Chapter # 6
Friction
[1]
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 & Sol.
(a) increse (a) esa o`f ) gksxh
B f = N N = mg – F cos q cos q
(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
Fcos
N
f
F
m f
2.
N
mg
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 &
` Sol.
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
B For smaller steps, normal force exerted by the ice is small. F = N
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 & (a) F = Mg
Sol.
(b) F = Mg
C If
T = 0 Fmin = Mg
If
T = T Fmax =
(c*) Mg F Mg 1 2 (d)Mg F Mg 1 2 1
F M
mg 2 T 2
= mg 1
2
T A Mg
mg F Mg 1 2 4.
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 (c) more than 500N throughout the journey (a) lEiw.kZ ;k=kk dky esa 500N U;wVu (c) lEiw.kZ ;k=kkdky esa 500N U;wVu ls vf/kd
(b*) less than 500N throughout the journey (d) > 500 N for time t1 and t3 and 500N for t2 . (b*) Eiw.kZ ;k=kk dky esa 500N U;wVu ls de
(d) t1 rFkk t3 le; ds fy;s > 500 rFkk t2 le; ds fy, 500N
manishkumarphysics.in
Chapter # 6
Friction
[2]
Sol.
For t2 constant velocity force exerted by the seat on the man = mg = 500 N for f1 & f3 friction force is also applied. So net force exerted by the seat on the man > 500 N
5.
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 &
Sol.
(a) is upward (c)is zero (a) Åij dh vksj gSA (c) 'kwU; gSA
(b) is downward (d*) the system cannot remain in equilibrium. (b) uhps dh vksj gSA (d*) fudk; dh lkE;koLFkk lEHko ugha gSA
D
0 A B
F
(mA+mB)g
Net vertical force is downwards. The system can not remain in equilibrium. 6.
Sol.
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
A
fA fBA A B
F
wA wB fBA
F is provide the normal force. Weight ofA& B in downward direction. So friction force fA & fBA (friction B due to A) is upwards direction. 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*) nksuksa dkjksa ds fy;s ,d leku 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
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Chapter # 6 Sol.
Friction
[3]
C Friction force on first car is = mg m1g acceleration due to friction force on first car = m = g 1 Friction force on second car is = m2g m 2g Acceleration due to friction force on speed car = m = g 2 Both acceleration are same & both initial speed are same (Given). S = ut + 1/2 at2 So, the minimum stopping distance is same for both cars.
8.
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 &
Sol.
9.
Sol.
10.
(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
In order to stop a car in shortest distance on a horizontal road, one should apply the brakes hard enough to just present slipping because hard brakes provide the enough normal & that provides the maximum friction. 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 izf rLFkkfir 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
D N = mg cos º fmax = N = mg cos º Just begins to slide mean mg sin = fmax mg sin º = mg cos º depends upon '' not depend on mass. Block sliding condition :mg sin – fmax = ma mg sin – mg cos = ma a = g (sin – cos ) a is depends upon '' & , not depend on mass.
N
mA
si mg
n
f
º
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
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Chapter # 6 Sol.
Friction
[4]
A F.B.D. of block M'
'
M T
'
N
Condition to present the sliding is fmax > T ' M'g > T ........ (1) F.B.D. boy M N
M'
T
f M'g
T f Mg
fmax = mg Condition to present the sliding is fmax > T mg > T Condition to present the sliding of the system (Block + Boy) is f ' > f (Block is not slide) ' M'g > mg ' M' > M < ' 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 Sol.
(a)
(b) (c) (d)
(b*) F > f
(c) FN > f
(d*) FN – f < F < FN + f
ABD System at equilibrium when F = f FN Net horizontal force is zero. f = FN F F > FN f Q f = FN and 0 1 So we can say that F >f, So net horizontal force is nonzero. F > f, Net horizontal force is zero. FN > f Fn > Fn <1 Here not given the relation between F & f so we can't say that net horizontal force is zero or nonzero. Fn – f < F < Fn + f f = Fn f f –f
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Chapter # 6 2.
Friction
[5]
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&
Sol.
3.
(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, ¡ [kqj njh gks ldrh gS] fdUrq os ,d nwljs ij fQlyrh ugha gSA
BD The contact force exerted by a body A on another body B is equal to the normal force between the bodies. We conclude that the force of friction between the bodies is zero and the bodies may be rough but they don't slip on each other. 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 &
Sol.
4.
Sol.
(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
BCD Coefficient of static friction is always greater than the coefficient of kinetic friction. Limiting friction is always greater than the kinetic & static friction. Maximum value of static friction is called the limiting friction. 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
CD Static friction force is a adjustable friction force. It adjust (equal) to applied force F upto limiting friction force than after it treat as a constant force. If F > limiting friction force at that time kinetic friction force is applied. Kinetic friction force is always less than the limiting friction force. f(friction force) Limiting friction force Kinetic friction force
45º
F (Applied Force)
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f
F
Chapter # 6 5.
Friction
[6]
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 &
Sol.
(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
AB If the vehicle is moving with a uniform velocity, so friction forces on the vehicle by the road is zero. If the vehicle is accelerating, the force is applied (due to tyre on the road) in west direction that cause net friction force applying east direction due to friction force car is moving in east direction.
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