Chapter # 6
Friction
SOLVED EXAMPLES 1. Sol.
A body of mass 400 g slides on a rough horizontal surface . If the frictional force is 3.0 N, find (a) the angle made by the contact force on the body with the vertical and (b) the magnitude of the force. Take g = 10 m/s2. Let the contact force on the block by the surface be F which makes an angle with the vertical (figure)
The component of F perpendicular to the contact surface is the normal force and the component of F parallel to the surface is the friction f. As the surface is horizontal is vertically upward. For vertical equilibrium, = Mg = (0.400 kg) (10 m/s2) = 4.0 N The friction force is f = 3.0 N (a) tan =
f N
=
3 4
3 = tan–1 = 37º 4 (b) The magnitude of the content force is
or,
F= = 2. Sol.
N 2 f2 (4.0 N) 2 (3.0 N) 2 = 5.0 N
A heavy box of mass 20 kg is pulled on a horizontal force. If the coefficient of kinetic friction between the box and the horizontal surface is 0.25, find the force of frictionless exerted by the horizontal surface on the box. The situation is shown in figure (6.6). In the vertical direction there is no acceleration , so = Mg
As the box slides on the horizontal surface exerts kinetic friction on the box. The magnitude of the kinetic friction is fk = k = k Mg = 0.25 × (20 kg) × (9.8 m/s2) = 49N. This force acts in the direction opposite to the pull. 3.
Sol.
A boy (30 kg) sitting on his horse whips it. The horse speeds up at an average acceleration of 2.0 m/s2.(a) If the boy does not slide back, what is the force of friction exerted by the hoerse on boy.(b) If the boy slides back during the acceleration , what can be said about the cofficientof static friction between the horse and the boy ? Take g = 10 m/s2. (a) The horse acting on the boy are (i) the weight Mg1 (ii) the normal contact force and (iii) the static friction fs.
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Chapter # 6
Friction
As the boy does not slide back, its acceleration a is equal to the acceleration of the horse. As friction is the only horizontal force , it must act along the acceleration and its magnitude is given by Newton’s second law fs = Ma = (30 kg) (2.0 m/s2) = 60 N (b) If the boy slides back , the horse could not exert a friction of 60 N on the boy. The maximum force of 60 N on the boy. The maximum force of static friction that the horse may exert on the boy. is fs = s
= s Mg = s (30 kg) (10 m/s2) = s 300N
where s is the coefficient of static friction. Thus,
s (300 N) < 60N s <
or, 4.
Sol.
60 = 0.20 . 300
A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually incresed. It is found that starts sliping when the plank makes an angle of 18º with the horizontal . However, once strated, the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficients of static and kinetic fraction between the block and the plank. The coeficient of static friction is s = tan 18° and the coefficient of kinetic friction is k = tan 15°
QUESTIONS
FOR
SHORT
ANSWER
1.
For most of the surfaces used in daily life, the friction cofficient is less than 1. Is it always necessary that the friction cofficient is less than 1?
2.
Why is it easier to push a heavy block from behind than to press it on the top and push ?
3.
What is the average friction force when a person has a usual 1 km walk ?
4.
Why is it difficult to walk on solid ice ?
5.
Can you acceleration a car on a frictionless horizontal road by putting more petrol in the engine?Can you stop a car going on a frictionless horizontal road by applying brakes?
6.
Spring fitted doors close by themselves when relesed.You want to keep the door open for a long time say for an hour. If you put a half kg stone in front of the door, it does not help. The stone slides with the door and the door gets closed . However , if you sandwitch a 20 g piece of wood in the small gap between the door and the floor, the door stays open. Explain why a much lighter piece of wood is able to keep the door open while the heavy stone fails.
7.
A classroom demonstration of Newton’s first law is as follows : A glass is covered with a plasstic card and a coin is placed on the card. The card is given a quick strike and the coin falls in the glass. (a) Should the friction cofficient between the card and the coin be small or large ? (b) Should the coin be light or heavy ? (c) Why does the experiment fail if the card is genntly pushed ?
8.
Can a tug of war be ever won on a frictionless surface ?
9.
Why do tyres have a better grip of the road while going on a level road than while going on an incline?
10.
You are standing with your bag in your hands on the ice in the middle of a pond. The ice is so slippery that it can offer no friction. How can you come out of the ice? manishkumarphysics.in
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Chapter # 6 11.
Friction
When two surfaces are polished, the friction cofficient between them decreases. But the friction cofficient increses and becomes very large if the surface are made highly smooth. Explain.
Objective - I 1.
In a situation the contact force by a rough horizontal surface on a body placed on it has constant magnitude. If the angle between this force and the vertical is decreased , the force and the vertical is decreased, the friction force between the surface and the body will
fdlh fof'k"V ifjfLfkfr esa ,d {ksfrt ,oa [kqjnjh lrg ij j[kh gqbZ oLrq ij yxk;s x;s lEidZ cy dkifjek.k fu;r jgrk gSa ;fn bl cy rFkk m/oZ js[kk ds e/; dks.k de gksrk gS rks lrg rFkk oLrq ds e/; ?k"kZ.k cy & (a) increse (a) esa o`f) gksxh 2.
(b*) decrease (b*) esa deh gksxhA
(c) remain the same (c) vifjofrZr jgsxkA
(d) may increase or decrease (d) de ;k vf/kd gks ldrk gSA
While walking on ice, one should take small steps to avoid slipping . This is because smaller steps to avoid sliping . This is because smaller steps ensure
cQZ ij pyrs le;. fQlyus ls cpus ds fy;s NksVs dne mBk;s tkrs gSA bldk dkj.k ;g gS fd NksVs dne lqj{kk djrs gS & ` 3.
(a) larger friction (b*) smaller friction (c) larger normal force (d) smaller normal force (a) vf/kd ?k"kZ.k ls (b*) vYi ?k"kZ.k ls (c) vf/kd vfYkyEcor~ cy ls (d) vYi vfHkyEcor~ cy ls A body of mass M is kept on a rough horizontal surface (friction cofficient = ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on A is F where ,d [kqjnjh {ksfrt lrg ij M nzO;eku dh oLrq A j[kh gqbZ gSA (?k"kZ.k xq.kkad = ) ,d O;fä bl oLrq ij {ksfrt fn'kk esa cy yxkdj bldks [khapuk pkgrk gS] fdUrq oLrq xfr'khy ugha gksrh gSA lrg ds }kjk A ij yxk;k x;k cy F gS] rks & (b) F = Mg
(a) F = Mg 4.
(c*) Mg F Mg 1 2 (d)Mg F Mg 1 2 1
A scooter starting from rest moves with a constant acceleration for a time t1, then with a constant deceleration for the next t2 and finally with a constant deceleration for the next t3 to come to rest . A 500N man sitting on the scooter behind the driver manges to stay at rest with respect to the scooter without touching any other part. The force exerted by the seat on the man is ,d LdwVj fojkekoLFkk ls pyuk izkjEHk djrk gS] ;g t1 le; ds fy;s fu;r Roj.k ls Rofjr gksrk gS] rRi'pkr~ vxys
t2 le; rd fu;r osx ls pyrk gS] ,oa vUr esa vxys t3 le; rd voeafnr gksdj fojkekoLFkk esa vk tkrk gSA LdwVj Mªkboj ds ihNs 500N U;wVu Hkkj dk O;fDr fdlh vU; Hkkx dks idM+s fcuk Lo;a dks LdwVj ds lkis{k fojkekoLFkk esa cuk;s j[krk gSA lhV }kjk O;fDr ij yxk;k x;k cy gS & (a) 500N throughout the journey (b*) less than 500N throughout the journey (c) more than 500N throughout the journey(d) > 500 N for time t1 and t3 and 500N for t2. (a) lEiw.kZ ;k=kk dky esa 500N U;wVu (b*) Eiw.kZ ;k=kk dky esa 500N U;wVu ls de (c) lEiw.kZ 500N 5.
;k=kkdky esa 500N U;wVu ls vf/kd
(d) t1 rFkk t3 le;
ds fy;s > 500 rFkk t2 le; ds fy,
Consider the situation shown in figure. The wall is smooth but the surface of A and B in contact are rough. The friction on B due to A in equilibrium fp=k esa iznf'kZr fLFkfr ds fy;s nhokj fpduh gS] fdUrq A rFkk B dh lEidZ lrgsa [kqjnjh gSA lkE;oLFkk esa A ds dkj.k B ij ?k"kZ.k cy &
(a) is upward (c)is zero (a) Åij dh vksj (c) 'kwU; gSA
gSA
(b) is downward (d*) the system cannot remain in equilibrium. (b) uhps dh vksj gSA (d*) fudk; dh lkE;koLFkk lEHko ugha gSA
manishkumarphysics.in
Page # 3
Chapter # 6
Friction
6.
Suppose all the surfaces in the previous problem are rough. The direction of friction on B due to A ekuk fd fiNys iz'u esa of.kZr leLr lrgsa [kqjnjh gSA A ds dkj.k B ij ?k"kZ.k cy dh fn'kk & (a*) is upward (b) is downward (c) is zero (d) depends on the masses of A and B. (a*) Åij dh vksj gSA (b) uhps dh vksj gSA (c) 'kwU; gSA (d) A rFkk B ds nzO;eku ij fuHkZj djrh gSA
7.
Two cars of unequal masses use simmilar tyres .If they are moving at the same intial speed, the minimum stopping distance
vleku nzO;eku dh nks dkjksa ds Vk;j ,d tSls gSA ;fn nksuksa ds vkjfEHkd osx ,d leku gS rks mudks jksdus dh U;wure nwjh & (a) is smaller for the heaviercar (c*) is same for both cars (a) Hkkjh dkj ds fy;s de gksxhA (c*) nksu ksa dkjksa ds fy;s ,d leku 8.
gksxhA
(b) is smaller for the lighter car (d) depends on the volume of the car. (b) gYdh dkj ds fy;s de gksxh (d) dkj ds vk;ru ij fuHkZj djsxhA
In order to stop a car in shortest distance on a horizontal road, one should
,d {ksfrt lM+d ij dkj dks U;wure nmwjh esa jksdus ds fy;s fdlh O;fä dks & (a) apply the brakes very hard so that the weels stop rotating (b*) apply the brakes hard enough to just prevent slipping (c) pump the brakes (press and release) (d) shut the engin off and not apply brakes . (a) bruh rkdr ls czsd yxkus pkfg;s fd ifg;s ?kweuk cUn dj nsaA (b*) i;kZIr rkdr ls czsd yxkus pkfg;s ftlls fQylus ls cpk jgsA (c) czsd iEi djus pkfg, (nck;s o NksM+s) (d) batu cUn djuk pkfg, o czsd ugha yxkus pkfg,A 9.
A block A kept on an inclined surface just begins to slide if the inclination is 30º. The block is replaced by another block B and it is found that it just begins to slide if the inclination is 40º. ;fn ur ry dk >qdko dks.k 30° gks rks ur ry ij j[kk gqvk CykWd A fQlyuk izkjEHk dj nsrk gSA bl CykWd dks ,d v; CykWd B ls izfrLFkkfir djkus ij] >qdko dks.k dk eku 40º gksus ij fQlyuk izkjEHk djrk gS & (a) mass of A > mass of B (b) mass of A < mass of B (c) mass of A = mass of B (d*) all the three are possible. (a) A dk nzO;eku > B dk nzO;eku (b) A dk nzO;eku < B dk nzO;eku (c) A dk nzO;eku = B dk nzO;eku (d*) rhuksa gh ifjLFkfr;k¡ lEHko gSA
10.
A boy of mass M is applying a horizontal force to slide a box of mass M’ on a rough horizontal surface . The cofficient of friction between the shoes of the boy and the floor is . In which of the following cases it is certainly not possible to slide the box ? ,d [kqjnjh {ksfrt lrg ij j[ks gq, M; nzO;eku ds ckWDl dks f[kldkus ds fy;s ,d M nzO;eku dk yM+dk ,d {ksfrt cy yxkrk gSA yM+ds ds twrksa rFkk Q'kZ ds e/; ?k"kZ.k xq.kkad ,oa ckWDl rFkk Q'kZ ds e/; ?k"kZ.k xq.kkad ' gSA fuEu
esa ls fdl ifjfLFkfr esa ckWDl f[kldkuk fu'fpr :i ls lEHko ugha gS & (a*) < ’, M < M’ (c) < ’, M > M’
(b) > ’, M < M’ (d) > ’, M > M
Objective - II 1.
Let F, FN and f denote the magnitudes of the contact force , normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero,
ekuk fd ,d nwljs ds lEidZ esa j[kh gqbZ lrgksa ds fy;s ,d oLrq ls nwljhoLrq ij yxk;k x;k lEidZ cy] vfHkyEcor~ cy rFkk ?k"kZ.k cy ds ifjek.k Øe'k% F, FN ,oa f gSA ;fn buesa ls dksbZ Hkh 'kwU; ugha gS rks & (a*) F > FN 2.
(b*) F > f
(c) FN > f
(d*) FN – f < F < FN + f
The contact force exerted by a body A on another body B is equal to the normal force between the bodies. We conclude that ,d oLrq A }kjk nwljh oLrq B ij yxk;k x;k lEidZ cy] nksuksa oLrqvksa ds e/; vfHkyEcor~ cy ds rqY; gSA bldk
rkRi;Z gS fd &
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Page # 4
Chapter # 6
Friction
(a) the surface must be frictionless (b*) the force of friction between the bodies is zero (c) the magniyude of normal force equals that of friction (d*) the bodies may be rough but they don’t slip on each other. (a) lrgsa ?k"k.kZ jfgr gSA (b*) nksuksa oLrqvksa ds e/; ?k"kZ.k cy 'kwU; gSA (c) vfHkyEcor~ cy dk ifjek.k ?k"kZ.k cy ds rqY; gSA (d*) oLrq,¡ [kqjnjh gks ldrh gS] fdUrq os ,d nwljs ij fQlyrh ugha 3.
gSA
Mark the correct statements about the friction between two bodies.
nks oLrqvksa ds e/; ?k"kZ.k ds fy;s lR; dFkuksa dks fpfUgr dhft;s & (a) Static friction is always greater than the kinetic friction. (b*) Cofficient of static friction is always greater than the cofficient of kinetic friction. (c*) Limiting friction is always greater than the kinetic friction. (d*) Limiting friction is never less than static friction. (a) LFkSfrd ?k"kZ.k lnSo] xfrd ?k"kZ.k ls vf/kd gksrk gSA (b*) LFkSfrd ?k"kZ.k xq.kkad lnSo xfrd ?k"kZ.k xq.kkad ls vf/kd gksrk gSA (c*) lhekar ?k"kZ.k xq.kkad lnSo xfrd ?k"kZ.k ls vf/kd gksrk gSA (d*) lhekar ?k"kZ.k dHkh Hkh LFkSfrd ?k"kZ.k ls de ugha gksrk gSA 4.
A block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them. ,d CykWd {ksfrt [kqjnjs Q'kZ ij j[kk gqvk gS rFkk bl ij ,d {ksfrt cy F yx jgk gsA F ds fofHkUu ekuksa ds fy;s Q'kZ }kjk CykWd ij yxk;s x;s ?k"kZ.k cy f ds ikB~;kad ysdj buds e/; ,d js[kkfp=k [khapk tkrk gS & (a) The graph is a straight line of slope 45º (b) The graph is straight line parallel to the F-axis. (c*) The graph is a straight line of slope 45º for small F and a straight line parallel to the F-axis for large F. (d*) There is a small kink on the graph. (a) js[kkfp=k 45º qdko okyh js[kk rFkk F ds vf/kd ekuksa ds fy;s F-v{k ds lekukUrj js[kk
gksxhA (d*) js[kkfp=k esa ,d NksVk lk mHkkj gksxkA 5.
Consider a vehicle going on a horizontal road toards east. Neglect any force by the air. The frictional forces on the vehicle by the road.
ekukfd {ksfrt lM+d ij ,d okgu iwoZ dh vksj tk jgk gSA ok;q ds dkj.k fdlh Hkh cy dks ux.; ekfu,A lM+d ds dkj.k okgu ij yxus okys ?k"kZ.k cy & (a*) is towards east if the vehicle is accelerating (b*) is zero if the vehicle is moving with a uniform velocity (c) must be towards east. (d) must be towards east. (a*) iwoZ dh vksj gksaxs] ;fn okgu Rofjr gSA (b*) 'kwU; gksaxs] ;fn okgu ,d leku osx ls xfr'khy gSA (c) iwoz dh vksj gh gksaxsA (d) if'pe dh vksj gh gksaxsA
WORKED OUT EXAMPLES 1.
Sol.
The coefficient of static friction between a block of mass ma nd an incline is 8 = 0.3. (a) What can be the maximum angle of the incline with the horizontal so that the block does not slip on the palne ? (b) If the incline makes an angle /2 with the horizontal, find the frictional force on the block. The situation is shown in figure. (a) the forces on the block are (i) the weight mg downward by the earth, (ii) the normal contact force by the incline, and (iii) the friction f parallel to the incline up the plane, by the incline.
manishkumarphysics.in
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Chapter # 6
Friction
As the block is at rest, these forces should add up to zero. Also, since is the maximum angle to prevent slipping, this is a case of limiting equilibrium and so f = m . Taking components prependicular to trhe incline, – mg cos = 0 or, f = mg sin .................... (i) Taking components parallel to the incline, f – mg sin = 0 or, f = mg sin or, sN= mg sin . ............... (ii) Dividing (ii) by (i) s = tan or, = tan–1 s = tan–1 (0.3). (b) If the angle of incline is reduced to /2, the equilibrium is not limiting, and hence the force of static friction f is less than s . To know the value of f, we proceed as in part (a) and get the equations. = mg cos (/2) and f = mg sin (/2). Thus, the force of friction is mg sin (/2). 2.
Sol.
A horizontal force of 20N is applied to a block of mass 4kg resting on a rough horizontal table. If the block does not move on the table, how much frictional force the table is applying on the block ? What can be said about the coefficient of static friction between the block and the table ? Take g = 10 m/s2. The situation is shown in figure. The force on the block are
(a) 4kg × 10 m/s2 = 40 N downward by the earth, (b) N upward by the table, (c) f = 20 N and N = 40 N (d) f towards left by the table (friction). As the block is at rest, these forces should add up to zero. Taking horizontal and veritical components, f = 20 N and = 40 N. Thus, the table exerts a firction (static) force of 20 N on the block in the direction opposite to the applied force. Since it is a case of static friction. f s , or, s f / or, s 0.5 3.
The coefficient of static fraction between the block of 2kg and the table shown in figure is s = 0.2. What should be the maximum value of m so that the blocks do not move ? Take g = 10 m/s2. The string and the pulley are light and smooth.
manishkumarphysics.in
Page # 6
Chapter # 6
Sol.
Friction
Consider the equilibrium of the block of mass m. The force on this block are (a) mg downward by the earth and (b) T upwrads by the string. Hence, T – mg = 0 Now consider the equilibrium of the two 2 kg block. The forces on this block are (a) T towards right by the string , (b) f towards left (friction) by the table, (c) 20 N downward (weight) by the earth and (d) upward (normal force) by the table. For vertical equilibrium of this block, = 20 N .....(ii) As m is the largest mass which can be used without moving the system , the friction is limiting. Thus, f = s For horizontal equilibrium of the 2 kg block, f =T Using equations (i) , (iii) and (iv) or, or,
.....(iii) ......(iv)
s N = mg 0.2 × 20 N = mg m
=
0.2 20 kg = 0.4 kg. 10
4.
The coefficient of static friction between the two blocks shown in figure is and table is smooth. What maximum horizontal force F can be applied to the block of mass M so that the blocks move together ?
Sol.
When the maximum force F is applied , both the blocks move together towards right. The horizontal force on the upper block of mass m is that due to the friction by the lower block of mass m is that due to the friction by the lower block of mass M. Hence this force on m should be towars right. The force on friction M by m should be towards left by Newton’s third law. As we are talking of maximum possible force that can be appled the friction is limiting and the hence f = N, where N is the normal force between the blocks. Consider the motion of m. The forces on m are (figure)
(a) mg downward by the earth (gravity), (b) N upward by the block M (normal force) and (c) f = N(friction) towards right by the block M. In the vertical direction , there is no acceleration. This gives N = mg manishkumarphysics.in
....(i) Page # 7
Chapter # 6
Friction
In the horizontal direction , let the acceleration be a , then N = ma mg = ma or, a = g Next, consider the motion of M (figure). The forces on M are or,
(a) mg downward by the earth (gravity) (b) N normal force by the incline and (c) f up the plane, (friction) by the incline. Taking components parallel to the incline and writing Newton’s second law, mg sin30º – f = mg / 4 or, f = mg / 4 There is no acceleration perpendicular to the incline . Hence = mg cos30º = mg.
3 2
As the block is slipping on the incline , friction is f = k
k =
So
=
mg 4 mg 3 / 2
=
1 2 3
5. Sol.
A block slides down on incline of angle 30° with an acceleration g/4. Find the kinetic frinction coefficient.
6.
A block of mass 2.5 kg is kept on a rough horizontal surface . It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is found that it takes 5 seconds to slide throught the first 10 m if a horizontal force of 15 N is applied and the block is genntly pushed to start the motion. Taking g = 10 m/s2, calculate the cofficients of static and kinetic friction between the block and the surface. The forces acting on the block are shown in figure. Here M = 2.5 kg and F = 15 N.
Sol.
When F = 15 N is applied to the block , the block remains in limiting equilibrium. The force is thus f =
. Applying Newton’s first law,, F =
so, that
and
= mg
F = s Mg
manishkumarphysics.in
Page # 8
Chapter # 6 or,
Friction
s =
15 N F = = 0.60 mg (2.5 kg) (10 m / s 2 )
When the block is gently pushed to start the motion, kinetic friction acts between the block and the surface . Since the blocks takes 5 seconds to slide throught the first 10 m, the acceleration a is given by 10 m = a=
1 a (5s)2 2 20 m/s2 = 0.8 m/s2. 25
The friction force is f = k = k Mg Applying Newton’s second law F – k Mg = Ma
k
=
F Ma Mg 15 N (2.5 kg) (0.8 m / s 2 )
= 7. Sol.
(2.5 kg) (10 m / s 2 )
= 0.52.
A block placed on a horizontal surface is being pushed by a force F making an angle with the vertical. If the friction force is µ, how much force is needed to get the block just started. Discuss the situation when < µ. will be equal to . For horizontal equilibrium, F sin = N For vertical equilibrium
For vertical equilibrium F cos + mg = N Eliminating N from these equations, F sin = F cos + mg or,
F
=
mg sin cos
If tan < we have (sin – cos ) < 0 and then F is negative . So for angles less than tan–1 , one cannot push the block ahead , however large the force may be . 8.
Find the maximum value of M / m in the situation shown in figure so that the system remains at restr. Friction coefficient at both the contacts is . Discuss the situation when tan < .
Sol.
Figure shows the forces acting on the two blocks . As we are looking for the maximum value of M / m, the manishkumarphysics.in
Page # 9
Chapter # 6
Friction
equilibrium is limiting . Hence , the frictional forces are equal to times the corresponding normal forces.
Equilibrium of the block m gives T = N1 and N1 = mg which gives T = mg ......(i) Next, consider the equilibrium of the block M. Taking components parallel to the incline T + N2 = Mg sin Taking components normal to the incline = Mg cos These give T = Mg(sin – cos ) .......(ii) From (i) and (ii) , mg = Mg(sin – cos ) or,
M/m =
sin cos
If tan < , (sin – cos ) < 0 and the system will not slide for any value of M/m. 9.
Consider the situation shown in figure. The coefficient of friction between the blocks is . Find the minimum and the maximum force F that can be applied in order to keep the smaller blocks at rest with respect to the bigger block. HCV_Ch-6_WOE_9 fp=k esa fn[kkbZ xbZ fLFkfr dks ekfu,A xqVdksa ds e/; ?k"kZ.k xq.kkad µ gSA cy F dk U;wure o vf/kdre eku Kkr dhft;s
rkfd NksVs xqVds] cM+s xqVds ds lkis{k fLFkj jg ldsA
Sol.
If no force is applied , the block A will slip on C towards right and the block B will move downward. Suppose the minimum force needed to prevent slipping is F. Taking A + B + C as the system, the only external horizontal force on the system is F. Hence, the acceleration of the system is
F M 2m Now take the block A as the system. The force on A are (figure), a=
....(i)
(i) tension T by the string towards right , (ii) friction f by the block C towards left, (iii) weigth mg downward and (iv) normal force upward For vertical equilibrium = mg As the minimum force needed to prevent slipping is applied , the friction is limiting . Thus, f = = mg As the block moves towards right with an acceleration a, T–f = ma manishkumarphysics.in
Page # 10
Chapter # 6
Friction T – mg = ma
or,
Now take the block B as the system. The forces are (figure 6-W14) (i) tension T upward, (ii) weight mg downward, (iii) normal force towards right with an acceleration a, (iv) friction f’ upward As the block moves towards right with an acceleration a, = ma As the friction is limiting , f’ = = ma For vertical equilibrium T + f’ = mg or, T + ma = mg Eliminating T from (ii) and (iii)
....(iii)
1 g. 1 When a large force is applied the block A slips on C towards left and the block B slips on C in the upward direction . The friction on A is toward right and that on B is downwards. Solving as above , the acceleration in this case is amin =
amax = Thus , a lies between
1 g 1
1 1 g and g 1 1
From (i) the force F should be between
1 1 (M + 2m) g and ( M + 2m )g. 1 1 10.
Figure shows two blocks connected by a light string placed on the two inclined parts of a triangular structure. The cofficients of static and kinetic friction are 0.28 and 0.25 respectively at each of the surfaces . (a) Find the minimum and maximum values of m for which the system ramains at rest.(b) Find the acceleration of either block if m is given the minimum value calculate in the first part (a) and is gently pushed up the incline for a short while.
f=kdks.kh;
2 kg
m Fixed 45º 45º Sol.
(a) Take the 2 kg block as the system. The forces on this block are shown in figure with M = 2 kg. It is assumed that m has its minimum value so that the 2 kg block has a tendency to slip down. As the block is in equilibrium , the resultant force should be zero.
T
N
T
N'
f f' mg
Mg Taking components to the incline = Mg cos 45º = Mg /
2
manishkumarphysics.in
Page # 11
Chapter # 6
Friction
Taking components | | to the incline T + f = Mg sin45º = Mg /
2
or, T = Mg Mg / 2 – f As its is a case of limiting equilibrium, = s
f
Mg Mg – s = (1 – s ) ...(i) 2 2 2 Now consider the other blocks as the system. The forcees acting on this block are shown in figure , Taking components to the incline , or,
T =
Mg
= mg cos45º =
Mg
2 Taking components || to the incline, T = mg sin45º + f’ =
Mg
2 As it is the case of limiting equilibrium Mg = s 2
f’ = s
Thus ,
T=
Mg 2
+ f’
(1 + s )
....(ii)
From (i) and (ii) m (1 – s ) = M (1 – s )
.....(iii)
(1 s ) 1 0.28 9 m = (1 ) M = × 2 kg = kg. 1 0.28 8 s When maximum possible value of m is supplied , the directions of friction are reserved because m has the tendency to slip down and 2 kg block to slip down and 2 kg block to slip up. Thus , the maximum value of m can be obtained from (iii) by or,
putting s = – 0.28. Thus , the maximum value of m is m= = (b) If m =
1 0.28 × 2 kg 1 0.28 32 kg. 9
9 kg and the sustem is gently pushed , kinetic friction will operate. Thus , 8
Mg f = k . 2
and
f’ =
k mg 2
,
where k = 0.25. If the accleration is a , Newton’s second law for M gives . Mg sin45º – T – f = Ma or,
Mg
k mg
= Ma 2 2 Applying Newton’s second law m figure (figure 6-W17), T – mg sin45º = f’= ma or,
T–
–T–
Mg 2
–
k mg 2
=ma
.....(iv)
.......(v)
Adding (iv) and (v)
Mg 2
(1 – s ) –
Mg 2
(1 + s ) = (M +m) a
manishkumarphysics.in
Page # 12
Chapter # 6 or,
Friction a =
a =
M (1 k ) m (1 k ) 2 (M m) 2 0.75 – 9 / 8 1.25 2 ( 2 9 / 8)
g
g
= 0.31 m/s2.
EXERCISE 1.
A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s2 . What is the cofficient of kinetic friction between the block and the plane ? [HCV_Chp_6_Ex_1] [Ans : 0.4] ,d [kqjnjs {ksfrt ry ij fQly jgh oLrq 4.0 eh-@ls-2 voeanu ls xfr'khy gSA oLrq rFkk ry ds e/; xfrd ?k"kZ.k
xq.kkad dk eku fdruk gSA [Ans : 0.4] 2.
A block is projected along a rough horizontal road with a speed of 10 m/s. If the cofficient of kinetic friction is 0.10 , how far will it travel before coming to rest ? [HCV_Chp_6_Ex_2] ,d [kqjnjh {ksfrt lM+d ds vuqfn'k ,d CykWd 10 eh-@ls-2 dh pky ls iz{ksfir fd;k tkrk gsA ;fn xfrd ?k"kZ.k xq.kkd 0.10 gS rks ;g iw.kZr;k fojkekoLFkk esa vkus ls iwo Z fdruh nwjh r; djsxk \ [Ans : 50 m]
3.
A block of mass m is kept on a horizontal table . If the static friction cofficient is , find the frictional force acting on the block. m nzO;eku dk ,d CykWd ,d {ksfrt Vscy ij j[kk gqvk gSA;fn LFkkSfrd ?k"kZ.k xq.kkad gS rks CykWd ij yxus okyk
?k"kZ.k cy Kkr dfj;sA Ans : zero] 4.
A block slides down an inclined surface of inclination 30º with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the cofficient of kinetic friction between two. {ksfrt ls 30º >qdko okys ur ry ij ,d CykWd uhps dh vksj fQlyrk gSA fLFkj voLFkk ls xfr izkjEHk djus ds izFke nks lsd.M i'pkr~ ;g 8 eh- nwjh r; djrk gsA nksuksa lrgksa ds e/; xfrd ?k"kZ.k xq.kkad Kkr dfj,A [Ans : 0.11]
5.
Suppose the block of the previous problem is pushed down the incline with a force of 4N. How far will the block move in the first two seconds after starting from rest? The mass of the block is 4 kg. ;fn fiNys iz'u esa of.kZr CykWd uhps dh vksj 4N cy ls /kdsyk tkrk gSA fojkekoLFkk ls xfr izkjEHk djus ds izFke nks lsd.M i'pkr~ ;g fdruh nwj foLFkkfir gksxk \ CykWd dk nzO;eku 4 fdxzk- gSA [Ans : 10m] A body of mass 2 kg is lyaing on a rough inclined plane of inclination 30º. Find the magnitude of the force parallel to the incline needed to make the block move. 30° >qdko okys ,d [kqjnjs ur ry ij 2 fdxzk- nzO;eku dh ,d oLrq fLFkr gsA CykWd dks xfr'khy djus ds fy;s
6.
ur ry ds lekukUrj vko';d cy Kkr dfj;s] ;fn CykWd foLFkkfir gksrk gSA (a) up the incline (b) down the incline. Cofficient of static friction = 0.2 (a) ur ry ij Åij dh vksj (b) ur ry ij LFkSfrd cy xq.kkad = 0.2 [Ans : (a) 13N (b) zero] 7.
8.
uhps dh vksjA
Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline. ;fn yxk;k x;k /kDdk {ksfrt fn'kk esa gS] fdUrq ur ry ds lekukUrj ugha gS rks iz'u 6 dk Hkkx (a) iqu% Ans : 17.5N
nksgjkbZ;sA
In a children-park an inclined plane is constructed with an angle of incline 45º in the part (figure). Find the acceleration of a boy slising on it if the friction cofficient between the cloth of the boy and the incline is 0.6 and g = 10 m/s2. cPpksa ds [ksyus ds ikdZ esa ,d ur ry bl izdkj cuk;k x;k gS fd blds e/; Hkkx esa >qdko dks.k 45º (fp=k)A ;fn yM+ds ds diM+ksa rfkk ur ry ds e/; ?k"kZ.k xq.kkad 0.6 rFkk g = 10 eh-@ls-2 gS rks fQly jgs yM+ds dk Roj.k Kkr
dfj;sA
manishkumarphysics.in
Page # 13
Chapter # 6
Friction
45º [Ans : 2 2 m/s2] 9.
A body starts sliping down an incline and moves half meter in half second . How long will it taketo move the next half meter ?
,d oLrq ur ry ij fQlyukizkjEhk djus ds i'pkr~ vk/ks lsd.M esa vk/kk ehVj nwjh r; djrh gSA vxyh vk/kk ehVj nwjh r; djus esa bldks fdruk le; yxsxk \ Ans : 0.21 s 10.
The angle between the resultant contract force and the normal force exerted by a body on the other is called the angle of friction. Show that, if be the angle of friction and the cofficient of static friction, tan–1
ifj.kkeh lEidZ cy rFkk ,d oLrq ls nwljh oLrq ij yxk;s x;s vfHkyEcor~ cy ds e/; dks.k] ?k"kZ.k dks.k dgykrk gSA ;fn ?k"kZ.k dks.k rFkk LFkSfrd ?k"kZ.k xq.kkad gS rks O;Dr dfj;s fd tan–1 . [HCV_Chp_6_Ex_10] 11.
Consider the situation shown in figure. In different cases 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 string atached to 0.50 kg. fp=k esa fn[kkbZ fLFkfr ij fopkj dhft;sA fofHkUu fLFkfr;ksa esa x.kuk dhft;sA (a) 1.0 fdxzk- xqVdksa dk Roj.k (b) 1 fdxzk xqVdksa dks tksM+us okyh jLlh esa ruko (c) 0.5 fdxzk- xqVds ls tqM+h jLlh esa rukoA Case I : m1 = 0.2 , m2 = 0.5 Case II : m1 = 0.2 , m2 = 0.3 Case III : m1 = 0.3 , m2 = 0.2 Case IV : m1 = 0.2 , m2 = 0.2 Case V : m1 = 0.5 , m2 = 0.2 (Take g = 10 m/s2) HCV_Ch-6_Ex._11
1.0 kg
1.0 kg
///////////////////////////////////////////////// µ1 µ2
12.
0.5 kg
[Ans : Case I : (a) 0 m/s2 , (b) 0 N , (c) 5 N ] Case II : (a) 0 m/s2 , (b) 2 N , (c) 5 N ] Case III : (a) 0 m/s2 , (b) 3 N , (c) 5 N ] Case IV : (a) 0.4m/s2 , (b) 2.4 N , (c) 4.8 N Case V : (a) 0 m/s2 , (b) 3 N , (c) 5 N ] If the tension in the string in figure is 16N and the acceleration of each block is 0.5 m/s2, find the friction cofficients at the two constants with the blocks. ;fn fp=k esa iznf'kZr Mksjh esaruko 16 U;wVu gS rFkk izR;sd CykWd dk Roj.k 0.5 eh-@ls-2 gS] CykWdksa dh lEidZ lrgksa
ds e/; ?k"kZ.k xq.kkad Kkr dfj;sA
manishkumarphysics.in
Page # 14
Chapter # 6
13.
14.
15.
Friction
[Ans : 1 = 0.75 , 2 = 0.06] The friction cofficient between the table and the block shown in figure is 0.2 . Find the tansions in the two strings. fp=k esa iznf'kZr Vscy rfk CykWd ds e/; ?k"kZ.k xq.kkad 0.2 gSA nksuksa Mksfj;ksa esa rukoksa dseku Kkr dfj;sA
[Ans : 96N is the left string and 68N in the right] The friction cofficient 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. ,d okgu ds Vk;jksa rFkk lM+d ds e/; ?k"kZ.k xq.kkad 4/3 gSA lM+d esa lEHko og vf/kdre >qdko Kkr djf;sa rkfd iwjh rkdr ls czd s yxkus ij tc ifg;s fQlyuk izkjEHk dj ns rks Hkh uhps dh vksj 36 fdeh@?k.Vk dh pky ls xfr'khy okgu dks 5 eh- nwjh esa jksdk tk ldsA [Ans :16º] The friction cofficient between an athelet’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 ? ,d /kod ds twrksa rFkk tehu ds e/; ?k"kZ.k xq.kkad 0.90 gSA ekukfd ,d lqijesu bu twrksa dks igurk gS rFkk 50 ehnwjh ij nkSM+rk gSA mlds rst xfr ls nkSM+us dh dksbZ lhek ugha gSA (a) og U;wure le; Kkr dfj;s ftlesa og fojkekoLFkk ls nkSM+uk izkjEHk djds 50 eh- nwjh r; dj ysxkA (b) eku fyft;s fd og Bhd bruk gh U;wure le; 50
eh- nwjh r; djus esa yxkrk gS rks og :dus esa U;wure fdruk le; yxk;sxk \ Ans :(a) 16.
10 10 s (b) s ] 3 3
A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30º (figure). The friction cofficient between the road and tyre is 1/2 3 . Show that no matter how hard the driver applies the brakes , the car will reach the bottom with a speed greater than 36 km/hr. Take g = 10 m/s2. tc ,d dkj ds lekus 12.8 eh- yEck o 30° dks.k dk
rFkk Vk;j ds e/; ?k"kZ.k xq.kkad 1/2 3 gSA O;ä dfj;s fd Mªkboj pkgs fdruk gh rkdr ls czsd yxk;s] dkj uhps 36 fdeh-@?k.Vk ls vf/kd pky ls igqapsxhA (g = 10 eh-@ls-2)
17.
A car starts from rest on a half kilometer long bridge . The cofficient of friction between the tyre and the road is 1.0. Show that one cannot drive through the bridge in less than 10s.
,d dkj vk/kk fdyksehVj yEcs lsrq ij fojkekoLFkk ls pyuk izkjEHk djrh gSA Vk;jksa rFkk lM+d ds e/; ?k"kZ.k xq.kkad 1.0 gSA O;ä dfj;s fd dksbZ Hkh lsrq dks 10 lsd.M ls de le; esa ikj ugha dj ldrk gSA 18.
Figure shows two blocks in contact sliding down an inclined surface of inclination 30º. The friction coefficient between the block of mass 2.0 k.g and the incline is 1 and that between the block of mass 4.0 kg and the incline is 2 . Calculate the acceleration of the 2.0 kg block if (a) 1 = 0.20 and 2 = 30.0, (b) 1 = 0.30 and
2 = 0.20. Take g = 10 m/s2. fp=kkuqlkj 30º >qdko okyh ur lrg
ij nks CykWd ,d nwljs ds lEidZ esa fQly jgs gS] 2.0 fdxzk nzO;eku okys CykWd
manishkumarphysics.in
Page # 15
Chapter # 6
Friction
,oa 4 fdxzk nzO;eku okys CykWd dk ur ry dh lrg ls ?k"kZ.k xq.kkad Øe'k% 1 o 2 gSA 2.0 fdxzk nzO;eku okys CykWd dk Roj.k Kkr dfj;sA ;fn (a) 1 = 0.20 rFkk 2 = 30.0, (b) 1 = 0.30 rFkk 2 = 0.20. (g = 10 eh-@ls-2)
19.
[Ans : 2.7 m/s2, 2.4 m/s2] Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle with with the horizontal. The friction cofficient at both the contacts is . Find the acceleration of the system and the force by the rod on one of the blocks . M1 rFkk M2 nks nzO;ekuksa dks ,d NM+ dh lgk;rk ls tksM+k x;k gS rFkk ;g fudk; >qdko okys [kqjnjs ur ry ij fQly jgk gSA nksuksa dh lEidZ lrgksa dk ?k"kZ.k xq.kkad gSA fudk; dk Roj.k Kkr dfj;s rFkk ,d CykWd ij NM+
}kjk yxk;k x;k cy Kkr dfj;sA
Ans : a = g(sin – cos ), zero] 20.
A block of mass M is kept on a rough horizontal surface . The cofficient of static friction between the block and the surface is . The block is to be pulled by applying a minimum force is needed to slide the block ? In which direction should this force act and what is its value. HCV_Ch-6_Ex_20 ,d M nzO;eku dk xqVdk [kqjnjs lrg ij j[kk gSA lrg rFkk xqVds ds e/; LFkSfrd ?k"kZ.k xq.kkad µ gSA xqVds dks U;wure
cy yxkdj [khapk tkrk gS fd xqVdk fQlyuk izkjEHk djs rks fdl fn'kk esa cy yxuk pkfg, vkSj bldk eku D;k gksxk \ [Ans :
21.
Mg 1 2
at an angle tan–1 with the horizontal]
The friction coefficient between the board and the floor shown in figure is .Find the maximum force that the man can exert on the rope so that the board does not slip on the floor. fp=k esa fn[kk, x, ry vkSj cksMZ ds chp ?k"kZ.k xq.kkad gSA jLlh ij yxk;s x;s cy dk vf/kdre eku Kkr djks ;fn cksMZ
ry ij ugha fQlyrk gSA
(M m)g 1/ 3 500 , (40 + 10) 10 = 125 N] 4/3 4 1 Board does not slip i.e. net force on tea board should be balanced i.e. T = N1 cksMZ fQly ugh jgk gS vFkkZr~ cksMZ ij dqy cy T = N1 gksuk pkfg,A [Ans :
Sol.
N1
N T m
T
N1 N
mg
N1 manishkumarphysics.in mg
Page # 16
Chapter # 6
Friction
by FBD of man T + N = Mg vkneh dk FBD T + N = Mg and block N1 = N + mg
.... (1) .... (1) .... (2)
rFkk CykWd dk
by (1) and (2)
N1 = Mg – T + mg N1 = (M + m)g – T
for block T = N1 CykWd ds fy, T = N1 T = [(M + m)g – T] T (1 + ) = (M + m)g T=
(M + m)g 1
by putting the value = =
1 , M = 40 , m = 10, g = 10 3
1 , M = 40 , m = 10, g = 10 dk 3
eku j[kus ij
1/ 3 500 (40 + 10) 10 125 N 4/3 4 A 2 kg block is placed over a 4 kg block and the both are placed on a smooth horizontal surface . The cofficient 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 m/s2. 4 fdxzk nzO;eku ds ,d CykWd ij 2 fdxzk nzO;eku dk CykWd j[kk gqvk gS rFkk nksuksa ,d fpduh {ksfrt Vscy ij j[ks gq, gSA CykWdksa ds e/; ?k"kZ.k xq.kkad 0.20 gSA nksuksa CykWdksa dk Roj.k Kkr dfj;s ;fn ,d {ksfrt 12 N cy yxk;k tkrk gSA (a) Åij okys CykWd ij (b) uhps okys CykWd ijA (g = 10 eh-@ls-2) [Ans :(a) upper block 4 m/s2, lower block 1 m/s2, (b) both blocks 2 m/s2]
T= 22.
23.
Find the accelerations a1, a2, a3 of the three blocks shown in figure. If a horizontal force of 10N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. (Take g = 10 m/s2) fp=k esa iznf'kZr rhu CykWdksa ds Roj.k a1, a2, a3 Kkr dfj;s] ;fn 10 N U;wVu dk ,d {ksfrt cy yxk;k tkrk gSA (a) 2 fdxzk okys CykWd ij (b) 3 fdxzk okys CykWd ij (c) 7 fdxzk okys CykWd ij (g = 10 eh-@ls-2)
5 m/s2 , (c)same as (b)] 6 The friction cofficient between the two blocks shown in figure is but 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. fp=k esa iznf'kZr nks CykWdksa ds e/; ?k"kZ.k xq.kkad gS] fdUrq Q'kZ ?k"kZ.k jfgr gSA (a) {ksfrt cy F dk vf/kdre eku fdrukgks fd fudk; dh lkE;oLFkk vizHkkfor jgs \ b) ekuk Hkkx (a) esa x.kuk fd;s x;s cy ls nqxuk {kSfrt cy ykx;k
[Ans : (a) a1 = 3 m/s2, a2 = a3 = 0.4 m/s2 24.
, (b) a1 = a2 = a3 =
tkrk gSA nksuksa nzO;ekuksa ds Roj.k dk eku Kkr dfj;sA
[Ans : (a) 2 mg 25.
(b) 2 mg in opposite direction ] M m
Suppose the entire system of the previous question is kept inside an elevator which is comming down with an acceleration a < g. Repeat parts (a) and (b). manishkumarphysics.in
Page # 17
Chapter # 6
Friction
fiNys iz'u dk lEiw.kZ fudk; ,d fy¶V esa j[k fn;k tkrk gS] tks fd uhps dh vksj Roj.k a (a < g) ls vk jgh gSA Hkkx (a) rFkk (b) dks gy dfj;sA [Ans : (a) 2 m(g–a) 26.
(b) 2 m(g a) m M
]
Consider the situation shown in figure. Suppose a small electric field E exists in the space in the a positively charge Q on its top surface .The friction cofficient between the two blocks is 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.] fp=k esa iznf'kZr O;oLFkk ij fopkj dfj;sA ekukfd Åij okys CykWd dh Åijh lrg ij /kukRed vkos'k Q gS rFkk vklikl ds {ks=k esa ,d vYi fo|qr {ks=k E Å/okZ/kj fn'kk esa gSA nksuksa CykWdksa ds e/; ?k"kZ.k xq.kkad gS rFkkQ'kZ fpduk
gSA lkE;oLFkk esa dksbZ ifjorZu fd;s fcuk vf/kdre fdruk {ksfrt cy yxk;k tk ldrk gS \ [ladsr % fo|qr {ks=k E ds dkj.k Q vkos'k ij cy F = QE, E dh fn'kk esa] Ans : 2 (mg–QE 27.
A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The cofficient of friction between the block and the table is . The table does not move on the floor. Find the total friction force applied by the floor on the legs of the table. Do you need the friction coefficint between the table and the floor or the mass of the table ? m nzO;eku ds ,d CykWd ij {ksfrt cy yxkus ij ;g ,d [kqjnjh {ksfrt Vscy fQlyrk gSA CykWd rFkk Vscy ds e/ ; ?k"kZ.k xq.kkad gSA Vscy Q'kZ ij xfr ugha djrh gSA Q'kZ ds }kjk Vscy ds iSjksa ij yxk;k x;k dqy ?k"kZ.k cy
KKr dfj;saA D;k vkidks Vscy rFkk Q'kZ ds e/; ?k"kZ.k xq.kkad dh rFkk Vscy ds nzO;eku dh tkudkjh pkfg;s \
28.
[Ans : mg ] Find the acceleration of the block of mass M in the situation of figure. The cofficient of friction between the two blocks is 1 and that between the bigger block and the ground is 2. fp=k esa iznf'kZr O;oLFkk ds fy;s M nzO;eku ds CykWd dk Roj.k Kkr dfj;sA nksuksa CykWdksa ds e/; ?k"kZ.k xq.kkad dk eku 1 rFkk cM+s CykWd ,oa tehu ds e/; 2 gSA
[2m 2 (M m)g] [Ans : M m[5 2( )] ] 1 2 29.
A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, cofficient of static friction being 0.5. Anotherhorizontal force of 15N is applied on the block in a direction parallel to the wall. Will the block move ? If yes , in which direction ? If no, find the frictional force exerted by the wall on the block. 2 fdxzk- nzO;eku dk ,d CykWd] [kqjnjh m/okZ/kj nhokj ij 40 U;wVu cy ls nck;k tk jgk gS] LFkSfrd ?k"kZ.k xq.kkad 0.5 gSA 15 U;wVCu dk ,d vU; {kSfrt cy nhokj ds lekukUrj fn'kk esa yxk;k tk jgk gSA D;k CykWd xfr djsxk \
;fn gk¡ rks fdl fn'kk esa \ ;fn ugha] nhokj }kjk CykWd ij yxk;k tkjgk ?k"kZ.k cy Kkr dfj;sA Ans : it will move at an angle of 53º with the 15N force manishkumarphysics.in
Page # 18
Chapter # 6 30.
Friction
A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall B by his back (figure). Assume that the friction cofficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts . (a) Show that the person pushes the two walls with equal force. (b) Find the normal force exerted by either wall on the peson. Take g = 101 m/s2. nks m/okz/kj nhokjksa A o B ds chp esa ,d O;fä nhokj A dks gkFkksa o iSjksa ls rFkk nhokj B dks vius i`"B Hkkx ls nckdj] Lo;a dks fLFkj cuk;s j[krk gSA (fp=k) ekukfd ml O;fDr ds 'kjhj rFkk nhokjksa ds e/; ?k"kZ.k xq.kkad 0.8 gS] ,oa lEidZ lrgksa lhekUr ?k"KZ.k yx jgk gS % (a) O;Dr dfj;s fd O;fDr nksuksa nhokjksa dks leku cy ls nck jgk gSA (b) fdlh ,d nhokj }kjk O;fDr ij yxk;k tk jgk vfHkyEcor~ cy Kkr dfj;sA (g = 101 m/s2)
[Ans : (b) 250N ] 31.
Figure shows a small block of mass m kept at the left end of a larger block of mass M and the length .The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and and the bigger block is and that between the blocks is /2. Find the time elapsed before the smaller block separates from the bigger block. fp=k esa n'kkZ, vuqlkj ,d yEckbZ o M nzO;eku ds cM+s xqVds ds cka;s fljs ij ,d m nzO;eku dk NksVk xqVdk j[kk gSA ;g fudk; {kSfrt lrg ij fQly ldrk gSA fudk; nka;h vksj çkjfEHkd osx v ls xfr djuk çkjEHk djrk gSA cM+s xqVds rFkk tehu ds e/; ?k"kZ.k xq.kkad rFkk nksuksa xqVdksa ds e/; ?k"kZ.k xq.kkad /2 gSA ml le; ds eku dh x.kuk
dhft, tc NksVk xqVdk cM+s xqVds ls vyx gks tk,xkA
HCV_Ch-6_Ex._31
m M
/////////////////////////////////// [Ans :
4M ] (M m) g
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