Friction

Concept Of Physics

H C Verma

1   1- A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s 2. What is the coefficient of kinetic friction between the block and the plane ?

Solution :

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Let $m$ = mass of the block from the freebody diagram, $\\$ $R$ - $mg$ = 0 $\Rightarrow$ $R$ = $mg$ ...(1) $\\$ Again $ma$ - $\mu$ $R$ = 0 $\Rightarrow$ $ma$ = $\mu$ $mg$ (from (1)) $\\$ $\Rightarrow$ $a$ = $\mu$g $\Rightarrow$ $4$ = $\mu$g $\Rightarrow$ $\mu$ =$\frac{4}{g}$ = $\frac{4}{10}$ = $0.4$ $\\$ The co-efficient of kinetic friction between the block and the plane is 0.4//

2   2- A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest ?

Due to friction the body will decelerate $\\$ Let the deceleration be 'a' $\\$ $R$ - $mg$ = 0 $\Rightarrow$ $R$ = $mg$ ...(1) $\\$ $ma$ - $\mu$ R = 0 $\Rightarrow$ $ma$ = $\mu$ $R$ = $\mu$ $mg$ (from(1))$\\$ $\Rightarrow$ $a$ = $\mu$$g = 0.1\times10 = 1m/sec^2 \\ Initial velocity u = 10m/s \\ Final velocity v = 0m/s \\ a = -1m/s^2(deceleration)\\ S = \frac{v^2-u^2}{2a}= \frac{0-10^2}{2(-1)}=\frac{100}{2}=50m\\ It will travel 50m before coming to rest. 3 3- A block of mass m is kept on a horizontal table. If the static friction coefficient is 1.t, find the frictional force acting on the block. Solution : Body is kept on the horizontal table. If no force is applied, no frictional force will be there\\ f\rightarrow frictional force\\F\rightarrow Applied force\\ From grap it can be seen that when applied force is zero, frictional force is zero. 4 4- A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two. Solution : From the free body diagram,\\R - mg cos \theta = 0 \Rightarrow R = mg cos \theta ...(1)\\For the block\\U = 0, s = 8m, t = 2sec.\\ \therefores = ut + \frac{1}{2}at^2 \Rightarrow 8 = 0 + \frac{1}{2} a 2^2 \Rightarrow a = 4\frac{m}{s^2}$$\\$ Again, $\mu$R + ma - mg sin $\theta$ = 0$\\$$\Rightarrow \mu mg cos \theta + ma - mg sin \theta = 0 [from(1)]\\m(\mug cos \theta + a - g sin\theta) = 0\\$$\Rightarrow$ $\mu$ $\times$ 10 $\times$ cos 30$^\circ$ = g sin 30 $^\circ$ - a$\\$ $\Rightarrow$ $\mu$ $\times$ 10 $\times$ $\sqrt{(\frac{3}{3})}$ = 10 $\times$ $(\frac{1}{2})$ - 4$\\$$\Rightarrow (\frac{5}{\sqrt{3}})$$\mu$ = 1 $\Rightarrow$ $\mu$ = $\frac{1}{(5/\sqrt{3})}$ = 0.11$\\$$\therefore Co-efficient of kinetic friction between the two is 0.11. 5 5- Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest ? The mass of the block is 4 kg. Solution : From the free body diagram\\4 - 4a - \muR + 4g sin 30° = 0 ...(1)\\R - 4g cos 30° = 0 ...(2)\\$$\Rightarrow$ R = 4g cos 30°$\\$Putting the values of R is & in equn. (1)$\\$4 - 4a - 0.11 $\times$ 4g cos 30° + 4g sin 30° = 0$\\$$\Rightarrow 4 - 4a - 0.11 × 4 × 10 × ( \sqrt3 / 2 ) + 4 × 10 × (1/2) = 0\\$$\Rightarrow$ 4 - 4a - 3.81 + 20 = 0 $\Rightarrow$ a $\approx$ 5 $\frac{m}{s^2}$$\\For the block u =0, t = 2sec, a = 5\frac{m}{s^2}$$\\$Distance s = ut + 1/2 a${t^2}$ $\Rightarrow$ s = 0 + (1/2) 5 × $2^2$ = 10m$\\$ The block will move 10m.

6   6- A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.

10   10- The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that, if X be the angle of friction and la the coefficient of static friction, $\lambda$ $\leq$ $tan^{-1}$$\mu Solution : f \rightarrow applied force\\F \rightarrow contact force\\F \rightarrow frictional force\\R \rightarrow normal reaction\\$$\mu$ = tan $\lambda$ = $\frac{F}{R}$$\\When F = \muR, F is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction (\muR)\\Before reaching limiting friction\\F < \muR\\$$\therefore$ tan $\lambda$ = $\frac{F}{R}$ $\leq$ $\frac{\mu R}{R}$ $\Rightarrow$ tan $\lambda$ $\leq$ $tan^{-1}$ $\mu$

11   11- Consider the situation shown in figure (6-E2). Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.

Solution :

From the free body diagram$\\$T + 0.5a – 0.5 g = 0 ...(1)$\\$$\muR + + 1a +T_1 - T = 0 ...(2)\\$$\mu$R + 1a - $T_1$ = 0$\\$ $\mu$R + 1a = $T_1$ ...(3)$\\$From (2) and (3) $\Rightarrow$ $\mu$R + a = T - $T_1$$\\ \therefore T - T_1 = T_1$$\\$$\Rightarrow T = 2T_1$$\\$Equation (2) becomes $\mu$R + a + $T_1$ - 2$T_1$ = 0$\\$$\Rightarrow \muR + a - T_1 = 0\\$$\Rightarrow$ $T_1$ = $\mu$R + a = 0.2g + a ...(4)$\\$Equation (1) becomes 2$T_1$ + 0/5a - 0.5g = 0 $\\$ $\Rightarrow$ $T_1$ = $\frac{0.5g - 0.5a}{2}$ = 0.25g - 0.25a ...(5)$\\$From (4) & (5) 0.2g + a = 0.25g – 0.25a$\\$$\Rightarrow a = \frac{0.05}{1.25}$$\times$ 10 = 0.04$\times$10 = 0.4$\frac{m}{s^2}$$\\a) Accln of 1kg blocks each is 0.4\frac{m}{s^2}$$\\$b) Tension $T_1$ = 0.2g + a + 0.4 = 2.4N$\\$c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N

12   12- If the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 $\frac {m}{s^2}$ , find the friction coefficients at the two contacts with the blocks.

Solution :

From the free body diagram$\\$ $\mu_1$ R + 1 - 16 = 0$\\$$\Rightarrow \mu_1 (2g) + (-15) = 0\\$$\Rightarrow$ $\mu_1$ = $\frac{15}{20}$ = 0.75$\\$$\mu_2 R_1 + 4 \times 0.5 + 16 - 4g sin 30^{\circ} = 0\\ = 0\\ \Rightarrow \mu_2 (20\sqrt{3}) + 2 + 16 - 20 = 0\\$$\Rightarrow$ $\mu_2$ = $\frac{2}{20\sqrt{3}}$ = $\frac{1}{17.32}$ = 0.057 $\approx$ 0.06$\\$ $\therefore$Coefficient of friction $\mu_1$ = 0.75 and $\mu_2$ = 0.06.

13   13- The friction coefficient between the table and the block shown in figure (6-E4) is 0.2. Find the tensions in the two strings.

From the free body diagram$\\$T + 15a - 15g = 0$\\$$\Rightarrow T = 15g -15a ...(i)\\$$T$ - ($T_1$ + 5a + $\mu$R) = 0$\\$ $\Rightarrow$ T= 5g + 10a +$\mu$R ...(ii)$\\$$T_1 - 5g -5a = 0\\$$\Rightarrow$ $T_1$ = 5g + 5a ...(iii)$\\$From (i) & (ii) 15g – 15a = 5g + 10a + 0.2 (5g)$\\$$\Rightarrow 25a = 90 \Rightarrow a = 3.6\frac{m}{s^2}$$\\$Equation (ii) $\Rightarrow$ T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10$\\$$\Rightarrow 96N in the left string\\Equation (iii) T_1 = 5g + 5a = 5 × 10 + 5 × 3.6 =68N in the right string. 14 14- The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m. Solution : s = 5m, \mu = \frac{4}{3}, g = 10\frac{m}{s^2}$$\\$u = 36km/h = 10m/s, v = 0 ,$\\$ a = $\frac{v^2 - u^2}{2s}$ = $\frac{0-10^2}{2\times5}$ = -10$\frac{m}{s^2}$$\\From the freebody diagrams, \\R – mg cos \theta = 0 ; g = 10\frac{m}{s^2}$$\\$$\Rightarrow = mg cos\theta...(i) ; \mu = \frac{4}{3}.\\Again, ma + mg sin \theta- \mu R = 0\\$$\Rightarrow$ a+ g sin $\theta$ - mg cos $\theta$ = 0$\\$$\Rightarrow30 + 30 sin\theta - 40cos\theta = 0\\$$\Rightarrow$ 4 cos$\theta$ - 3 sin$\theta$ = 3$\\$$\Rightarrow 4\sqrt{1-sin^2\theta} = 3 + 3 sin \theta$$\\$$\Rightarrow 16 (1 - sin^2\theta) = 9 + 9 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} 15 15- The friction coefficient between an athelete's shoes and the ground is 0.90. Suppose a superman wears these shoes and races for 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time that he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop ? Solution : to reach in minimum time, he has to move with maximum possible acceleration. \\Let, the maximum acceleration is ‘a’\\$$\therefore$ ma – $\mu$R = 0 $\Rightarrow$ ma = $\mu$ $mg$$\\$$\Rightarrow$ a = $\mu$ g = 0.9 $\times$ 10 = 9$\frac{m}{s^2}$$\\a) Initial velocity u = 0, t = ?\\a = 9\frac{m}{s^2} , s = 50m\\ s = ut + \frac{1}{2}at^2 \Rightarrow 50 = 0 + (\frac{1}{2}) 9 t^2 \Rightarrow t = \sqrt{\frac{100}{9}} = \frac{10}{3}$$\\$b) After overing 50m, velocity of the athelete is$\\$V = u + at = 0 + 9 ×$\frac{10}{3}$ = 30$\frac{m}{s}$$\\He has to stop in minimum time. So deceleration ia –a = –9\frac{m}{s^2} (max)\\[R = ma\\ma = \muR((max frictional force)\\$$\Rightarrow$ a = $\mu$g = 9$\frac{m}{s^2}$( Deceleration )]$\\$$u^1 = 30\frac{m}{s}, v^1 = 0\\t = \frac{v^1 - u^1}{a} = \frac{0 - 30}{-a} = \frac{-30}{-a} = \frac{10}{3}sec. 16 16- A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30° (figure 6-E5). The friction coefficient between the road and the tyre is 1/2\sqrt{3}. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 \frac{km}{hr}. Take g = 10 \frac{m}{s} . Solution : Hardest brake means maximum force of friction is developed between car’s type & road. \\Max frictional force = \muR \\From the free body diagram\\R - mg cos \theta = 0\\$$\Rightarrow$ R = mg cos $\theta$ ...(i)$\\$and $\mu$R + ma - mg sin $\theta$ = 0 ...(ii)$\\$$\Rightarrow \mumg cos \theta + ma - mg sin\theta = 0\\ \Rightarrow a = 5 - {1 - (2\sqrt{3})} \times 10 (\frac{\sqrt3}{2}) = 2.5 \frac{m}{s^2}$$\\$When, hardest brake is applied the car move with acceleration 2.5$\frac{m}{s^2}$$\\S = 12.8m, u = 6\frac{m}{s}$$\\$SO, velocity at the end of incline$\\$V = $\sqrt{u^2 + 2as}$ = $\sqrt{6^2 + 2(2.5)(12.8)}$ = $\sqrt{36+64}$ = 36$\frac{km}{h}$$\\Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 36\frac{km}{h}. 17 17- A car starts from rest on a half kilometer long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one cannot drive through the bridge in less than 10 s. Solution : Let, a maximum acceleration produced in car.\\ \therefore ma = \mu [For more acceleration, the tyres will slip]\\$$\Rightarrow$ ma = $\mu$ mg $\Rightarrow$ a = $\mu$g = 1 × 10 = 10$\frac{m}{s^2}$$\\For crossing the bridge in minimum time, it has to travel with maximum acceleration u = 0,s = 500m, a = 10\frac{m}{s^2}$$\\$s = ut + $\frac{1}{2} a t^2$$\\$$\Rightarrow$ 500 = 0 + ($\frac{1}{2}$) 10$t^2$ $\Rightarrow$ t = 10sec.$\\$If acceleration is less than 10$\frac{m}{s^2}$ , time will be more than 10sec. So one can’t drive through the bridge in less than 10sec.

18   18- Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is $\mu_1$ and that between the block of mass 4.0 kg and the incline is $\mu_2$. Calculate the acceleration of the 2.0 kg block if (a) $\mu_1$ = 0.20 and $\mu_2$, = 0.30, (b) $\mu_1$ = 0.30 and $\mu_2$ = 0.20. Take g = 10 $\frac{m}{s^2}$.

From the free body diagram$\\$R = 4g cos 30$^{\circ}$ = 4 $\times$ 10 $\times$ $\frac{\sqrt3}{2}$ = 20$\sqrt3$ ...(i)$\\$$\mu_2 R + 4a - P - 4g sin 30° = 0 \Rightarrow 0.3 (40) cos 30° + 4a - P - 40 sin 20° = 0 ...(ii)\\$$\\$P + 2a + $\mu_1$ $R_1$ – 2g sin 30° = 0 ...(iii)$\\$$R_1 = 2g cos 30° = 2 \times 10 \times \frac{\sqrt3}{2} = 10\sqrt{3}$$\\$ ...(iv)$\\$Equn. (ii) 6 $\sqrt3$ + 4a - P - 20 = 0$\\$Equn (iv) P + 2a + 2 $\sqrt3$ - 10 = 0$\\$From Equn (ii) & (iv) 6 $\sqrt3$ + 6a - 30 + 2 $\sqrt3$ = 0$\\$$\Rightarrow 6a = 30 - 8 \sqrt3 = 30 - 13.85 = 16.15\\$$\Rightarrow$ a = $\frac{16.15}{6}$ = 2.69 = 2.7$\frac{m}{s^2}$$\\ b) can be solved. In this case, the 4 kg block will travel with more acceleration because, coefficient of friction is less than that of 2kg. So, they will move separately. Drawing the free body diagram of 2kg mass only, it can be found that, a = 2.4\frac{m}{s^2} . 19 19- Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle \theta with the horizontal. The friction coefficient at both the contacts is pt. Find the acceleration of the system and the force by the rod on one of the blocks. Solution : R_1 = M_1 g cos \theta ...(i)\\$$R_2$ = $M_2$ g cos $\theta$ ...(ii)$\\$T + $M_1$ g sin $\theta$ - $m_1$ a - $\mu$ $R_1$ = 0 ...(iii)$\\$T - $M_2$ - $M_2$ a + $\mu R_2$ = 0 ...(iv)$\\$Equn (iii) $\Rightarrow$ T + $M_1$g sin$\theta$ - $M_1$ a - $\mu$ $M_1$g cos$\theta$ = 0$\\$Equn (iv) $\Rightarrow$T - $M_2$ g sin $\theta$ + $M_2$ a + $\mu$ $M _2$ g cos $\theta$= 0 ...(v)$\\$Equn (iv) & (v) $\Rightarrow$ g sin $\theta$ ($M_1$ + $M_2$ ) - a($M_1$ + $M_2$ ) - $\mu$ g cos $\theta$ ($M_1$ + $M_2$ ) = 0$\\$$\Rightarrow a (M_1 + M_2) = g sin \theta (M_1 + M_2) - \mu g cos \theta (M_1 + M_2)\\ \Rightarrow a = g(sin\theta - \mu cos\theta)\\$$\therefore$ The blocks (system has acceleration g(sin $\theta$ – $\mu$ cos $\theta$)$\\$The force exerted by the rod on one of the blocks is tension.$\\$Tension T = - $M_1$ g sin $\theta$ + $M_1$ a + $\mu$ $M_1$ g sin $\theta$$\\$$\Rightarrow$ T = -$M_1$g sin$\theta$ + $M_1$(g sin $\theta$ - $\mu$ g cos $\theta$) + $\mu$ $M_1$g cos$\theta$$\\ \Rightarrow T = 0. R_1 = M_1 g cos \theta ...(i)\\$$R_2$ = $M_2$ g cos $\theta$ ...(ii)$\\$T + $M_1$ g sin $\theta$ - $m_1$ a - $\mu$ $R_1$ = 0 ...(iii)$\\$T - $M_2$ - $M_2$ a + $\mu R_2$ = 0 ...(iv)$\\$Equn (iii) $\Rightarrow$ T + $M_1$g sin$\theta$ - $M_1$ a - $\mu$ $M_1$g cos$\theta$ = 0$\\$Equn (iv) $\Rightarrow$T - $M_2$ g sin $\theta$ + $M_2$ a + $\mu$ $M _2$ g cos $\theta$= 0 ...(v)$\\$

20   20- A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is $\mu$. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block ? In which direction should this force act ?

Solution :

Let ‘p’ be the force applied to at an angle $\theta$$\\From the free body diagram\\R + P sin \theta - mg = 0\\$$\Rightarrow$ R = - P sin$\theta$ + mg ...(i)$\\$$\muR - p cos \theta ...(ii)\\Equn. (i) is \mu(mg - P sin \theta) - P cos \theta = 0\\$$\Rightarrow$ $\mu$ mg = $\mu$ P sin$\theta$ - P cos $\theta$$\Rightarrow P = \frac{\mu mg}{\mu sin\theta + cos\theta}$$\\$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$$\frac{d}{d\theta}$ ($\mu$ sin $\theta$ + cos $\theta$) = 0 $\\$$\Rightarrow \mu cos\theta - sin\theta = 0 \Rightarrow \theta = tan^{1} \mu$$\\$So, P = $\frac{\mu mg}{\mu sin \theta + cos \theta}$ = $\frac{\mu mgsec\theta}{1 + \mu tan\theta}$ = $\frac{\mu mgsec\theta}{1 + tan^2\theta}$$\\= \frac{\mu mg}{sec \theta} = \frac{\mu mg}{\sqrt{(1 + tan^2 \theta)}} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$$\\$Minimum force is $\frac{\mu mg}{\sqrt{1 + \mu^2}}$ at an angle $\theta$ = $tan^{1} \mu$.

21   21- The friction coefficient between the board and the floor shown in figure (6-E7) is $\mu$. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

Let, the max force exerted by the man is T. $\\$From the free body diagram$\\$R + T - Mg = 0$\\$$\Rightarrow R = Mg - T ...(i)\\$$R_1$ = R + mg ...(iii)$\\$And T - $\mu$ $R_1$ = 0$\\$$\Rightarrow T - \mu(R+ mg) = 0\\ [From equn.(ii)]\\$$\Rightarrow$ T - $\mu$ R - $\mu$ mg = 0$\\$$\Rightarrow T - \mu (Mg + T) - \mu mg = 0 [from (i)]\\$$\Rightarrow$ T = $\frac{\mu(M + m)g}{1+\mu}$$\\Maximum force exerted by man is \frac{\mu(M + m)g}{1 + \mu} 22 22- A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take g= 10 \frac{m}{s^2}. Solution : R_1 - 2g = 0\\$$\Rightarrow$ $R_1$ = 2 $\times$ 10 = 20$\\$2a + 0.2 $R_1$ - 12 = 0$\\$$\Rightarrow 2a = 12 - 4 = 8\\$$\Rightarrow$ a = 4$\frac{m}{s^2}$$\\4a_1 - \mu R_1 = 0\\$$\Rightarrow$ 4$a_1$ = $\mu$$R_1 = 0.2(20)\\$$\Rightarrow$ 4$a_1$ = 4$\\$$\Rightarrow a_1 = 1\frac{m}{s^2}$$\\$2kg block has acceleration 4m/s & that of 4 kg is 1$\frac{m}{s^2}$$\\(ii) R_1 = 2g = 20\\Ma - \mu R_1 = 0\\$$\Rightarrow$ 2a = 0.2 (20) = 4$\\$$\Rightarrow a = 2\frac{m}{s^2}$$\\$4a + 0.2 $\times$ 2 $\times$ 10 - 12 = 0$\\$$\Rightarrow 4a + 4 = 12\\$$\Rightarrow$ 4a = 8$\\$$\Rightarrow a = 2 \frac{m}{s^2}. 23 23- Find the accelerations a_1 , a_2 ,a_3 of the three blocks shown in figure (6-E8) if a horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take g= 10 \frac{m}{s^2} . Solution : a) When the 10N force applied on 2kg block, it experiences maximum frictional force\\$$\mu$$R_1 = mu \times 2kg = (0.2) \times 20 = 4N from the 3kg block.\\So, the 2kg block experiences a net force of 10 - 4 = 6N\\So, a_1 = \frac{6}{2} = 3 \frac{m}{s^2}$$\\$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 7 kg block is$\\$$\mu_2$$R_2$ = (0.3) $\times$ 5kg = 15N$\\$So, the 3kg block cannot move relative to the 7kg block. The 3kg block and 7kg block both will have same acceleration ($a_2$ = $a_3$) which will be due to the 4N force because there is no friction from the floor.$\\$$\therefore$$a_2$ = $a_3$ = $\frac{4}{10}$ = 0.4$\frac{m}{s^2}$$\\b) 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$ $a_1$ = $a_2$ = $a_3$ = $\frac{10}{12}$ = ($\frac{5}{6}$)$\frac{m}{s^2}$$\\c) Similarly, it can be proved that when the 10N force is applied to the 7kg block, all the three blocks will move together.\\Again a_1 = a_2 = a_3 = (\frac{5}{6})\frac{m}{s^2} 24 24- The friction coefficient between the two blocks shown in figure (6-E9) is \mu but the floor is smooth. (a) What maximum horizontal force F can be applied without disturbing the equilibrium of the system ? (b) Suppose the horizontal force applied is double of that found in part (a). Find the accelerations of the two masses. Solution : Both upper block & lower block will have acceleration 2 \frac{m}{s^2}$$\\$$R_1 = mg ...(i)\\F - \mu$$R_1$ - T = 0 $\Rightarrow$ F - $\mu$mg - T = 0 ...(ii)$\\$T - $\mu$$R_1 = 0\\$$\Rightarrow$ T = $\mu$mg$\\$$\\$$\therefore$ F = $\mu$ mg + $\mu$ mg = 2 $\mu$ mg [putting T = $\mu$mg]$\\$b) 2F - T - $\mu$ mg - ma = 0 ...(i)$\\$$\\T - Ma - \mu mg = 0\\$$\Rightarrow$ T = Ma + $\mu$ mg$\\$Putting value of T in (i)$\\$2f - Ma - $\mu$ mg - $\mu$ mg - ma = 0$\\$$\Rightarrow 2(2\mumg) - 2 \mu mg = a(M + m) [Putting F = 2 \mumg]\\$$\Rightarrow$ 4$\mu$ mg - 2 $\mu$ mg = a (M + m)$\\$$\Rightarrow a = \frac{2\mu mg}{M+m}$$\\$Both blocks move with this acceleration ‘a’ in opposite direction.

25   25- Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b).

Solution :

$R_1$ + ma - mg =0$\\$$\Rightarrow R_1 = m(g-a) = mg - ma ...(i)\\T - \mu R_1 = 0 \Rightarrow T = m (mg - ma) ...(ii)\\Again, F - T - \mu R_1 = 0\\$$\Rightarrow$ F - {$\mu$(mg - ma)} - $\mu$(mg - ma) = 0$\\$$\Rightarrow F - \mu mg + \mu ma - \mumg + \mu ma = 0\\ F - \mu mg + \mu ma - \mu mg + \muma = 0\\$$\Rightarrow$ F = 2 $\mu$ mg - 2$\mu$ ma $\Rightarrow$ F = 2$\mu$ m(g-a)$\\$b) Acceleration of the block be $a_1$$\\$$R_1$ = mg – ma ...(i)$\\$2F - T - $\mu$$R_1 - ma_1 = 0 ...(ii)\\$$T_1$ - $\mu$$R_1 - M a_1 = 0\\$$\Rightarrow$ T = $\mu$$R_1 + M a_1$$\\$$\Rightarrow T = \mu mg - \mu ma + M a_1$$\\$Subtracting values of F & T, we get$\\$2(2$\mu$m(g – a)) – 2($\mu$mg – $\mu$ma + M$a_1$ ) – $\mu$mg + $\mu$ ma – $\mu$ $a_1$ = 0$\\$$\Rightarrow 4\mu mg – 4 \mu ma – 2 \mu mg + 2\mu ma = ma_1 + M a_1$$\\$$\Rightarrow a_1 = \frac{2\mu m(g-a)}{M+m}$$\\$Both blocks move with this acceleration but in opposite directions.

26   26- Consider the situation shown in figure (6-E9). Suppose a small electric field E exists in the space in the vertically upward direction and the upper block carries a positive charge Q on its top surface. The friction coefficient between the two blocks is g but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium ? [Hint : The force on a charge Q by the electric field E is F = QE in the direction of E.]

$R_1$ + QE - mg = 0$\\$$R_1 = mg - QE ...(i)\\F - T - \mu$$R_1$ = 0$\\$$\Rightarrow F - T \mu(mg – QE) = 0\\$$\Rightarrow$ F - T - $\mu$ mg + $\mu$QE = 0 ...(2)$\\$T - $\mu$ R 1 = 0$\\$$\Rightarrow T = \mu R_1 = \mu (mg - QE) = \mu mg - \muQE\\Now equation (ii) is F – mg + \mu QE – \mu mg + \mu QE = 0\\$$\Rightarrow$ F - 2 $\mu$ mg + 2$\mu$QE = 0$\\$$\Rightarrow F = 2\mumg - 2\mu QE \\ F = 2\mu(mg - QE)\\Maximum horizontal force that can be applied is 2\mu(mg – QE). 27 27- A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient of friction between the block and the table is g. The table does not move on the floor. Find the total frictional force applied by the floor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table ? Solution : Because the block slips on the table, maximum frictional force acts on it.\\From the free body diagram\\R = mg\\$$\therefore$ F - $\mu$ R = 0 $\Rightarrow$ F = $\mu$R = $\mu$mg$\\$But the table is at rest. So, frictional force at the legs of the table is not $\mu$ $R_1$ . Let be f, so form the free body diagram.$\\$ $f_0$ - $\mu$ $R_1$ = 0 $\Rightarrow$ $f_0$ = $\mu$R = $\mu$ mg$\\$Total frictional force on table by floor is $\mu$ mg.

28   28- Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the two blocks is $\mu_1$, and that between the bigger block and the ground is $\mu_2$.