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Question: 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 $\mathrm{g}$. 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: When block slips, the limiting friction force acts. $f f_{\text {table }}=\mu N$ $f...

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Question: Consider the situation shown in figure. Suppose a small electric field Eexists 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 Fcan be applied without disturbing the equilibrium? [Hint: The force on a charge $Q$ by the electric field $E$ is $F=Q E$ in the direction of $E$.] Solution: For block $\mathrm{m}$, $...

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Question: Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration $\mathrm{a}$ g.Repeat parts (a) and (b). Solution: In elevator, moving down with acceleration a, effective acceleration will be $g_{\text {eff }}=g-a$ So, replace $g$ by $g_{\text {eff }}$ in above answers....

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Question: The friction coefficient between the two blocks shown in figure 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: (a)For block m; $\mathrm{N}=\mathrm{mg}$ $\mathrm{ff}=\mu \mathrm{N}=\mu \mathrm{mg}$ $\mathrm{F}=\mathrm{T}+\mathrm{ff}(\therefore$ acceleration $=\mu)$-(i) For ...

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Question: Find the accelerations $a_{1}, a_{2}, a_{3}$ of the three blocks shown in figure (6-E8) if a horizontal force of $10 \mathrm{~N}$ is applied on (a) $2 \mathrm{~kg}$ block, (b) $3 \mathrm{~kg}$ block, (c) $7 \mathrm{~kg}$ block. Take $\mathrm{g}=10^{\mathrm{m}} / \mathrm{s}^{2}$. Solution: For $2 \mathrm{Kq}$; $N_{1}=2 g=20 \mathrm{~N} ;$ $f f_{1}=\mu_{1} N_{1}=0.2 \times 20=4 \mathrm{~N}$ For $3 \mathrm{Ka}$ : $N_{2}=N_{1}+3 g=20+30=50 N$ $f f_{2}=\mu_{2} N_{2}=0.3 \times 50=15 \mathrm...

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Question: A $2 \mathrm{~kg}$ block is placed over a $4 \mathrm{~kg}$ 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 $12 \mathrm{~N}$ is applied to (a) the upper block, (b) the lower block. Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$. Solution: (a) Normal contact for $2 \mathrm{Kg}=2 \mathrm{~g}=20 \mathrm{~N}$. Maximum friction force between $2 \mathrm{Kg}$ and $4 \m...

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Question: The friction coefficient between the board and the floor shown in figure is IA. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor. Solution: For man $\mathrm{N}+\mathrm{T}=\mathrm{Mg}$ For board $N_{g}=N+m g$ (vertical equilibrium) $T=f f=\mu N_{g}$ (Horizontal equilibrium) $T=\mu(N+m g)$ $T=\mu(M g-T+m g)$...

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Question: A block of mass Mis kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is $p$. 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 minimum force required be $\mathrm{P}$ at an angle $\theta$. $N+P \sin \theta=m g$ (vertical equilibrium) -(i) $P \cos \theta=\mu N$ (Body is about to move) -(ii) From (i) $P \cos \theta=\mu(m g-P \s...

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Question: Two masses $M_{1}$ and $M_{2}$ 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 $\mu$. Find the acceleration of the system and the force by the rod on one of the blocks. Solution: Both blocks are connected by rod. So, there acceleration must be same. $M_{1} g \sin \theta-T-f f_{1}=M_{1} a$ $f f_{1}=\mu N_{1}=\mu M_{1} g \cos \theta$ Add (1) and (3) $M_{1} g \sin \theta+M_...

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Question: Figure shows two blocks in contact sliding down an inclined surface of inclination $30^{\circ}$. The friction coefficient between the block of mass $2.0 \mathrm{~kg}$ and theincline is $p$ and that between the block of mass $4.0 \mathrm{~kg}$ and the incline is $\mu$. Calculate the acceleration of the $2.0 \mathrm{~kg}$ block if $(\mathrm{a}) \mu 1=0.20$ and $\mu 2=0.30$, (b) $\mu 1=0.30$ and $\mu 2=0.20$. Take $g=10^{m} / \mathrm{s}^{2}$. Solution: $P$ is the contact force. $N_{1}=4 g...

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Question: A car starts from rest on a half kilometre long bridge. The co-efficient of friction between the tyre and the road is $1.0$. Show that one cannot drive through the bridge is less than $10 \mathrm{~s}$. Solution: The maximum acceleration for car is given by, $a=\mu g=1 \times 10=10 \mathrm{~m} / \mathrm{s}^{2}$ $u=0$ $s=500 \mathrm{~m}$ .$s=u t+\frac{1}{2} a t^{2}$ $t=10 \mathrm{sec}$...

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Question: A car is going at a speed of $21.6 \mathrm{~km} / \mathrm{hr}$ when it encounters a $12.8 \mathrm{~m}$ long slope of angle $30^{\circ}$. The friction co-efficient between the road and the tyre is $\frac{\sqrt{3}}{2}$. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than $36 \mathrm{~km} / \mathrm{hr}$. Take $\mathrm{g}=10^{\mathrm{m} / \mathrm{s}^{2}}$ Solution: When driver applies brakes frictional force will be opposite t...

Question: The friction coefficient between an athelete's shoes and the ground is $0.90$. Suppose a superman wears these shoes and races for $50 \mathrm{~m}$. 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 completingthe $50 \mathrm{~m}$ starting from rest. (b) Suppose he takes exactly this minimum time to complete the $50 \mathrm{~m}$, what minimum time will he take to stop? Solution: (a) frictional force exerted by ground...

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Question: The friction coefficient between a road and the type 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 $36 \mathrm{~km} / \mathrm{hr}$ is stopped within $5 \mathrm{~m}$. Solution: $u=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}$ $v=0 ; s=5 \mathrm{~m}$ $v^{2}=u^{2}+2 a s$ $0^{2}=(10)^{2}+2(a)(5)$ $a=-10 \mathrm{~m} / \mathrm{s}^{2}$ Now, by translatory motion...

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Question: The friction coefficient between the table and the blockshown in figure is $0.2$. Find the tensions in the two strings. Solution: $15 g-T=15 a \quad-(i)$ $T-T_{1}-\mu N=5 a \ \mu N=0.2 \times 5 g=g$ $T-T_{1}-g=5 a-(i i)$ $T_{1}-5 g=5 a \quad-(i i i)$ Adding (i), (ii) and (iii) $15 g-g-5 g=15 a+5 a+50$ $9 q=25 a$ $a=\frac{9 g}{25}$ Now, put in eq.(i) $T=15 g-15\left(\frac{9 g}{25}\right)$ $\mathrm{T}=96 \mathrm{~N}$ From eq.(3) $T_{1}=5 g+5 \times \frac{9 g}{25}$ $T_{1}=68 \mathrm{~N}$...

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Question: If the tension in the string in figure is $16 \mathrm{~N}$ and the acceleration of each block is $0.5 \mathrm{~m} / \mathrm{s}^{2}$, find the friction coefficients at the two contacts with the blocks. Solution: For $2 \mathrm{Kg}$ block $T-\mu_{1} N_{1}=2 \times a$ $16-\mu_{1}(20)=2 \times(0.5)$ $15=20 \mu_{1}$ $\mu_{1}=0.75$ For $4 \mathrm{Ka}$ block $40 \sin 30-T-\mu_{2} N_{2}=4 a$$40 \sin 30-T-\mu_{2} N_{2}=4 a$ $40 \times\left(\frac{1}{2}\right)-T-\mu_{2}(40 \cos 30)=4 a$ $20-16-\m...

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Question: Consider the situation shown in figure. Calculate (a) the acceleration of the $1.0 \mathrm{~kg}$ blocks, (b) the tension in the string connecting the $1.0 \mathrm{~kg}$ blocks and (c) the tension in the string attached to $0.50 \mathrm{~kg}$. Solution: For mass $0.5 \mathrm{Kg}$ $m g-T=m a$ $0.5 g-T=0.5 a-(i)$ for $1 \mathrm{Kg}$ night mass $T-T_{1}-\mu N=m a$ $T-T_{1}-0.2 \times 10=1 \times a$ $T-T_{1}=2+a \quad-(i i)$ For $1 \mathrm{Kg}$ left mass $T_{1}-\mu N=m a$ $T_{1}-0.2 \times ...

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Question: 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 $\lambda$ be the angle of friction and $\mu$ the coefficient of static friction, $\lambda$ $\leq \tan -1 \mu$ Solution: Angle of friction, $\tan \lambda=\frac{f f}{N}$ The value of friction force depends upon external force applied. If the body does not move then $(f f=F)(f f(I m)=\mu N)$ When body is about to move or moves then $\mathrm{ff}=f_...

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Question: A body starts slipping down an incline and moves half meter in half second. How long will it take to move the next half meter? Solution: In the first half metre, $\mathrm{u}=0 \mathrm{~m} / \mathrm{s}, \mathrm{s}=0.5 \mathrm{~m}, \mathrm{t}=0.5 \mathrm{~s}$ $v=0+(0.5 \times 4)=2 \mathrm{~m} / \mathrm{s}$ $s=u t+1 / 2 a t^{2}$ $0.5=0+12(a)(0.5)^{2}$ $\mathrm{v}=\mathrm{u}+\mathrm{at}$ For the next half metre, $u=2 \mathrm{~m} / \mathrm{s}, \mathrm{a}=4 \mathrm{~m} / \mathrm{s}^{2}, \mat...

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Question: In a children-park an inclined plane is constructed with an angle of incline $45^{\circ}$ in the middle part. Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is $0.6$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ . Solution: $N=m g \cos 45^{\circ}$ ........(perpendicular to incline) $m g \sin 45-f f=m a$ .(along the incline) $m g \sin 45-\mu N=m a$ $g \sin 45-\mu g \cos 45=a$ $a=2 \sqrt{2} \mathrm{~m} / \mathrm{s}...

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Question: Repeat part (a) of the problem 6 if the push is applied horizontally and not parallel to the incline. Solution: Since the block is just to move up the incline so frictional force will act in downward direction. Since, block is in equilibrium. $N=m g \cos 30^{\circ}+F \sin 30^{\circ}$ .........(perpendicular to incline) $F \cos 30^{\circ}=m g \sin 30^{\circ}+f f$ ..........(along the incline) Now, $F \cos 30^{\circ}=m g \sin 30^{\circ}+\mu N$ $F \cos 30^{\circ}=m g \sin 30^{\circ}+\mu\l...

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Question: A body of mass $2 \mathrm{~kg}$ is lying on a rough inclined plane of inclination $30^{\circ}$. 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 $=0.2$. Solution: a) Applied force must be greater than net force which is acting downwards to make to move up. $F_{\text {req }}=\mu \mathrm{N}+\mathrm{mg} \sin 30^{\circ}$ $\mathrm{N}=\mathrm{mg} \cos 30^{\circ}$ $M=2 k g, g=9.8 \math...

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Question: Suppose the block of the previous problem is pushed down the incline with a fame of $4 \mathrm{~N}$. How far will the block move in the first two seconds after starting from rest? The mass of the block is $4 \mathrm{~kg}$. Solution: Along inclined plane $F_{N}=m a$ $F+m g \sin 30-f f_{k}=m a$ $4+4(10)\left(\frac{1}{2}\right)-\mu_{k} m g \cos 30^{\circ}=m a$ $24-0.11 \times 4 \times 10 \times \frac{\sqrt{3}}{2}=4 a$ $a=5 \mathrm{~m} / \mathrm{s}^{2}$ Now, $u=0 \frac{m}{s} ; t=2 s ; a=5 ...

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Question: A block slides down an inclined surface of inclination $30^{\circ}$ with the horizontal. Starting from rest it covers $8 \mathrm{~m}$ in the first two seconds. Fins the coefficient of kinetic friction between the two. Solution: $N=m g \cos 30^{\circ}$ $F_{N}=m a$ $m g \sin 30-f f=m a$ $m g \sin 30-\mu_{k} N=m a$ $m g \sin 30-\mu_{k} m g \cos 30=m a$ $g \sin 30-\mu_{k} g \cos 30=a$ Now, $u=0 ; s=8 m ; t=2 s$ $s=u t+\frac{1}{2} a t^{2}$ $8=0+\frac{1}{2}(a)(2)^{2}$ $a=4 \frac{m}{s^{2}}$ F...

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Question: A block is projected along a rough horizontal road with a speed of $10 \mathrm{~m} / \mathrm{s}$. If the coefficient of kinetic friction is $0.10$, how far it will travel before coming to rest? Solution: Since no driving force is present to move the block. So, frictional force will be zero....