SOLUTIONS TO CONCEPTS CHAPTER 6 Let m = mass of the block From the freebody diagram, R-mg=0>R=mg ‘Again ma — u R= 0=> ma img (from (1)) eee, wee = a= 19 4=pg=fi=Alg=4lt0=0.4 ‘The co-efficient of kinetic fiction between the block and the plane is 0.4 ‘ Due to friction the body will decelerate Let the deceleration be ‘a a >R=mg (1) >ma= 1 R= wmg (rom (1)) 410 tmis® mae] Initial velocity u = 10 mis Final velocity v= 0 mis ‘ ‘a= “Amis? (deceleration) 40? _ 100 2 4) 2 it wil travel 50m before coming to rest Body is kept on the horizontal table. Ifno force is applied, no frictional force will be there t ieecbes je | F Applied force : From grap it can be seen that when applied force is zero, Tr frictional foreeis zero. a From the free body diagram, R-mgcos@=0->R=mgcos® .(1) For the block U=0, =8m,t=2s00, stutt %at’ > 8=0+%a2' a= 4mis* ‘Again, wR + ma—mg sin 0 =0 = img cos © + ma—mg sin 0 = 0 from (1) = mlug cos 4 + a~g sin 6) = 4 10 * cos 30° =g sin 30° —a =x 10 (G73) = 10% (112)-4 = (V3) u=1 > p= 1(6/¥3) = 0.11 Coreffcient of kinetic friction between the two is 0.11 From the free body diagram 44a — uR + 4g sin 30° = 0 ) R-4g cos 30°= 0 2) > R= 4g cos 30° R Putting the values of Ris & in equn. (1) 4= 4a ~0.11 xg cos 30° + 4g sin 30 = 4—da-0.11 x4 x 10 (V3 12) +4 x 10 * (1/2) =0 3 4-4a-381+20=0-a=5mis? Forthe block u=0, t=2sec, a= Smis? m9 Distance s= ut + % at? = $= 0+ (1/2) 5 x2®= 10m The block will move 10m. = 60m etChapter 6 To make the block move up the incine, the force should be equal and opposite fo the net force acting down the incine = wR-+2. sin 30° =02%(28)/32198% (12) [rom(t) =3.99+9.8=19N \With tis minimum force the body movo up te inane witha constant valocty as net YO force on tis zero R WR 2 b) Net force acting down the inetina is given by, {boy moving down) F=2q sin 30° — pR Rg = 298 (1/2)-339=641N £ Due to F =641N the body will move down the incline wih acceleration No extemal force is required Force required is zero. wR YN From the fee ody diagram > 9= 10m m=2kg, © 0= 30", u=02 R—mg.os 0-F sin 0=0 > R=mg cos 0+F sin» ..(1) ‘And mg sin 6 + uR-F 0s @ = 0 = mg sin 6 + (mg cos 0 + F sin 6) ~F cos 0 = 0 = mg sin 0+ jimg cos 0 + 4 F sin OF cos _(Mgsin0—ymgcos0) ‘(sind cose) 10 x(12)+0.2%2%10x(V5I2) _ 13.464 0.2x(1/2)=(V5/2) 0.76 sm mass of child R, R—mg cos 45°= 0 = R=mg.cos 45" = mg i (1) Net force acting on the boy due fo which t sides down is mg sin 45" pR = mg sin 45° - mg cos 45° = mx 10 (1/'¥2) =0.6 x mx 10% (1/y2) = m[(6/ V2 )-0.6 x (5 /v2)} = m(242) a, SF =F = 17.7N = 17.5N lis. fg tera A ore om Siete, he Bay nee le eer meee oat From the free body diagram aR (ay ma + mg sin 6 -wR=0 + mg(sind - cos) Fe 7g (sin 8-H 60s 8) ms Forthe first half mt. u=0, s=0.5m, t= 0.5 sec. So, v=u+al=0+(0.5)4=2mis S=ut+ %at®05= For the next half metre W=2mis, a=4mis?, — s=05. 3 05=2t+ (124? > 2042-05 + %a (06 > a= Amis? (2) 62Chapter 6 Time taken to cover next half meter is 0.21sec. 10. {> applied force Fi contact force ma F frictional force ‘ R— normal reaction t F pe tan. =FIR 2 Wi When F = uR, F is the limiting friction (max friction). When applied force increase) force of friction increase upto limiting friction (4) Before reaching limiting friction mo Feu Fcuon F sth tana sus astan” yp ‘ tan R 11. Frog the fee body clegiih t T+05a-05 pR+1a+T;- wR+ta-T,=0 R+1a=T, (3) From (2) & (3) > pR+a=T-T, 2T-T2T, ' 3 1 =2T, Equation (2) becomes p+ a +T,— 27, =0 Jatt] 7 = uR+a—T; ca STauR+a=029+8 ..(4) ahg 3 Equation (1) becomes 2T; + 0/5a - 0.5g = 0 950-059 9255_0250 1) From (4) & (5) 0.29 + a = 0,25g — 0.262 05 125 a) Accln of 1kg blocks each is 0.4m/s* 'b) Tension T, = 0.2g + a+ 0. 4N o) T=059- 05a =05* 10-0.5%04-48N 12, From the free body diagram wR4+1-16=0 > ws (29) + -15)=8 > ms = 15/20 = 0.75 wa Ry #405 + 164g sin 30! wR =z (20 V3 ) +2 + 16 - 20=0 sms 2. = 1 = 0057 ~0.06 208 68 408 17.32 Fe (a) (2) Y059 089 st % 10 = 0.04 | 10 = 0.4mis* 0.75 & up = 0.06, esChapter 6 3 14 16. 8 Tog ah ‘ T aR 53 5% 150 rag"? From the free body diagram, T+ 18-159 =0 T=(T, + Sat uR) 0 T,259-5a=0 >T=159—18@ fl) > T- (6g +5a+ Sat pR)=0 ST.H59 + Sa...) T= 59+ 10a+yR ..(i) From () & (il 159 - 15a= 59 + 10a + 0.2 (59) > 25a = 90 > a =3.6mis? Equation (i) = T= 10+ 10x36 +0.2%5x 10 = 96N in the left string Equation (ii) T; = 5g + Sa = 3=5m, u=6kmih % 10+ 5 x 3.6 =68N in the right string g= 10m" From the freebody diagrams, R-mg cos 8=0; 9 = 10mis" => R=mg cos @...(); w= 4/3, ‘Again, ma + mg sin 0- w R= 0 = ma + mg sin 8 —umg cos =a +gsin 6—mg cos 8 = 0 = 10+ 10 sin @- (4/3) x 10 cos 8 = 0 = 30+ 30 sin 8-40 cos 6 =0 3 3+3sin0—4.cos > 400s 0-3sino= =o 24, 1-sin?@ =3+3 sine 16 (1-sin?e)=9+9sin?a+ 18 sin@ 4 = 21882 14 9 28 (raking +ve sign oni Bin 0 = #4 =0.28 [Taking +ve sign only] > = sin” (0.28) = 16° Maximum incline is to reach in minimum time, he has to move with maximum possible acceleration. Lt, the maximum acceleration is ‘at ma-uR=0= mal= mg anu g= 0.9% 10= 9m 2) Initial velocity u = 0, a= mis’, $= 50m s=utt Kat > 60=0+ (1/290 t= b) After overing 50m, velocity of the athelete is V=utat=0+9 x (10/3) = 30mis He has to stop in minimum time, So deceleration ia ~a = ~9m/s* (max) 6aChapter 6 R=ma IR(max frictional force) .9= 9m/s*(Deceleration) 30mis, vi=0 nul _ 0-30 _ -30 _ 10.0 a a a 16, Hardest brake means'riaximum force of friction is developed between cat's type & road. Max frictional foroe = wR From the free body diagram and + ma ~mg sin = 1mg cos 0 + ma—mg sin 0 =0 = 1g cos 0 + a= 10 (1/2) =a =5—{1-(2V3)) x 10 (v3 /2)=25 mis* re When, hardest brake is applied the car move with acceleration 2 Sm/s* S=12.8m,u=6mis 0, velocity at the end of incline v= Wu? +2as = 6? ~2(2.5)(12.8) = 36464 = 10m/s = 36km/h Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 36kmih. 17. Let, , a maximum acceleration produced in car ma = pR [For more acceleration, the tyres will lip] =. = ma=ymg= a= For crossing the bridge in minimum time, it has to travel with maximum AR acceleration u=0, $=500m, a= 10mis* s=utt Kat = 500 = 0 + (1/2) 10 => t= 10 sec. ii If acceleration is less than 10mi/s*, time will be more than 10sec. So one can't drive through the bridge in less than 10sec. 418) From the free body diagram R= 4g cos 30°=4« 10 xV/3/2=20V3 (i) iz R+ 4a P—dg sin 30° = 0 = 0.3 (40) cos 30° + 4a P—40 sin 20 P28 u, Ry =2g sin 30° Ai) Ri= 29008 30° =2* 10% 3/2 = 10V3 (iv) Equn. (i) 63 #4a-P -20=0 a Equn (iv) P+ 2a +2¥3 =10=0 1a P From Equn (i) & (iv) 6V3 +6a-30 +243 =0 > 6a = 30-8 V3 =30- 13.85= 16.15 ax 18:15 0) ii) = 2.69 = 2.7mis* 4 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 = 2.4mis*. osChapter 6 19. From the free body diagram we 8 ‘ & : “ < e aS ° + 1 nemaoet cml t : Me ae) T+Mgsina T + Mig sin@—M, a—y M,g cos 8 = 0 Equn (iv) = T- Mig sin @+M, a+ u Mig cos 0=0 ..(v) Equn (iv) & (v) => sin 6 (My + M,)— a(M, + Mz) ~ jg cos @ (M, + M;) =0 =a (M, + M2) =g sin 6 (M; + M,)~ 1.9 cos @ (My + Mz) = a= g(sin 0—y cos 6) The blocks (system has acceleration g(sin 0 — p cos 0) ‘The force exerted by the rod on one of the blocks is tension. Tension T =~ Mig sin 0+ Mya + w Mig sin © Mag sin 6 + My(g sin & 1g cos 6) + . Mig cos & 20 be the force applied to at an angle 0 From the free body diagram . R+Psino—mg ° = R=-Psind+mg (i) uR=p cos 6 Ai) ce Equn. (is plmg - P sin 0) ~P cos 0=0 - = Eng Suman p sin dP cosd-> p= — ema 4 ‘Applied force P should be minimum, when j sin 6 + cos @ is maximum ‘Again, 1 sin 8+ cos @ is maximum when its derivative is zero. ide (x sin 8 + cos 8) = 0 = 1008 0-sind=0— 0=tan*y So, p= ima umg/coso__ mgsecd _ pmgseco P= Teind=-cos8 USING, COsO~ T+ytan® — ttane cost * cose mg, uma. See It tante teu? Minimum fores 18/49 at an angle @= tan“ w raw 21. Let, the max force exerted by the man is T. From the free body diagram R+T-Mg=0 ft % i) mR T A 8 And T= Ry =0 66Chapter 6 22, 23, T= (R+mg)=0 [From equn.(i)] = T-4R-wmg=0 = T-u(Mg+T)—nmg=0 [from (i) ST (1 +H) = phy + mg sr Hema a Maximum force exerteaby man is HMT=m8 =H 1 Pea Bie 2 fg] wo cee 4 o2R, 4s sg vy 1s ie 40 Ri-2g9=0 23 = Ry=2%10=20 2a+02R;—12=0 => 2a+0.2(20)= 12 =day=4 > 2a=12-4=8 San 4ms? 2kg block has acceleration dims? & that of 4 kg is tmis® fw R Big] y-02 = ie) “a ee Rs 49 * 25 (Ri=29 4a+0.2%2%10-12=0 Ma Ry=0 = 4a+4=12 = 2a=0.2(20)=4 > 4a=8 = a=2ms? = a=2mis? on [0 fh wets ET) oy LE Ik Tres a) When the 10N'force applied on 2kg block, it experiences maximum frictional force Ry = w* 2kg = (0.2) 20 = 4N from the 3kg block. So, the 2kg block experiences amet force of 10— 4 - 6N So, a; = 6/2= 3 mis 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 eRe = (0.3) x Skg = 15N So, the 3kg block cannot move relative to the 7kg block. The 3kq block and 7kg block both will have ‘same acceleration (a, = as) which will be due to the 4N force because there is no friction from the floor. 10 = O.4mis* eTChapter 6 24, 25. ots 3g LON, al Reg b) When the 10N force is applied to the kg block, it can experience maximum frictional force of 15 + 4 = 19N from the 2kg block & 7kg block So, itcan not move with respect to them. AAs the floors frictionless, all the three bodies will move together a 10/12 = (6/6)mis* ©) Similarly it can be proved that when the 10N force is applied to the 7kg block, all the thrée blocks will move together. ‘Again ay ~ ag = as = (5/6)mis* Both upper block & lower block will have acceleration 2mis* r R AR a Teer a 9 By Ry Ry=mg FapR;-T=0-F -ymg-T=0 ..(i) a F=ymg+umg=2ymg {putting T= 1 mg} R, — ae Ed To me yi tn ma tr. b)2F-T=wmg—ma=0 ...) ToMa—jmg=0 [2 Ri=mg] ST=Ma+ img Putting value of Tin (i) 21 —Ma— jumg— mg —ma=0 = 22umg)-2mg=a(M+m) [Putting F= 2 umg] ; > 4umg—2 19 8 (Mm) saz 2m bs ce 4 z : = ‘s iB =m (mg-ma) Ry + ma ~mg =0 = Ry =m(g-a)=mg—ma (i) T-pRi=05T=m(mg—ma) (i) Again, FT 1 R, = osChapter 6 26 27. 28 > F —{u(mg —ma)}— ufmg — ma) = 0 = F—1mg + nma—jimg + ma=0 = F=2ymg-2u ma = F = 2y m(g-2) b) Acceleration of the block be a; 1 na | Re TT hg man m ig t Rs Ry = mg-ma Ai) TR; -May=0 2F-T=uR; - maj =0 = T=pRi + May Sor i) T= p(mg—ma) + Ma; = T=1mg-yma+Ma, mg + ja may Subtracting values of F & T, we get 2(2um(g—a)) ~ 2(jsmg =ma + May)— mg + wma ~ ua =0 hums Aume-2um2ume=me, Ma; y= 29-8) Both blocks move with this acceleration but in opposite directions. R,+QE=mg=0 Ry=mg-QE Ai) F-T-yR,=0 F = F-T mg —OE => F-T-mg+ pOE =0 ...(2) FeQe T-uRi=0 mn ® $5 T= Ry =p (mg— QE) = mg — .OE ie 3 : Now equation (ii) is F—mg+HQE-pmg+QE=O | Tur * => F-2mg+2uQE=0 3 erect ; r = Fe 2u(mg - QE) a Maximum horizontal force that can be applied is 21(mg — QE). Because the block sips on the table, maximum frictional force acts on it ue £ From the free body diagram R= mg 3 F=WR=0-5F=uR=mg y But the tabe is at rest. So, rctional force atthe legs ofthe table isnot wR Letbe | mah) [_o® {30 form the free body diagram, : fo- KR=O0=fo= pR= mg. ee Total frictional fores on table by floor is j mg. ° Letine acceleration of Bock Mi towards rt So, the block’ must down wit an acclaraon 2 ; Re wal Ry me wel] MS, + 204) (rao) ‘As the block ‘m’is in contact with the block ‘Mit will also have acceleration ‘a’ towards right. So, it will ‘experience two inertia forces as shown in the free body diagram-1 From free body diagram -1 CrChapter 6 29, 30, 3 Ry-ma=05R,=ma ..() Again, 2ma+T—mg + wR, = T=mg- (2-p)ma From free body diagram-2 T+ pik + mg—Re=0 > Ry=T +; ma+Mg [Putting the value of R; from ()] = (mg - 2ma—p, ma)+j4ymaj#Mg [Putting the value of T from (i)] Ra=Mg+mg-2ma (ii) Again, form the free body diagram -2 T+T-R=Ma-.R,=0 = ZT2MA=mA~ya(Mgtmg-2ma)=0 [Putting the valueB of Rs and Re rom) and (i) > 2T = (M + m) + Ha(Mg + mg - 2ma) iv) From equation (i) and (v) BT = 2mg=2(2 + wslmg = (M+ mma + ps{Mg + mg— 2m) = 2mg ~ ye(M + m)g = a (M +m —2yom + 4m + 2u;m) a= L2m=n,Memia M+mf5 + (1; -H2)) Net force = *(202 + (15)2— (0.5) * 4¢ stan @ = 20/15 = 4/3 = y= tan“\(4/3) = 53" 0, the block wil move at an angle 63 * with an 18N force a) Mass of man = 50kg. = 10 mis® Frictional force developed between hands, legs & back side with the wall the ut of frail Soho terete eet t ° oll. He gives equal force on both tne walls so gets equal reaction R from both the wall. if he applies unequal forces R should be different he cant rest between the walls Frictional force 2uR balance his wt. ai From the ree body diagram bR+uR=409 > 2yR=40%10 =R one =250N b) The normal fore is 250 N Leta; and aa be the accelerations of ma and M respectively. tee Here, a)> ar 80 that m moves on M — ‘Suppose, after time 't m separate from M In this time, m covers vt + %4 a;f*and Sy = vt + at? For‘m’to mto'm separate fromM. vt+ sa,t=vt+ arty (1) ‘Again from free body diagram rn May + wl2R = 0 (2) ma =~ (u2)m* 10-9 = 5 “Ch Maz + 1 (M+ mg ~ (Wl2)m: = 2Ma, + 2u (M+ mjg— wimg outs ™ = 2Maz= mg - 2uMg—2 umg umg-24Mg 2M Putting values of a, & a, in equation (1) we can find that [ 4m ) (wembg. >a 1 saaae 610