Physics 5 Marks Questions

$\text{R}_1-\text{ma}=0\Rightarrow\text{R}_1=\text{ma} \ ...(\text{i})$

$2\text{ma + T}-\text{mg}+\mu_1\text{R}_1=0$

$\Rightarrow\text{T = mg}-(2-\mu_1)\text{ma} \ ...(\text{ii})$

From free body diagram-2

$\text{T}+\mu_1\text{R}_1+\text{mg}-\text{R}_2=0$

$\Rightarrow\text{R}_2=\text{T}+\mu_1\text{ma + Mg}$ [Putting the value of R₁ from (i)]

$=(\text{mg}-2\text{ma}-\mu_1\text{ma})+\mu_1\text{ma + Mg}$ [Putting the value of T from (ii)]

$\therefore\text{R}_2=\text{Mg + mg}-2\text{ma} \ ...(\text{iii})$

Again, form the free body diagram-2

$\text{T + T}-\text{R}-\text{Ma}-\mu_2\text{R}_2=0$

$\Rightarrow2\text{T}-\text{MA}-\text{mA}-\mu_2(\text{Mg + mg}-2\text{ma})=0$ [Putting the values of R₁ and R₂ from (i) and (iii)]

$\Rightarrow2\text{T}=\text{(M + m})+\mu_2(\text{Mg + mg}-2\text{ma}) \ ...(\text{iv})$

From equation (ii) and (iv)

$2\text{T = 2mg}-2(2+\mu_1)\text{mg}$

$= \text{(M + m)a}+\mu_2(\text{Mg + mg}-2\text{ma})$

$\Rightarrow2\text{mg}-\mu_2(\text{M + m)g}$

$= \text{ a}\text{(M + m}-2\mu_2\text{m}+4\text{m}+2\mu_1\text{m})$

$\Rightarrow\text{a}=\frac{[2\text{m}-\mu_2(\text{M + m})]\text{g}}{\text{M + m}[5+2(\mu_1-\mu_2)]}$

$\text{R}_1=\text{mg} \ ...(\text{i})$

$\text{F}-\mu\text{R}_1-\text{T}=0\Rightarrow\text{F}-\mu\text{mg}-\text{T}=0 \ ...(\text{ii})$

$\therefore\text{F}=\mu\text{mg}+\mu\text{mg}=2\mu\text{mg}$ [putting $\text{T}=\mu\text{mg}$]

$\text{T}-\mu\text{R}_1=0$

$\Rightarrow\text{T}=\mu\text{mg}$

$2\text{F}-\text{T}-\mu\text{mg}-\text{ma}=0 \ ...(\text{i})$

$\text{T}-\text{Ma}-\mu\text{mg}=0$ [$\therefore$ R₁ = mg]

$\Rightarrow\text{T = Ma}+\mu\text{mg}$

$2\text{f}-\text{Ma}-\mu\text{mg}-\mu\text{mg}-\text{ma}=0$

$\Rightarrow2(2\mu\text{mg})-2\mu\text{mg = a(M + m)}$ [Putting $\text{F}=2\mu\text{mg}$]

$\Rightarrow4\mu\text{mg}-2\mu\text{mg = a(M + m)}$

$\Rightarrow\text{a}=\frac{2\mu\text{mg}}{\text{M + m}}$

Because the block slips on the table, maximum frictional force acts on it.

From the free body diagram

$\text{R = mg}$

$\therefore\text{F}-\mu\text{R}=0\Rightarrow\text{F}=\mu\text{R}=\mu\text{mg}$

But the table is at rest. So, frictional force at the legs of the table is not $\mu\text{R}_1.$ Let be f, so form the free body diagram.

$\text{f}_{\text{o}}-\mu\text{R}=0\Rightarrow\text{f}_{\text{o}}=\mu\text{R}=\mu\text{mg}.$

Total frictional force on table by floor is $\mu\text{mg}.$

Suppose, the body is accelerating down with acceleration ‘a’.

From the free body diagram

$\text{R}-\text{mg}\cos\theta=0$

$\Rightarrow\text{R}=\text{mg}\cos\theta \ ...(1)$

$\text{ma + mg}\sin\theta-\mu\text{R}=0$

$\Rightarrow\text{a}=\frac{\text{mg}(\sin\theta-\mu\cos\theta)}{\text{m}}=\text{g}(\sin\theta-\mu\cos\theta)$

For the first half mt. u = 0, s = 0.5m, t = 0.5sec.

So, v = u + at = 0 + (0.5)4 = 2m/s

$\text{S = ut}+\frac{1}{2}\text{at}^2\Rightarrow0.5=0+\frac{1}{2}\text{a}\Big(\frac{0}{5}\Big)^2$

$\Rightarrow\text{a}=4\text{m/s}^2 \ ...(2)$

For the next half metre

$\text{u}=2\text{m/s},\text{ a}=4\text{m/s}^2,\text{ s}=0.5.$

$\Rightarrow0.5=2\text{t}+\Big(\frac{1}{2}\Big)4\text{t}^2\Rightarrow2\text{t}^2+2\text{t}-0.5=0$

$\Rightarrow4\text{t}^2+4\text{t}-1=0$

$\therefore=\frac{-4\pm\sqrt{16+16}}{2\times4}=\frac{1.656}{8}=0.207\text{sec}$

Time taken to cover next half meter is 0.21sec.

To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline $=\mu\text{R}+2\text{g}\sin30^{\circ}$

$=0.2\times(9.8)\sqrt{3}+219.8\times\Big(\frac{1}{2}\Big)$ [from(1)]

$=3.39+9.8=13\text{N}$

With this minimum force the body move up the incline with a constant velocity as net force on it is zero.

2. Net force acting down the incline is given by,

$\text{F}=2\text{g}\sin30^{\circ}-\mu\text{R}$

$=2\times9.8\times\Big(\frac{1}{2}\Big)-3.39=6.41\text{N}$

Due to F = 6.41N the body will move down the incline with acceleration.

No external force is required.

$\therefore$ Force required is zero.

$\text{R}_1+\text{ma}-\text{mg}=0$

$\Rightarrow\text{R}_1=\text{m(g}-\text{a) = mg}-\text{ma} \ ...(\text{i})$

$\text{T}-\mu\text{R}_1=0\Rightarrow\text{T = m(mg}-\text{ma}) \ ...(\text{ii})$

Again, $\text{F}-\text{T}-\mu\text{R}_1=0$

$\Rightarrow\text{F}-\{\mu(\text{mg}-\text{ma})\}-\text{u}(\text{mg}-\text{ma})=0$

$\Rightarrow\text{F}-\mu\text{mg}+\mu\text{ma}-\mu\text{mg}+\mu\text{ma}=0$

$\Rightarrow\text{F}=2\mu\text{mg}-2\mu\text{ma}\Rightarrow\text{F}=2\mu\text{m(g}-\text{a})$

2. Acceleration of the block be a₁

$\text{R}_1=\text{mg}-\text{ma} \ ...(\text{i})$

$2\text{F}-\text{T}-\mu\text{R}_1-\text{ma}_1=0$

$\Rightarrow2\text{F}-\text{t}-\mu\text{mg}+\mu\text{a}-\text{ma}_1=0 \ ...(\text{ii})$

$\text{T}-\mu\text{R}_1-\text{Ma}_1=0$

$\Rightarrow\text{T}=\mu\text{R}_1+\text{Ma}_1$

$\Rightarrow\text{T}=\mu(\text{mg}-\text{ma})+\text{Ma}_1$

$\Rightarrow\text{T}=\mu\text{mg}-\mu\text{ma + Ma}_1$

Subtracting values of F & T, we get

$2(2\mu\text{m(g}-\text{a}))-2(\mu\text{mg}-\mu\text{ma}+\text{Ma}_1)\\-\mu\text{mg}+\mu\text{ma}-\mu\text{a}_1=0$

$\Rightarrow4\mu\text{mg}-4\mu\text{ma}-2\mu\text{mg}+2\mu\text{ma = ma}_1+\text{Ma}_1$

$\Rightarrow\text{a}_1=\frac{2\mu\text{m(g}-\text{a})}{\text{M + m}}$

Both blocks move with this acceleration but in opposite directions.

R₁ - 2g = 0

⇒ R₁ = 2 × 10 = 20

2a + 0.2 R₁ - 12 = 0

⇒ 2a + 0.2(20) = 12

⇒ 2a = 12 - 4 = 8

⇒ a = 4m/s²

$4\text{a}_1-\mu\text{R}_1=0$

$\Rightarrow4\text{a}_1=\mu\text{R}_1=0.2(20)$

⇒ 4a₁ = 4

⇒ a₁ = 1m/s²

2kg block has acceleration 4m/s² & that of 4kg is 1m/s²

R₁ = 2g = 20

$\text{Ma}-\mu\text{R}_1=0$

⇒ 2a = 0.2(20) = 4

⇒ a = 2m/s²

4a + 0.2 × 2 × 10 - 12 = 0

⇒ 4a + 4 = 12

⇒ 4a = 8

⇒ a = 2m/s²

From the free body diagram

$\text{T}+0.5\text{a}-0.5\text{g}=0 \ ...(1)$

$\mu\text{R}+1\text{a}+\text{T}_1-\text{T}=0 \ ...(2)$

$\mu\text{R}+1\text{a}-\text{T}_1=0$

$\mu\text{R}+1\text{a}=\text{T}_1 \ ...(3)$

From (2) & (3)

$\Rightarrow\mu\text{R + a = T}-\text{T}_1$

$\therefore\text{T}-\text{T}_1=\text{T}_1$

$\Rightarrow\text{T}=2\text{T}_1$

Equation (2) becomes

$\mu\text{R + a + T}_1-2\text{T}_1=0$

$\Rightarrow\mu\text{R + a}-\text{T}_1=0$

$\Rightarrow\text{T}_1=\mu\text{R + a}=0.2\text{g + a} \ ...(4)$

Equation (1) becomes

$2\text{T}_1+\frac{0}{5\text{a}}-0.5\text{g}=0$

$\Rightarrow\text{T}_1=\frac{0.5\text{g}-0.5\text{a}}{2}=0.25\text{g}-0.25\text{a} \ ...(5)$

From (4) & (5) $0.2\text{g + a}=0.25\text{g}-0.25\text{a}$

$\Rightarrow\text{a}=\frac{0.05}{1.25}\times10=0.04/10=0.4\text{m/s}^2$

1. Accln of 1kg blocks each is 0.4m/s²
2. Tension T₁ = 0.2g + a + 0.4 = 2.4N
3. T = 0.5g - 0.5a = 0.5 × 10 - 0.5 × 0.4 = 4.8N

From the free body diagram,

$\text{R}-\text{mg}\cos\theta=0\Rightarrow\text{R = mg}\cos\theta \ ...(1)$

For the block

U = 0, s = 8m, t = 2sec.

$\therefore\text{s = ut}+\frac{1}{2}\text{at}^2\Rightarrow8=0+\frac{1}{2}\text{a}2^2\Rightarrow\text{a}=4\text{m/s}^2$

Again, $\mu\text{R}+\text{ma}-\text{mg}\sin\theta=0$

$\Rightarrow\mu\text{ mg}\cos\theta+\text{ma}-\text{mg}\sin\theta=0$ [from(1)]

$\Rightarrow\text{m}(\mu\text{g}\cos\theta+\text{a}-\text{g}\sin\theta)=0$

$\Rightarrow\mu\times10\times\cos30^{\circ}=\text{g}\sin30^{\circ}-\text{a}$

$\Rightarrow\mu\times10\times\sqrt{\Big(\frac{3}{3}\Big)}=10\times\Big(\frac{1}{2}\Big)-4$

$\Rightarrow\Big(\frac{5}{\sqrt{3}}\Big)\mu=1\Rightarrow\mu=\frac{1}{\big(\frac{5}{\sqrt{3}}\big)}=0.11$

$\therefore$ Co-efficient of kinetic friction between the two is 0.11.

From the free body diagram

$\mu_1\text{R}+1-16=0$

$\Rightarrow\mu_1(2\text{g})+(-15)=0$

$\Rightarrow\mu_1=\frac{15}{20}=0.75$

$\mu_2\text{R}_1+4\times0.5+16-4\text{g}\sin30^{\circ}=0$

$\Rightarrow\mu_2\Big(\frac{20}{\sqrt{3}}\Big)+2+16-20=0$

$\Rightarrow\mu_2=\frac{2}{20\sqrt{3}}=\frac{1}{17.32}=0.057\approx0.06$

$\therefore$ Co-efficient of friction $\mu_1=0.75$ and $\mu_2=0.06$

From the free body diagram

$\text{g}=10\text{m/s}^2,\text{m}=2\text{kg},\theta=30^{\circ},\mu=0.2$

$\text{R}-\text{mg}\cos\theta-\text{F}\sin\theta=0$

$\Rightarrow\text{R = mg}\cos\theta+\text{F}\sin\theta \ ...(1)$

And $\text{mg}\sin\theta+\mu\text{R}-\text{F}\cos\theta=0$

$\Rightarrow\text{mg}\sin\theta+\mu(\text{mg}\cos\theta+\text{F}\sin\theta)-\text{F}\cos\theta=0$

$\Rightarrow\text{mg}\sin\theta+\mu\text{ mg}\cos\theta+\mu\text{F}\sin\theta-\text{F}\cos\theta=0$

$\Rightarrow\text{F}=\frac{(\text{mg}\sin\theta+\mu\text{mg}\cos\theta)}{(\mu\sin\theta-\cos\theta)}$

$\Rightarrow\text{F}=\frac{2\times10\times\big(\frac{1}{2}\big)+0.2\times2\times10\times\big(\frac{\sqrt{3}}{2}\big)}{0.2\times\big(\frac{1}{2}\big)-\big(\frac{\sqrt{3}}{2}\big)}$

$=\frac{13.464}{0.76}=17.7\text{N}\approx17.5\text{N}$

1. From the free body diagram

$\text{R = 4g}\cos30^{\circ}=4\times10\times\frac{\sqrt{3}}{2}=20\sqrt{3} \ ...(\text{i})$

$\mu_2\text{R + 4a}-\text{P}-4\text{g}\sin30^{\circ}=0$

$\Rightarrow0.3(40)\cos30^{\circ}+4\text{a}-\text{P}-40\sin20^{\circ}=0 \ ...(\text{ii})$

$\text{P + 2a}+\mu_1\text{R}_1-2\text{g}\sin30^{\circ}=0 \ ...(\text{iii})$

$\text{R}_1=2\text{g}\cos30^{\circ}=2\times10\times\frac{\sqrt{3}}{2}=10\sqrt{3} \ ...(\text{iv})$

Equn. (ii) $6\sqrt{3}+4\text{a}-\text{P}-20=0$

Equn (iv) $\text{P + 2a}+2\sqrt{3}-10=0$

From Equn (ii) & (iv) $6\sqrt{3}+6\text{a}-30+2\sqrt{3}=0$

$\Rightarrow6\text{a}=30-8\sqrt{3}=30-13.85=16.15$

$\Rightarrow\text{a}=\frac{16.15}{6}=2.69=2.7\text{m/s}^2$

2. can be solved. In this case, the 4kg block will travel with more acceleration because, coefficient of friction is less than that of 2kg. So, they will move separately. Drawing the free body diagram of 2kg mass only, it can be found that, a = 2.4m/s².

$\text{s = 5m}, \ \mu=\frac{4}{3},\text{g}=10\text{m/s}^2$

$\text{u}=36\text{km/h}=10\text{m/s},\text{ v}=0,$

$\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}=\frac{0-10^2}{2\times5}=-10\text{m/s}^2$

From the freebody diagrams,

$\text{R}-\text{mg}\cos\theta=0;\text{g}=10\text{m/s}^2$

$\Rightarrow\text{R = mg}\cos\theta \ ...(\text{i});\mu=\frac{4}{3}.$

Again, $\text{ma + mg}\sin\theta-\mu\text{R}=0$

$\Rightarrow\text{ma + mg}\sin\theta-\mu\text{mg}\cos\theta=0$

$\Rightarrow\text{a + g}\sin\theta-\text{mg}\cos\theta=0$

$\Rightarrow10+10\sin\theta-\Big(\frac{4}{3}\Big)\times10\cos\theta=0$

$\Rightarrow30+30\sin\theta-40\cos\theta=0$

$\Rightarrow3+3\sin\theta-4\cos\theta=0$

$\Rightarrow4\cos\theta-3\sin\theta=3$

$\Rightarrow4\sqrt{1-\sin^2\theta}=3+3\sin\theta$

$\Rightarrow16(1-\sin^2\theta)=9+9\sin^2\theta+18\sin\theta$

$\sin\theta=\frac{-18\pm\sqrt{18^2-4(25)(-7)}}{2\times25}\\=\frac{-18\pm32}{50}=\frac{14}{50}=0.28$ [Taking +ve sign only]

$\Rightarrow\theta=\sin^{-1}(0.28)=16^{\circ}$

Maximum incline is $\theta=16^{\circ}$

Let a₁ and a₂ be the accelerations of ma and M respectively.

Here, a₁ > a₂ so that m moves on M

Suppose, after time ‘t’ m separate from M.

In this time, m covers $\text{vt}+\frac{1}{2}\text{a}_1\text{t}^2$ and $\text{S}_{\text{M}}=\text{vt}+\frac{1}{2}\text{a}_2\text{t}^2$

For ‘m’ to m to ‘m’ separate from M. $\text{vt}+\frac{1}{2}\text{a}_1\text{t}^2=\text{vt}+\frac{1}{2}\text{a}_2\text{t}^2+\ell \ ...(1)$

Again from free body diagram

$\text{Ma}_1+\frac{\mu}{2}\text{R}=0$

$\Rightarrow\text{ma}_1=-\Big(\frac{\mu}{2}\Big)\text{mg}$

$=-\Big(\frac{\mu}{2}\Big)\text{m}\times10\Rightarrow\text{a}_1=-5\mu$

Again,

$\text{Ma}_2+\mu(\text{M + m)g}-\Big(\frac{\mu}{2}\Big)\text{mg}=0$

$\Rightarrow2\text{Ma}_2+2\mu(\text{M + m)g}-\mu\text{mg}=0$

$\Rightarrow2\text{Ma}_2=\mu\text{mg}-2\mu\text{Mg}-2\mu\text{mg}$

$\Rightarrow\text{a}_2\frac{-\mu\text{mg}-2\mu\text{Mg}}{2\text{M}}$

Putting values of a₁ & a₂ in equation (1) we can find that

$\text{T}=\sqrt{\Big(\frac{4\text{ml}}{(\text{M + m})\mu\text{g}}\Big)}$

$\text{R}_1=\text{M}_1\text{g}\cos\theta \ ...(\text{i})$

$\text{R}_2=\text{M}_2\text{g}\cos\theta \ ...(\text{ii})$

$\text{T + M}_1\text{g}\sin\theta-\text{m}_1\text{a}-\mu\text{R}_1=0 \ ...(\text{iii})$

$\text{T}-\text{M}_2-\text{M}_2\text{a}+\mu\text{R}_2=0 \ ...(\text{iv})$

Equn (iii) $\Rightarrow\text{T + M}_1\text{g}\sin\theta-\text{M}_1\text{a}-\mu\text{M}_1\text{g}\cos\theta=0$

Equn (iv) $\Rightarrow\text{T}-\text{M}_2\text{g}\sin\theta+\text{M}_2\text{a}+\mu\text{M}_2\text{g}\cos\theta=0 \ ...(\text{v})$

Equn (iv) & (v) $\Rightarrow\text{g}\sin\theta(\text{M}_1+\text{M}_2)-\text{a}(\text{M}_1+\text{M}_2)\\-\mu\text{g}\cos\theta(\text{M}_1+\text{M}_2)=0$

$\Rightarrow\text{a}(\text{M}_1+\text{M}_2)=\text{g}\sin\theta(\text{M}_1+\text{M}_2)\\-\mu\text{g}\cos\theta(\text{M}_1+\text{M}_2)$

$\Rightarrow\text{a = g}(\sin\theta-\mu\cos\theta)$

$\therefore$ The blocks system has acceleration $\text{g}(\sin\theta-\mu\cos\theta)$

The force exerted by the rod on one of the blocks is tension.

Tension $\text{T}=-\text{M}_1\text{g}\sin\theta+\text{M}_1\text{a}+\mu\text{M}_1\text{g}\sin\theta$

$\Rightarrow\text{T}=-\text{M}_1\text{g}\sin\theta+\text{M}_1(\text{g}\sin\theta-\mu\text{g}\cos\theta)\\+\mu\text{M}_1\text{g}\cos\theta$

$\Rightarrow\text{T}=0$

f → applied force

F_i → contact force

F → frictional force

R → normal reaction

$\mu=\tan\lambda=\frac{\text{F}}{\text{R}}$

When $\text{F}=\mu\text{R, F}$ is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction $(\mu\text{R})$

Before reaching limiting friction

$\text{F}<\mu\text{R}$

$\therefore\tan\lambda=\frac{\text{F}}{\text{R}}\leq\frac{\mu\text{R}}{\text{R}}\Rightarrow\tan\lambda\leq\mu\Rightarrow\lambda\leq\tan^{-1}\mu$

From the free body diagram

$4-4\text{a}-\mu\text{R}+4\text{g}\sin30^{\circ}=0 \ ...(1)$

$\text{R}-4\text{g}\cos30^{\circ}=0 \ ...(2)$

$\Rightarrow\text{R}=4\text{g}\cos30^{\circ}$

Putting the values of R is & in equn. (1)

$4-4\text{a}-0.11\times4\text{g}\cos30^{\circ}+4\text{g}\sin30^{\circ}=0$

$\Rightarrow4-4\text{a}-0.11\times4\times10\times\Big(\frac{\sqrt{3}}{2}\Big)\\+4\times10\times\Big(\frac{1}{2}\Big)=0$

$\Rightarrow4-4\text{a}-3.81+20=0\Rightarrow\text{a}\approx5\text{m/s}^2$

For the block $\text{u}=0,\text{t}=2\text{sec},\text{a}=5\text{m/s}^2$

Distance $\text{s = ut}+\frac{1}{2}\text{at}^2\Rightarrow\text{s}=0+\Big(\frac{1}{2}\Big)5\times2^2=10\text{m}$

The block will move 10m.

$\text{R}_1+\text{QE}-\text{mg}=0$

$\text{R}_1=\text{mg}-\text{QE} \ ...(\text{i})$

$\text{F}-\text{T}-\mu\text{R}_1=0$

$\Rightarrow\text{F}-\text{T}=\mu(\text{mg}-\text{QE}) = 0$

$\Rightarrow\text{F}-\text{T}=\mu\text{mg}+\mu\text{QE}=0 \ ...(\text{ii})$

$\text{T}-\mu\text{R}_1=0$

$\Rightarrow\text{T}=\mu\text{R}_1=\mu(\text{mg}-\text{QE})=\mu\text{mg}-\mu\text{QE}$

Now equation (ii) is $\text{F}-\text{mg}+\mu\text{QE}-\mu\text{mg}+\mu\text{QE}=0$

$\Rightarrow\text{F}-2\mu\text{mg}+2\mu\text{QE}=0$

$\Rightarrow\text{F}=2\mu\text{mg}-2\mu\text{QE}$

$\Rightarrow\text{F}=2\mu(\text{mg}-\text{QE})$

Maximum horizontal force that can be applied is $2\mu(\text{mg}-\text{QE}).$

Let ‘p’ be the force applied to at an angle $\theta$

From the free body diagram

$\text{R + P}\sin\theta-\text{mg}=0$

$\Rightarrow\text{R}=-\text{P}\sin\theta+\text{mg} \ ...(\text{i})$

$\mu\text{R}-\text{p}\cos\theta \ ...(\text{ii})$

Equn. (i) is

$\mu(\text{mg}-\text{P}\sin\theta)-\text{P}\cos\theta=0$

$\Rightarrow\mu\text{mg}=\mu\rho\sin\theta-\text{P}\cos\theta\rightarrow\rho=\frac{\mu\text{mg}}{\mu\sin\theta+\cos\theta}$

Applied force P should be minimum, when $\mu\sin\theta+\cos\theta$ is maximum.

Again, $\mu\sin\theta+\cos\theta$ is maximum when its derivative is zero.

$\therefore\frac{\text{d}}{\text{d}\theta}(\mu\sin\theta+\cos\theta)=0$

$\Rightarrow\mu\cos\theta-\sin\theta=0\Rightarrow\theta\tan^{-1}\mu$

So, $\text{p}=\frac{\mu\text{mg}}{\mu\sin\theta+\cos\theta}=\frac{\frac{\mu\text{mg}}{\cos\theta}}{\frac{\mu\sin\theta}{\cos\theta}+\frac{\cos\theta}{\cos\theta}}$

$=\frac{\mu\text{mg}\sec\theta}{1+\mu\tan\theta}=\frac{\mu\text{mg}\sec\theta}{1+\tan^2\theta}$

$=\frac{\mu\text{mg}}{\sec\theta}=\frac{\mu\text{mg}}{\sqrt{(1+\tan^2\theta)}}=\frac{\mu\text{mg}}{\sqrt{1+\mu^2}}$

Minimum force is $\frac{\mu\text{mg}}{\sqrt{1+\mu^2}}$ at an angle $\theta=\tan^{-1}\mu.$

1. When the 10N force applied on 2kg block, it experiences maximum frictional force $\mu\text{R}_1=\mu\times2\text{kg}=(0.2)\times20=4\text{N}$ from the 3kg block.

So, the 2kg block experiences a net force of 10 - 4 = 6N

So, $\text{a}_1=\frac{6}{2}=3\text{m/s}^2.$

But for the 3kg block, (fig-3) the frictional force from 2kg block (4N) becomes the driving force and the maximum frictional force between 3kg and 7kg block is

$\mu_2\text{R}_2=(0.3)\times5\text{kg}=15\text{N}$

So, the 3kg block cannot move relative to the 7kg block. The 3kg block and 7kg block both will have same acceleration (a₂ = a₃) which will be due to the 4N force because there is no friction from the floor.

$\therefore\text{a}_2=\text{a}_3=\frac{4}{10}=0.4\text{m/s}^2$

2. When the 10N force is applied to the 3kg block, it can experience maximum frictional force of 15 + 4 = 19N from the 2kg block & 7kg block.

So, it can not move with respect to them.

As the floor is frictionless, all the three bodies will move together

$\therefore\text{a}_1=\text{a}_2=\text{a}_3=\frac{10}{12}=\Big(\frac{5}{6}\Big)\text{m/s}^2$

3. Similarly, it can be proved that when the 10N force is applied to the 7kg block, all the three blocks will move together.

Again $\text{a}_1=\text{a}_2=\text{a}_3=\Big(\frac{5}{6}\Big)\text{m/s}^2$

From the free body diagram

$\text{T + 15a}-15\text{g}=0$

$\Rightarrow\text{T = 15g}-15\text{a} \ ...(1)$

$\text{T}-\text{(T}_1+5\text{a}+\mu\text{R})=0$

$\Rightarrow\text{T}-(5\text{g + 5a + 5a}+\mu\text{R})=0$

$\Rightarrow\text{T = 5g + 10a}+\mu\text{R} \ ...(\text{ii})$

$\text{T}_1-5\text{g}-5\text{a}=0$

$\Rightarrow\text{T}_1=5\text{g + 5a} \ ...(\text{iii})$

From (i) & (ii)

$15\text{g}-15\text{a = 5g + 10a + 0.2 (5g)}$

$\Rightarrow25\text{a}=90\Rightarrow\text{a}=3.6\text{m/s}^2$

Equation (ii)

$\Rightarrow\text{T}=5\times10+10\times3.6+0.2\times5\times10$

$\Rightarrow96\text{N}$ in the left string

Equation (iii) $\text{T}_1=5\text{g + 5a}=5\times10+5\times3.6=68\text{N}$ in the right string.

Let, the max force exerted by the man is T.

From the free body diagram

$\text{R + T}-\text{Mg}=0$

$\Rightarrow\text{R = Mg}-\text{T} \ ...(\text{i})$

$\text{R}_1-\text{R}-\text{mg}=0$

$\Rightarrow\text{R}_1=\text{R}+\text{mg} \ ...(\text{ii})$

And $\text{T}-\mu\text{R}_1=0$

$\Rightarrow\text{T}-\mu(\text{R + mg})=0$ [From equn. (ii)]

$\Rightarrow\text{T}-\mu\text{R}-\mu\text{mg}=0$

$\Rightarrow\text{T}-\mu(\text{Mg + T})-\mu\text{mg}=0$ [from (i)]

$\Rightarrow\text{T}(1+\mu)=\mu\text{Mg}+\mu\text{mg}$

$\Rightarrow\text{T}=\frac{\mu(\text{M + m)g}}{1+\mu}$

Maximum force exerted by man is $\frac{\mu(\text{M + m)g}}{1+\mu}$