therefore a_1=g $

When surface is rough

$d = {1 \over 2}\left( {g\,\sin \,\theta - \mu g\,\cos \theta } \right)t_2^2$

${t_2} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}} $

According to question, ${t_2} = n{t_1}$

$n\sqrt {{{2d} \over {g\,\sin \,\theta }}} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}} $

$n = {1 \over {\sqrt {1 - {\mu _k}} }}$ ( as $\cos \,{45^ \circ } = \sin \,{45^ \circ } = {1 \over {\sqrt 2 }}$ )

${n^2} = {1 \over {1 - {\mu _k}}}$

or $1 - {\mu _k} = {1 \over {{n^2}}}$

or ${\mu _k} = 1 - {1 \over {{n^2}}}$

The spring ${S_1}$ is massless. Therefore it exerts a force of $Mg$ on the spring of spring balance ${S_2}$ which shows the reading of $Mkgf.$

With respect to the man standing on the ground the acceleration of the ball

${\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}$

$ \Rightarrow {a_{bm}} = g - 0$

$ = g$

Adding those equations we get

$a = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}$

$\therefore$ ${g \over 8} = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}$

$ \Rightarrow {1 \over 8} = {{{{{m_1}} \over {{m_2}}} - 1} \over {{{{m_1}} \over {{m_2}}} + 1}}$

$ \Rightarrow$ ${{{m_1}} \over {{m_2}}} + 1 =$ ${8\left( {{{{m_1}} \over {{m_2}}} - 1} \right)}$

$ \Rightarrow$ ${{{m_1}} \over {{m_2}}} = {9 \over 7}$

When a simple pendulum swings, the tension in the string is not just balancing the component of the gravitational force but also provides the necessary centripetal force for the circular motion of the bob.

Forces Acting on the Pendulum Bob:

At any angular displacement $ \theta $:

Gravitational Force ($ mg $): Acts downward.

Tension in the String ($ T $): Acts along the string towards the pivot.

Radial Component of Gravitational Force: $ mg \cos \theta $

Tangential Component of Gravitational Force: $ mg \sin \theta $

Net Radial Force:

The net radial force provides the centripetal acceleration required for circular motion:

$ T - mg \cos \theta = m \frac{v^2}{L} $

Where:

$ T $ is the tension in the string,

$ v $ is the speed of the bob at angle $ \theta $,

$ L $ is the length of the pendulum.

Calculating the Speed $ v $ at $ \theta = 20^\circ $:

Using conservation of mechanical energy between the highest point ($ \theta_{\text{max}} = 40^\circ $) and the point at $ \theta = 20^\circ $:

Potential Energy at $ \theta_{\text{max}} $:

$ U_{\text{max}} = mgL (1 - \cos 40^\circ) $

Potential Energy at $ \theta = 20^\circ $:

$ U = mgL (1 - \cos 20^\circ) $

Kinetic Energy at $ \theta = 20^\circ $:

$ K = \frac{1}{2} m v^2 = U_{\text{max}} - U $

Compute the difference in potential energy:

$ \Delta U = mgL (\cos 20^\circ - \cos 40^\circ) $

Using known values:

$ \cos 20^\circ \approx 0.9397 $

$ \cos 40^\circ \approx 0.7660 $

Calculate $ \Delta U $:

$ \Delta U = mgL (0.9397 - 0.7660) = mgL (0.1737) $

Set $ \Delta U = \frac{1}{2} m v^2 $ and solve for $ v^2 $:

$ \frac{1}{2} m v^2 = mgL (0.1737) \implies v^2 = 2gL (0.1737) $

Calculating the Centripetal Acceleration:

$ a_{\text{centripetal}} = \frac{v^2}{L} = \frac{2gL (0.1737)}{L} = 2g (0.1737) = 0.3474g $

Calculating the Tension $ T $:

$ T = mg \cos \theta + m a_{\text{centripetal}} = mg \cos 20^\circ + m (0.3474g) $

Plugging in $ \cos 20^\circ $:

$ T = mg (0.9397) + m (0.3474g) = mg (0.9397 + 0.3474) = mg (1.2871) $

Conclusion:

The tension $ T $ is greater than $ mg \cos 20^\circ $ because it includes the additional centripetal force component.

Therefore, at an angular displacement of $ 20^\circ $, the tension in the string is indeed greater than $ mg \cos 20^\circ $.

Therefore, the statement is TRUE.