Chpter 4 0. While the engines operte, their totl upwrd thrust eceeds the weight of the rocket, nd the rocket eperiences net upwrd fce. his net fce cuses the upwrd velocit of the rocket to increse in mgnitude (speed). he upwrd thrust of the engines is constnt, but the remining mss of the rocket (nd hence, the downwrd grvittionl fce weight) decreses s the rocket consumes its fuel. hus, there is n incresing net upwrd fce cting on diminishing mss. his ields n ccelertion tht increses in time.. R R R mg mg () (c) n mg f In the free-bod digrms give bove, R ur represents fce due to ir resistnce, ur is fce due to the thrust of the rocket engine, n r is nml fce, r f is friction fce, nd the fces lbeled mg r re grvittionl fces. 4. If the brkes lock, the cr will trvel frther thn it would trvel if the wheels continued to roll, becuse the coefficient of kinetic friction is less thn tht of sttic friction. Hence, the fce of kinetic friction is less thn the mimum fce of sttic friction. PROBLEM SOLUIONS 4. tons w = 000 lbs 4.448 N 4 ton = 0 N lb 4. From v = v + t, the ccelertion given to the footbll is v hen, from Newton s v 0 v v 0 m s 0 = = = 50 m s t 0.0 s 0 nd lw, we find F = m v = 0.50 kg 50 m s = 5 N Pge 4.38
Chpter 4 4.3 () F m Σ = = 6.0 kg.0 m s = N ΣF N 3.0 m s = = = m 4.0 kg 4.4 () Action: he hnd eerts fce to the right on the spring. Rection: he spring eerts n equl mgnitude fce to the left on the hnd. Action: he wll eerts fce to the left on the spring. Rection: he spring eerts n equl mgnitude fce to the right on the wll. Action: Erth eerts n downwrd grvittionl fce on the spring. Rection: he spring eerts n equl mgnitude grvittionl fce upwrd on the Erth. (c) (d) Action: he hndle eerts fce upwrd to the right on the wgon. Rection: he wgon eerts n equl mgnitude fce downwrd to the left on the hndle. Action: Erth eerts n upwrd contct fce on the wgon. Rection: he wgon eerts n equl mgnitude downwrd contct fce on the Erth. Action: Erth eerts n downwrd grvittionl fce on the wgon. Rection: he wgon eerts n equl mgnitude grvittionl fce upwrd on the Erth. Action: he pler eerts fce upwrd to the left on the bll. Rection: he bll eerts n equl mgnitude fce downwrd to the right on the pler. Action: Erth eerts n downwrd grvittionl fce on the bll. Rection: he bll eerts n equl mgnitude grvittionl fce upwrd on the Erth. Action: M eerts grvittionl fce to the right on m. Rection: m eerts n equl mgnitude grvittionl fce to the left on M. (e) Action: he chrge +Q eerts n electrosttic fce to the right on the chrge q. Rection: he chrge q eerts n equl mgnitude electrosttic fce to the left on the chrge +Q. (f) Action: he mgnet eerts fce to the right on the iron. Rection: he iron eerts n equl mgnitude fce to the left on the mgnet. 4.5 he weight of the bg of sugr on Erth is w E 4.448 N = mge = ( 5.00 lbs) =. N lb If g M is the free-fll ccelertion on the surfce of the Moon, the rtio of the weight of n object on the Moon to its weight when on Erth is so w mg g = = w mg g M M M E E E Pge 4.39
Chpter 4 w M g. M = we g E Hence, the weight of the bg of sugr on the Moon is w M = (. N) = 3.7 N 6. On Jupiter, its weight would be w J gj = we = (. N)(.64) = 58.7 N g E he mss is the sme t ll three loctions, nd is given b E ( 5.00 lb)( 4.448 N lb) we m = = = g 9.80 m s.7 kg 4.6 Σ F = = = 5.0 0 m s m 7.5 05 N.5 07 kg nd v = v0 + t gives t v v0 80 km h 0 0.78 m s min = = = 5.0 0 m s km h 60 s 7.4 min 4.7 Summing the fces on the plne shown gives.0 m/s Σ F = m 0 N f = ( 0.0 kg) (.0 m s ) f F 0 N From which, f = 9.6 N 4.8 () he sphere hs lrger mss thn the fether. Hence, the sphere eperiences lrger grvittionl fce Fg = mg thn does the fether. (c) (d) he time of fll is less f the sphere thn f the fether. his is becuse ir resistnce ffects the motion of the fether me thn tht of the sphere. In vcuum, the time of fll is the sme f the sphere nd the fether. In the bsence of ir resistnce, both objects hve the free-fll ccelertion g. In vcuum, the totl fce on the sphere is greter thn tht on the fether. In the bsence of ir resistnce, the totl fce is just the grvittionl fce, nd the sphere weighs me thn the fether. Pge 4.40
Chpter 4 4.9 he verticl ccelertion of the slmon s it goes from v 0 = 3.0 m s (underwter) to v = 6.0 m s (fter moving upwrd.0 m /3 of its bod length) is ( 6.0 m s) ( 3.0 m s) v 0 v = = = 3.5 m s Δ.00 m Appling Newton s second lw to the verticl lep of this slmon hving mss of 6 kg, we find Σ F = m F mg = m m m 3 ( ) F = m + g = 6 kg 3.5 + 9.8 =.4 0 N s s 4.0 he ccelertion of the bullet is given b hen, ( Δ) v v 0 30 m s 0 = = 0.8 m ( 30 m s) ( 0.8 m) Σ F = m = ( ) = 5.0 0 3 kg 3. 0 N 4. () From the second lw, the ccelertion of the bot is + + f = 800 N F = 000 N ΣF 000 N 800 N = = = m 000 kg he distnce moved is 0.00 m s ur F ur f Δ = 0 + = 0 + 0.00 m s 0.0 s = 0.0 m ## v t t (c) he finl velocit is## v v t = 0 + = 0 + 0.00 m s 0.0 s =.00 m s. Pge 4.4
Chpter 4 4. () Choose the positive -is in the fwrd direction. We resolve the fces into their components s Fce -component -component 400 N 00 N 346 N 450 N 78. N 443 N Resultnt Σ = N Σ = 790 N F F F 450 N 0 30 F 400 N he mgnitude nd direction of the resultnt fce is F FR = Σ F + Σ F = 799 N tn θ Σ = = 8.77 to right of -is ΣF hus, Fr R = 799 N t 8.77 to the right of the fwrd direction he ccelertion is in the sme direction s F r R nd hs mgnitude F R 799 N = = = m 3 000 kg 0.66 m s 4.3 () At terminl speed, the mgnitude of the ir resistnce fce is equl to the weight of the skdiver ( F drg = mg). herefe, drg v v terminl ( 65.0 kg ) ( 9.80 m s ) F mg k = = = = ( 55.0 m s) 0. kg m At the speed v = v, the upwrd fce due to ir resistnce is less thn the downwrd grvittionl terminl fce cting on the skdiver, leving net downwrd fce of mgnitude Fnet = mg Fdrg. hus, the skdiver will hve downwrd ccelertion of mgnitude F net mg kv k vterminl = = = g m m m 0. kg m 55.0 m s = 9.80 m s = 7.35 m s 65.0 kg 4.4 () With the wind fce being hizontl, the onl verticl fce cting on the object is its own weight, mg. his gives the object downwrd ccelertion of Pge 4.4
Chpter 4 ΣF mg = = = g m m he time required to undergo verticl displcement Δ = h, strting with initil verticl velocit v 0 = 0, is found from Δ = v t + t s 0 g h = 0 t t = h g he onl hizontl fce cting on the object is tht due to the wind, so Σ F = F nd the hizontl ccelertion will be ΣF = = m F m (c) With v =, the hizontl displcement the object undergoes while flling verticl distnce h is given b 0 0 o Δ = v t + t s F h Fh Δ = 0 + = m g mg (d) he totl ccelertion of this object while it is flling will be = + = F m + g = F m + g 4.5 Strting with v = nd flling 30 m to the ground, the velocit of the bll just befe it hits is v 0 0 = v + Δ = 0 + 9.80 m s 30 m = 4 m s 0 On the rebound, the bll hs v = 0 fter displcement Δ =+ 0 m. Its velocit s it left the ground must hve been v =+ v Δ =+ 0 9.80 m s 0 m =+ 0 m s hus, the verge ccelertion of the bll during the.0-ms contct with the ground ws v v + 0 m s 4 m s 4 v = = = +. 0 m s upwrd Δ t.0 0 3 s he verge resultnt fce cting on the bll during this time intervl must hve been Fne + = + 4 4 t = mv = 0.50 kg. 0 m s. 0 N F ur. 04 net = N upwrd Pge 4.43
Chpter 4 4.6 Since the two fces re perpendiculr to ech other, their resultnt is: F R = ( 80 N) + ( 390 N) = 430 N, t 390 N tn hus, θ = = 65. N of E 80 N F R 430 N = = =.59 m s m 70 kg r =.59 m s t 65. N of E. 4.7 () Since the burglr is held in equilibrium, the tension in the verticl cble equls the burglr s weight of 600 N Now, consider the junction in the three cbles:## 37.0 Σ F = 0, giving sin 37.0 600 N = 0 w 600 N 600 N = = 997 N in the inclined cble sin 37.0 Also, Σ = 0 which ields cos 37.0 = 0, F = ( 997 N) cos 37.0 = 796 N in the hizontl cble If the left end of the iginll hizontl cble ws ttched to point higher up the wll, the tension in this cble would then hve n upwrd component. his upwrd component would suppt prt of the weight of the ct burglr, thus. decresing the tension in the cble on the right Pge 4.44
Chpter 4 4.8 Using the reference is shown in the sketch t the right, we see tht Σ F = cos4.0 cos4.0 = 0 nd Σ F = sin4.0 sin4.0 = sin4.0 hus, the mgnitude of the resultnt fce eerted on the tooth b the wire brce is R = Σ F + Σ F = 0 + sin4.0 = sin4.0 4 4 R = ( 8.0 N) sin4.0 = 8.7 N 4.9 From Σ = 0, cos 30.0 cos 60.0 = 0 F =.73 [] hen Σ F = 0 becomes ( ) sin 30.0 +.73 sin 60.0 50 N = 0 which gives = 75.0 N in the right side cble. Finll, Eqution [] bove gives = 30 N in the left side cble. 60.0 30.0 50 N 4.0 If the hip eerts no fce on the leg, the sstem must be in equilibrium with the three fces shown in the free-bod digrm. hus Σ F = 0 becomes w cos α = 0 N cos 40 From Σ F = 0, we find [] w sin α = 0 N 0 N sin 40 [] 0 N 40 w 0 N α Dividing Eqution [] b Eqution [] ields Pge 4.45
Chpter 4 0 N 0 N sin 40 α = = 0 N cos 40 tn 6 hen, from either Eqution [] [], w =.7 0 N. 4. () Free-bod digrms of the two blocks re shown t the right. Note tht ech block eperiences downwrd grvittionl fce F = 3.50 kg 9.80 m s = 34.3 N g Also, ech hs the sme upwrd ccelertion s the elevt, in this cse =+.60 m s. Appling Newton s second lw to the lower block: Σ F = m F = m g = Fg + m = 34.3 N + ( 3.50 kg) Net, ppling Newton s second lw to the upper block: Σ F = m F = m g.60 m s = 39.9 N = + F + m = 39.9 N + 34.3 N + 3.50 kg.60 m s = 79.8 N g Note tht the tension is greter in the upper string, nd this string will brek first s the ccelertion of the sstem increses. hus, we wish to find the vlue of when = 85.0 N. Mking use of the generl reltionships derived in () bove gives: = + Fg + m = Fg + m + Fg + m = Fg + m F 85.0 N 34.3 N = = m 3.50 kg g =.34 m s Pge 4.46
Chpter 4 4. () Free-bod digrms of the two blocks re shown t the right. Note tht ech block eperiences downwrd grvittionl fce Fg = mg. Also, ech hs the sme upwrd ccelertion s the elevt, =+. Appling Newton s second lw to the lower block: Σ F = m F = m = m g + m = m( g + ) g Net, ppling Newton s second lw to the upper block: Σ F = m F = m g = + F + m = mg + m + mg + m = mg + m = g Note tht =, so the upper string breks first s the ccelertion of the sstem increses. (c) When the upper string breks, both blocks will be in free-fll with = g. hen, using the results of prt (), = m g + = m g g = 0 nd = = 0. 4.3 m =.00 kg nd mg = 9.80 N α 0.00 m = tn = 0.458 5.0 m Since = 0, require tht Σ F = sin α + sin α mg = 0, giving sinα = mg, 9.80 N = 63 N sinα = 5.0 m 5.0 m 0.00 m mg 4.4 he resultnt fce eerted on the bot b the people is 3 = 600 N cos 30.0.04 0 N in the fwrd direction If the bot moves with constnt velocit, the totl fce cting on it must be zero. Hence, the resistive fce eerted on the bot b the wter must be Pge 4.47
Chpter 4 r f =.04 03 N in the rerwrd direction 4.5 he fces on the bucket re the tension in the rope nd the weight of mg = 5.0 kg 9.80 m s = 49 N. Choose the positive the bucket, direction upwrd nd use Newton s second lw: Σ F = m 49 N = ( 5.0 kg)( 3.0 m s ) = 64 N mg 4.6 () From Newton s second lw, we find the ccelertion s ΣF 0 N = = = 0.33 m s m 30 kg o find the distnce moved, we use Δ = v 0t + t = 0 + 0.33 m s 3.0 s =.5 m F 0 N If the shopper plces her 30 N ( 3. kg) child in the crt, the new ccelertion will be ΣF 0 N = = = 0.30 m s m 33 kg totl nd the new distnce trveled in 3.0 s will be Δ = 0 + ( 0.30 m s )( 3.0 s) =.4 m Pge 4.48
Chpter 4 4.7 () he verge ccelertion is given b v v v 5.00 m s 0.0 m s = = = 3.75 m s Δt 4.00 s 0 he verge fce is found from Newton s second lw s F m 3 v = v = 000 kg 3.75 m s = 7.50 0 N he distnce trveled is: 5.00 m s + 0.0 m s = vv ( Δ t) = ( 4.00 s) = 50.0 m 4.8 Let m,, nd = 0.0 kg m = 5.00 kg θ = 40.0. Appling the second lw to ech object gives nd m = mg m = m gsinθ [] [] f n m g m m m m g g Adding these equtions ields so m + m = mg + m m sin θ mg g = m + m 0.0 kg 5.00 kg sin 40.0 = ( 9.80 m s ) = 4.43 m s 5.0 kg hen, Eqution [] ields = m g = 0.0 kg 9.80 4.43 m s = 53.7 N 4.9 () he resultnt eternl fce cting on this sstem, consisting of ll three blocks hving totl mss of 6.0 kg, is 4 N directed hizontll towrd the right. hus, the ccelertion produced is ΣF 4 N = = = m 6.0 kg 7.0 m s hizontll to the right Drw free bod digrm of the 3.0-kg block nd ppl Newton s second lw to the hizontl fces cting on this block: Pge 4.49
Chpter 4 Σ F = m 4 N ( 3.0 kg) ( 7.0 m s ) =, nd therefe = N (c) he fce ccelerting the.0-kg block is the fce eerted on it b the.0-kg block. herefe, this fce is given b: F m (.0 kg)( 7.0 m s ) = =, F r = 4 N hizontll to the right. 4.30 he ccelertion of the mss down the incline is given b his gives Δ = vt 0 + t, 0.80 m = 0 + ( 0.50 s) ( 0.50 s) 0.80 N = = 6.4 m s hus, the net fce directed down the incline is F m = =.0 kg 6.4 m s = 3 N. 4.3 () Assuming frictionless pulles, the tension is unifm through the entire length of the rope. hus, the tension t the point where the rope ttches to the leg is the sme s tht t the 8.00-kg block. Prt () of the sketch t the right gives freebod digrm of the suspended block. Recognizing tht the block hs zero ccelertion, Newton s second lw gives Σ F = mg = 0 = mg = 8.00 kg 9.80 m s = 78.4 N Prt of the sketch bove gives free-bod digrm of the pulle ner the foot. Here, F is the mgnitude of the fce the foot eerts on the pulle. B Newton s third lw, this is the sme s the mgnitude of the fce the pulle eerts on the foot. Appling the second lw gives: Σ F = + cos 70.0 F = m = 0 F ( cos 70.0 ) ( 78.4 N) ( cos ) = + = + 70.0 = 05 N Pge 4.50
Chpter 4 4.3 () Note tht the blocks move on hizontl surfce with = 0. hus, the net verticl fce cting on ech block nd on the combined sstem of both blocks is zero. he net hizontl fce cting on the combined sstem consisting of both m nd m is Σ F = F P + P = F. (c) Looking t just m, Σ F = 0 s eplined bove, while Σ F = F P. (d) Looking t just m, we gin hve Σ = 0, while Σ F = + P F (e) F : Σ F = m F P = m. m F : m Σ F = m P = m. (f) Substituting the second of the equtions found in (e) bove into the first gives the following: m = F P = F m m + m = F ( ) nd = F ( m + m) hen substituting this result into the second eqution from (e), we hve F P = m = m m + m m P = F m + m (g) Relize tht ppling the fce to rther thn m would hve the effect of interchnging the roles of m m nd m. We m esil find the results f tht cse b simpl interchnging the lbels m nd m in the results found in (f) bove. his gives = F + m (the sme result s in the first cse) nd m P = m m F + m We see tht the contct fce, P, is lrger in this cse becuse m > m. Pge 4.5
Chpter 4 4.33 king the downwrd direction s positive, ppling Newton s second lw to the flling person ields Σ F = mg f = m, f 00 N = g = 9.80 m s 8.6 m s m = 80 kg hen, v = v 0 + Δ ( ) gives the velocit just befe hitting the net s v = v0 + Δ = 0 + 8.6 m s 30 m = 3 m s 44.34 () First, consider sstem consisting of the two blocks combined, with mss m + m. F this sstem, the onl eternl hizontl fce is the tension in cd A pulling to the right. he tension in cd B is fce one prt of our sstem eerts on nother prt of our sstem, nd is therefe n internl fce. Appling Newton s second lw to this sstem (including onl eternl fces, s we should) gives A Σ F = m = m + m [] Now, consider sstem consisting of onl m. F this sstem, the tension in cd B is n eternl fce since it is fce eerted on block b block (which is not prt of this sstem). Appling Newton s second lw to this sstem gives Σ F = m = m [] B Compring equtions [] nd [], nd relizing tht the ccelertion is the sme in both cses (see Prt below), it is cler tht: Cd A eerts lrger fce on block thn cd B eerts on block Since cd B connecting the two blocks is tut nd unstretchble, the two blocks st fied distnce prt, nd the velocities of the two blocks must be equl t ll times. hus, the rtes t which the velocities of the two blocks chnge in time re equl, the the two blocks must hve equl ccelertions. (c) Yes. Block eerts fwrd fce on Cd B, so Newton s third lw tells us tht Cd B eerts fce of equl mgnitude in the bckwrd direction on Block. Pge 4.5
Chpter 4 4.35 () When the ccelertion is upwrd, the totl upwrd fce must eceed the totl downwrd fce 500 kg 9.80 m s.47 04 N w = mg = = When the velocit is constnt, the ccelertion is zero. he totl upwrd fce nd the totl downwrd fce w must be equl in mgnitude. (c) If the ccelertion is directed downwrd, the totl downwrd fce w must eceed the totl upwrd fce. (d) Σ F = m mg m Yes, > w. = + = 500 kg 9.80 m s +.50 m s =.85 04 N (e) Σ F = m mg m Yes, = w. 500 kg 9.80 m s 0.47 04 N = + = + = (f) Σ = F m = mg + m = 500 kg 9.80 m s.50 m s =.5 04 N Yes, < w. 4.36 Note tht if the cd connecting the two blocks hs fied length, the ccelertions of the blocks must hve equl mgnitudes, even though the differ in directions. Also, observe from the digrms, we choose the positive direction f ech block to be in its direction of motion. (First consider the block moving long the hizontl. he onl fce in the direction of movement is. hus, Σ F = m = (5.00 kg) Net consider the block which moves verticll. he fces on it re the tension nd its weight, 98.0 N. Σ F = m 98.0 N = ( 0.0 kg) [] [] 5.00 kg 0.0 kg () mg 98.0 N Equtions [] nd [] cn be solved simultneousl to give: 98.0 N ( 5.00 kg) = ( 0.0 kg) = 98.0 N 5.0 kg = 6.53 m s nd Pge 4.53
Chpter 4 = 5.00 kg 6.53 m s = 3.7 N 4.37 n riler 300 kg w n c Cr 000 kg w c F R cr q F n c Choose the + direction to be hizontl nd fwrd with the + verticl nd upwrd. he common ccelertion of the cr nd triler then hs components of = +.5 m s nd = 0. () he net fce on the cr is hizontl nd given b ( F ) F mcr cr 000 kg.5 m s.5 03 N fwrd Σ = = = = he net fce on the triler is lso hizontl nd given b ( F ) mtriler Σ = + = = 300 kg.5 m s = 645 N fwrd triler (c) Consider the free-bod digrms of the cr nd triler. he onl hizontl fce cting on the triler is = 645 N fwrd, nd this is eerted on the triler b the cr. Newton s third lw then sttes tht the fce the triler eerts on the cr is 645 N towrd the rer. (d) he rod eerts two fces on the cr. hese re F nd nc shown in the free-bod digrm of the cr. From prt (), F.5 03 N 645 N.5 03 N.80 03 N = + = + = + Also,, so Σ F = n w = m = 0 nc = wc = mcr g = 9.80 03 N cr c c cr he resultnt fce eerted on the cr b the rod is then R = F + n = cr (.80 0 N) + ( 9.80 0 ) 3 c n c t θ 3 4 N =.0 0 N = tn = tn 3.5 = 74. F bove the hizontl nd fwrd. Newton s third lw then sttes tht the resultnt fce eerted on the rod b the cr is.0 04 N t 74. below the hizontl nd rerwrd Pge 4.54
Chpter 4 44.38 First, consider the 3.00-kg rising mss. he fces on it re the tension,, nd its weight, 9.4 N. With the upwrd direction s positive, the second lw becomes 9.4 N = ( 3.00 kg) [] Rising Mss m 3.00 kg Flling Mss m 5.00 kg he fces on the flling 5.00-kg mss re its weight nd, nd its ccelertion hs the sme mgnitude s tht of the rising mss. Choosing w 9.4 N w 49.0 N the positive direction down f this mss, gives ( 49.0 N = 5.00 kg) [] () Solving Eqution () f nd substituting into [] gives 3.00 kg 9.4 N = ( 49.0 N 5.00 kg ).60 = 58.8 N nd the tension is = 36.8 N (c) Eqution () then gives the ccelertion s 49.0 N 36.8 N = = 5.00 kg.44 m s Consider the 3.00-kg mss. We hve Δ = v 0t + t = 0 +.44 m s.00 s =. m 4.39 When the block is on the verge of moving, the sttic friction fce hs mgnitude m Since equilibrium still eists nd the pplied fce is 75 N, we hve Σ F = 75 N f = 0 ( f ) m = 75 N s In this cse, the nml fce is just the weight of the crte, n ( f ) ( f ) μ s m s m 75 N s = n = mg = = 0.38 s ( 0 kg )( 9.80 m s ) f = f = μ n. s s s = mg. hus, the coefficient of sttic friction is After motion eists, the friction fce is tht of kinetic friction, fk = μ n. Since the crte moves with constnt velocit when the pplied fce is 60 N, we find tht f k = 60 N. herefe, the coefficient of kinetic friction is μ k = fk fk 60 N n = mg = = ( 0 kg )( 9.80 m s ) 0.3 k Σ F = 60 N f = 0 k Pge 4.55
Chpter 4 4.40 () he sttic friction fce ttempting to prevent motion m rech mimum vlue of ( f s) μsn μsmg m = = = 0.50 0 kg 9.80 m s = 49 N his eceeds the fce ttempting to move the sstem, w = mg = 39 N. Hence, the sstem remins t rest nd the ccelertion is = 0 Once motion begins, the friction fce retrding the motion is fk = μkn = μkmg = 0.30 0 kg 9.80 m s = 9 N his is less thn the fce tring to move the sstem, w = mg. Hence, the sstem gins speed t the rte totl ( 0 kg ) F 4.0 kg 0.30 9.80 m s net m g μ m g = = = = m m + m 4.0 kg + 0 kg k 0.70 m s 4.4 () Since the crte hs constnt velocit, = = 0. Appling Newton s second lw: Σ F = F cos 0.0 f = m 0 f = ( 300 N) cos 0.0 = 8 N k = k nd Σ F = n Fsin 0.0 w = 0 n = 3 ( 300 N) sin 0.0 + 000 N =.0 0 N he coefficient of friction is then fk 8 N μ k = = = n.0 03 N 0.56 In this cse, Σ F = n + Fsin 0.0 w = 0, so n = w Fsin 0.0 = 897 N he friction fce now becomes f = μ n = k k 0.56 897 N = 30 N w herefe, Σ F = F cos 0.0 f = m = g k nd the ccelertion is ( F cos 0.0 f ) g ( 300 N) cos 0.0 30 N ( 9.80 m s ) = = w k 000 N = 0.509 m s 4.4 () v v 6.00 m s.0 m s = = =.0 m s t 5.00 s 0 Pge 4.56
Chpter 4 From Newton s second lw, Σ F = f = m, f = m. k k he nml fce eerted on the puck b the ice is n = mg, so the coefficient of friction is fk μk = = n m m (.0 m s ) ( 9.80 m s ) = 0. v + v 6.00 m s +.0 m s Δ = = 5.00 s 45.0 m v = = 0 (c) ( v ) t t 4.43 When the lod on the verge of sliding fwrd on the bed of the slowing truck, the rerwrd directed sttic friction fce hs its mimum vlue m = f μ n = μ m g s s s lod fs m lod g n Since slipping is not et occurring, this single hizontl fce must give the lod n ccelertion equl to tht the truck. hus, F m μs mlod Σ = g = mlod truck truck = μsg. () If slipping is to be voided, the mimum llowble rerwrd ccelertion of the truck is seen to be = μ g, nd v = v + ( Δ) gives the minimum stopping distnce s 0 truck s 0 v 0 v0 Δ = = min μ g If v0 m s nd μs 0.500 ( ) truck m = =, then s (.0 m s) Δ = = min 0.500 9.80 m s 4.7 m Emining the clcultion of Prt () shows tht neither mss is necessr 4.44 () he free-bod digrm of the crte is shown t the right. Since the crte hs no verticl ccelertion ( = 0), we see tht Σ F = m n mg = 0 n = mg he onl hizontl fce present is the friction fce eerted on the crte b the truck bed. hus, Σ F = m f = m Pge 4.57
Chpter 4 If the crte is not to slip (i.e., the sttic cse is to previl), it is necessr tht the required friction fce not eceed the mimum possible sttic friction fce, ( f llowble ccelertion s: ( fs ) μ n μ ( mg) s ) m = μ n. From this, we find the mimum f m = = = = ( 0.350)( 9.80 m s ) = 3.43 m s m m m m m m s s = μ sg = s Once slipping hs strted, the kinetic friction cse previls nd f = fk = μkn. he ccelertion of the crte in this cse will be ( mg) f μkn μk = = = = μ kg = ( 0.30)( 9.80 m s ) = 3.4 m s m m m 4.45 () he ccelertion of the sstem is found from Δ = v 0t + t,.00 m = 0 + (.0 s) which gives =.39 m s Using the free bod digrm of, the second lw gives m ( 5.00 kg)( 9.80 m s ) ( 5.00 kg) (.39 m s ) = f n m g m m m g m g = 4. N hen ppling the second lw to the hizontl motion of m 4. N f = 0.0 kg.39 m s f = 8. N Since n = mg = 98.0 N, we hve f 8. N μ k = = = n 98.0 N 0.87 4.46 () Since the puck is on hizontl surfce, the nml fce is verticl. With = 0, we see tht Σ F = m n mg = 0 n = mg Once the puck leves the stick, the onl hizontl fce is friction fce in the negtive -direction (to oppose the motion of the puck). he ccelertion of the puck is ( mg) ΣF fk μkn μk = = = = = μkg m m m m Pge 4.58
Chpter 4 hen v = v 0 + ( Δ ) gives the hizontl displcement of the puck befe coming to rest s v v0 0 v0 v0 Δ = = = μ g ( μ g) k k 4.47 () he crte does not ccelerte perpendiculr to the incline. hus, Σ F = m = 0 n = F + mgcosθ F n he net fce tending to move the crte down the incline is Σ F = mg sinθ f, where fs is the fce of sttic friction s between the crte nd the incline. If the crte is in equilibrium, then mg sinθ fs = 0 so fs = Fg sin θ But, we lso know f μ n = μ ( F + mgcosθ) s s s herefe, we m write mg sin θ μ ( F mg cosθ) +, s mg q 35.0 fs F sinθ sin 35.0 mg cosθ = ( 3.00 kg)( 9.80 m s ) cos 35.0 = 3. N μ 0.300 s 4.48 () Find the nml fce n r on the 5.0 kg bo: Σ F = n+ 80.0 N sin 5.0 45 N = 0 n = N Now find the friction fce, f, s k f = μ n = 0.300 N = 63.4 N From the second lw, we hve Σ F = m, 80.0 N cos 5.0 63.4 N= 5.0 kg which ields = 0.366 m s. f n 5.0 F 80.0 N mg 45 N When the bo is on the incline, Σ F = n + 80.0 N sin 5.0 45 N cos0.0 = 0 giving n = 07 N he friction fce is f μ n = = 0.300 07 N = 6. N he net fce prllel to the incline is then k Σ = 80.0 N cos 5.0 45 N sin0.0 6. N=-3.3 N F f n 45 N 5.0 0.0 F 80.0 N Pge 4.59
Chpter 4 hus, ΣF 3.3 N = = =.9 m s m 5.0 kg.9 m s down the incline 4.49 () he object will fll so tht m = mg bv, = ( mg bv) m f bv where the downwrd direction is tken s positive. 0) Equilibrium ( = is reched when m v ( 50 kg)( 9.80 m s ) mg v = v terminl = = = b 5 kg s 33 m s mg If the initil velocit is less thn 33 m/s, then 0 nd 33 m/s is the lrgest velocit ttined b the object. On the other hnd, if the initil velocit is greter thn 33 m/s, then 0 nd 33 m/s is the smllest velocit ttined b the object. Note lso tht if the initil velocit is 33 m/s, then = 0 nd the object continues flling with constnt speed of 33 m/s. 4.50 () he fce of friction is found s f = μ n = μ ( mg ) k k Choose the positive direction of the -is in the direction of motion nd ppl Newton s second lw. We hve f m Σ F = f = m = = k μ g From v = v 0 + Δ, with v = 0, v0 = 50.0 km h = 3.9 m s, we find ( 3.9 m s) ( ) = + μ ( Δ ) 3.9 m s 0 k g Δ = [] g μ k With μ k = 0.00, this gives ( 3.9 m s) ( ) Δ = = 0.00 9.80 m s 98.6 m With μ k = 0.600, Eqution () bove gives ( 3.9 m s) ( ) Δ = = 0.600 9.80 m s 6.4 m Pge 4.60
Chpter 4 4.5 () Δ = v 0 t + t = 0 + t gives: ( Δ ) t (.50 s).00 m = = =.78 m s Considering fces prllel to the incline, Newton s second lw ields ( 9.4 N) sin 30.0 fk ( 3.00 kg) (.78 m s ) Σ F = = n f k m k n f k = 9.37 N Perpendiculr to the plne, we hve equilibrium, so Σ F = n 9.4 N cos 30.0 = 0 n = 5.5 N 30.0 30.0 w mg 9.4 N hen, f k 9.37 N μ k = = = 0.368 n 5.5 N (c) From prt bove, f k = 9.37 N (d) Finll, v = v + ( Δ ) gives 0 v= v0 + Δ = 0 +.78 m s.00 m =.67 m s 4.5 When the minimum fce F r is used, the block tends to slide down r the incline so the friction fce, f is directed up the incline. While the block is in equilibrium, we hve [] nd s Σ F = F cos 60.0 + f 9.6 N sin 60.0 = 0 s Σ F = n Fsin 60.0 9.6 N cos 60.0 = 0 [] 30.0 n 30.0 f k m k n w mg 9.4 N F minimum F (impending motion), f = ( f ) = μ n = ( 0.300) s s m s n [3] Eqution [] gives n = 0.866F + 9.80 N [4] Pge 4.6
Chpter 4 f = 0.300 0.866F + 9.80 N = 0.60 F +.94 N, so () Eqution [3] becomes: s Eqution [] gives 0.500F + 0.60F +.94 N 7.0 N = 0 F = 8.5 N. Finll, Eqution (4) gives the nml fce n = 0.866( 8.5 N) + 9.80 N= 5.8 N. 4.53 b) First, tking downwrd s positive, ppl Newton s second lw to the.0 kg block: (.0 kg) (.0 kg) Σ F = g = F the 7.00 kg block, we hve nd f ( =.0 kg 9.80 m s ) [] Σ F = 0 n = 68.6 N cos 37.0 = 54.8 N k ( 0.50) ( 54.8 N) = μ n = = 3.7 N.0 kg 7.00 kg w 8 N f n 37.0 w 68.6 N king up the incline s the positive direction nd ppling Newton s second lw to the 7.00-kg block gives Σ F = f 68.6 N sin 37.0 = 7.00 kg, 7.00 kg = 3.7 N 4.3 N () Substituting Eqution [] into [] ields ( 7.00 kg+.0 kg) 6.7 N = = 3.30 m s 4.54 () Both objects strt from rest nd hve ccelertions of the sme mgnitude,. his mgnitude cn be determined b ppling Δ = v t + t to the motion of : ( Δ) t ( 4.00s).00 m = = = 0 0.5 m s m Pge 4.6
Chpter 4 Consider the free-bod digrm of m nd ppl Newton s nd lw: Σ F = m m g =m + m ( )( 9. ) = m g + = 4.00 kg 80 m s + 0.5 m s = 39.7 N m g (c) Considering the free-bod digrm of m : Σ F = m n mgcosθ = 0 n= mgcosθ, so n = 9.00 kg 9.80 m s cos 40.0 = 67.6 N. k Σ F = m m g sinθ f = m + hen = ( θ ) fk m gsin, f k q m m g n f k = 9.00 kg 9.80 m s sin 40.0 0.5 m s 39.7 N = 5.9 N he coefficient of kinetic friction is fk 5.9 N μ k = = = n 67.6 N 0.35 4.55 n ground w/ 85.0 lb.0.0 F F n tip f w 70 lb F 45.8 lb.0 Free-Bod Digrm of Person Free-Bod Digrm of Crutch ip From the free-bod digrm of the person, F F F F = F = F Σ = sin.0 sin.0 = 0, which gives hen, Σ F = F cos.0 + 85.0 lbs 70 lbs = 0 ields F = 45.8 lb () Now consider the free-bod digrm of crutch tip. Σ F = f 45.8 lb sin.0 = 0, Pge 4.63
Chpter 4 f = 7. lb Σ F = n 45.8 lb cos.0 = 0, tip which gives n tip = 4.5 lb F minimum coefficient of friction, the crutch tip will be on the verge of slipping, so m f = f = μ n s s tip f 7. lb nd μ s = = = n 4.5 lb tip 0.404 As found bove, the compression fce in ech crutch is F = F = F = 45.8 lb 4.56 he ccelertion of the bll is found from ( Δ) v v 0 0.0 m s 0 = = = 33 m s.50 m From the second lw, Σ F = F w = m, so F = w + m =.47 N + 0.50 kg 33 m s =.5 N m f bv mg v 4.57 () Note tht the suspended block on the left,, is hevier thn tht on the right, m. hus, if the sstem m4 overcomes friction nd moves, the center block will move right to left with ech block s ccelertion being in the directions shown bove. Pge 4.64
Chpter 4 First, consider the center block,, which hs no verticl ccelertion. hen, m Σ F = n mg = 0 n = m g = (.00 kg) ( 9.80 m s ) = 9.80 N his mens the friction fce is: k f = μ n = 0.350 9.80 N = 3.43 N Assuming the cds do not stretch, the speeds of the three blocks must lws be equl. hus, the mgnitudes of the blocks ccelertions must hve common vlue,. uur uur uur 4 = = = king the indicted direction of the ccelertion s the positive direction of motion f ech block, we ppl Newton s second lw to ech block s follows F : mg m m4 4 4 = m ( g ) ( g ) F m : f = F : m g m m = 4 = 4.00 kg [] m =.00 kg + 3.43 N [] = m ( g ) ( g ) = + =.00 kg + [3] Substituting equtions [] nd [3] into eqution [], nd solving f ields the following: 4.00 kg g.00 kg g + =.00 kg + 3.43 N ( ) 4.00 kg.00 kg 9.80 m s 3.43 N = = 4.00 kg +.00 kg +.00 kg.3 m s (c) Using this result in equtions [] nd [3] gives the tensions in the two cds s: nd = 4.00 kg g = 4.00 kg 9.80.3 m s = 30.0 N =.00 kg g + =.00 kg 9.80 +.3 m s = 4. N (e) From the finl clcultion in prt, observe tht if the friction fce hd vlue of zero (rther thn 3.53 N), the ccelertion of the sstem would increse in mgnitude. hen, observe from equtions [] nd [3] tht this would men would decrese while would increse Pge 4.65
Chpter 4 4.58 he sketch t the right gives n edge view of the sil (hev line) s seen from bove. he velocit of the wind, v r wind, is directed to the est nd the fce the wind eerts on the sil is perpendiculr to the sil. he mgnitude of this fce is N F sil = 550 wind ms vr where v r wind is the component of the wind velocit perpendiculr to the sil. nth 30 30 F sil v wind est When the sil is iented t 30 from the nth-south line nd the wind speed is v = 7 knots, we hve wind F sil N N = 550 wind = 550 7 knots ms ms v r 0.54 m s cos 30 4. 03 = N knot he estwrd component of this fce will be counterblnced b the fce of the wter on the keel of the bot. Befe the silbot hs significnt speed (tht is, befe the drg fce develops), its ccelertion is provided b the nthwrd component of F ur sil. hus, the initil ccelertion is F ur sil 3 ( 4. 0 N) sin 30 nth = = =.6 m s m 800 kg 4.59 () he hizontl component of the resultnt fce eerted on the light b the cbles is R he resultnt component is: R =Σ F = 60.0 N cos 45.0 60.0 N cos 45.0 = 0 =Σ F = 60.0 N sin 45.0 + 60.0 N sin 45.0 = 84.9 N 60.0 N 60.0 N 45.0 45.0 Hence, the resultnt fce is 84.9 N verticll upwrd. he fces on the trffic light re the weight, directed downwrd, nd the 84.9 N verticll upwrd fce eerted b the cbles. Since the light is in equilibrium, the resultnt of these fces must be zero. hus, w = 84.9 N. Pge 4.66
Chpter 4 4.60 () F the suspended block, Σ F = 50.0 N = 0, so the tension in the rope is = 50.0 N. hen, considering the hizontl fces on the 00-N block, we find Σ F = f = 0, f = = 50.0 N s s If the sstem is on the verge of slipping, m the minimum cceptble coefficient of friction is f = f = μ n. herefe, s s s fs 50.0 N μ s = = = n 00 N 0.500 (c) If μ k = 0.50, then the friction fce cting on the 00-N block is f k = μ n = 0.50 00 N = 5.0 N k Since the sstem is to move with constnt velocit, the net hizontl fce on the 00-N block must be zero, Σ F = f = 5.0 N = 0. he required tension in the rope is = 5.0 N. Now, considering k the fces cting on the suspended block when it moves with constnt velocit, Σ F = w = 0, giving the required weight of this block s w = = 5.0 N. 4.6 On the level surfce, the nml fce eerted on the sled b the ice equls the totl weight, n = 600 N. hus, the friction fce is k f = μ n = 0.050 600 N = 30 N Hence, Newton s second lw ields Σ F = f = m, ( 30 N )( 9.80 m s ) f f = = = = 0.49 m s m w g 600 N he distnce the sled trvels on the level surfce befe coming to rest is ( ) ( ) v v 0 0 7.0 m s Δ = = = 0.49 m s 50 m 4.6 Consider the verticl fces cting on the block: Σ F = 85.0 N sin 55.0 39. N n = m = 0, so the nml fce is n = 69.6 N 39. N = 30.4 N Now, consider the hizontl fces: Σ F = 85.0 N cos 55.0 f = m = 4.00 kg ( 6.00 m s ) k f = 85.0 N cos 55.0 4.0 N = 4.8 N k f k 55.0 85.0 N n 4.00 kg mg 39. N Pge 4.67
Chpter 4 fk 4.8 N he coefficient of kinetic friction is then μ k = = = 0.84. n 30.4 N 4.63 () he fce tht ccelertes the bo is the friction fce between the bo nd the truck bed. he mimum ccelertion the truck cn hve befe the bo slides is found b considering the mimum sttic friction fce the truck bed cn eert on the bo: ( f ) = μ n = μ ( mg ) s m s s hus, from Newton s second lw, ( f ) μ s ( mg) μ g m s m = = = s = 0.300 9.80 m s =.94 m s m m 4.64 Let m = 5.00 kg, m = 4.00 kg, nd m = 3.00 kg. Let be the tension in the string between m nd m, nd 3 the tension in the string between m nd m. 3 () We m ppl Newton s second lw to ech of the msses. f m: m = mg [] f : m= + mg [] m f : m= mg [3] m 3 Adding these equtions ields 3 3 ( m ) ( m m m ) m m 3 3 + + = + + g, so m + m + m = g = m + m + m 3 3.00 kg.0 kg ( 9.80 m s ) =.63 m s From Eqution [], m ( g ) = + = 5.00 kg.4 m s = 57. N, nd from Eqution [3], m ( g ) = 3 = 3.00 kg 8.7 m s = 4.5 N Pge 4.68
Chpter 4 4.65 When n object of mss m is on this frictionless incline, the onl fce cting prllel to the incline is the prllel component of weight, mg sinθ directed down the incline. he ccelertion is then F mg sinθ = = = gsinθ = ( 9.80 m s ) sin 35.0 = 5.6 m s (directed down the incline) m m () king up the incline s positive, the time f the sled projected up the incline to come to rest is given b v v0 0 5.00 m s t = = = 0.890 s 5.6 m s he distnce the sled trvels up the incline in this time is v + v0 0 + 5.00 m s Δ s = vv t = t = ( 0.890 s) =. m he time required f the first sled to return to the bottom of the incline is the sme s the time needed to go up, tht is, t = 0.890 s. In this time, the second sled must trvel down the entire 0.0 m length of the incline. he needed initil velocit is found from Δ s = vt 0 + t s v 0 ( 5.6 m s ) ( 0.890 s) Δs t 0.0 m = = t 0.890 s = 8.74 m s 8.74 m s down the incline. 4.66 Befe she enters the wter, the diver is in free-fll with n ccelertion of 9.80 m s downwrd. king downwrd s the positive direction, her velocit when she reches the wter is given b v = v 0 + Δ = 0 + 9.80 m s 0.0 m = 4.0 m s his is lso her initil velocit f the.00 s fter hitting the wter. Her verge ccelertion during this.00 s intervl is v v0 0 ( 4.0 m s) v = = = 7.00 m s t.00 s Continuing to tke downwrd s the positive direction, the verge upwrd fce b the wter is found s Σ F = F, v + mg = mv F = g = v m = F 3 v =.8 0 N upwrd v 70.0 kg 7.00 m s 9.80 m s.8 0 3 N Pge 4.69
Chpter 4 4.67 () Free-bod digrms f the two blocks re given t the right. he coefficient of kinetic friction f luminum on steel is μ = 0.47 while tht f copper on steel is μ = 0.36. Since = n hus, nd 0 f ech block, = w nd n = w cos 30.0. f = μn = 0.47 9.6 N = 9. N f = μn = 0.36 58.8 N cos 30.0 = 8.3 N F the luminum block: Σ F = m f = m + = f + m giving 9. N (.00 kg) = + [] f f Aluminum.00 kg Copper 6.00 kg 30 n w 9.6 N n w 58.8 N F the copper block: ( 58.8 N) sin 30.0 8.3 N ( 6.00 kg) Σ F = m =. N = 6.00 kg) [] Substituting Eqution [] into Eqution [] gives (. N 9. N.00 kg = 6.00 kg =.86 N 8.00 kg = 0.3 m s ( From Eqution [] bove, ) = 9. N +.00 kg 0.3 m s = 9.68 N. 4.68 In the verticl direction, we hve Σ F = cos 4.0 mg = 0, = In the hizontl direction, the second lw becomes: Σ F = sin 4.0 = m, so mg cos 4.0 sin 4.0 = = gtn 4.0 = 0.69 m s m 4.0 mg Pge 4.70
Chpter 4 4.69 Figure is free-bod digrm f the sstem consisting of both blocks. he friction fces re = μk = μk ( g ) nd f = μk ( mg) f n m F this sstem, the tension in the connecting rope is n internl fce nd is not included in second lw clcultions. he second lw gives Σ F = 50 N f f = m + m f n n m m m g f m g 50 N which reduces to 50 N = μkg m + m [] Figure gives free-bod digrm of m lone. F this sstem, the tension is n eternl fce nd must be included in the second lw. We find: Σ F = f = m, ( μ g) = m + [] k If the surfce is frictionless, μ k = 0. hen, Eqution [] gives 50 N 50 N = 0 = =.7 m s m + m 30 kg nd Eqution [] ields = 0 kg.7 m s + 0 = 7 N. If μ k = 0.0, Eqution [] gives the ccelertion s 50 N = 30 kg 0.0 9.80 m s = 0.69 m s while Eqution () gives the tension s = 0 kg 0.69 m s + 0.0 9.80 m s = 7 N 4.70 () Pge 4.7
Chpter 4 No. In generl, the sttic friction fce is less thn the mimum vlue of ( fs ) m = μ n. It is equl to this s mimum vlue onl when the coin is on the verge of slipping, t the criticl ngle θ. m F θ θ, f f = μ n. c s s c (c) Recognize tht when the is is chosen perpendiculr to the incline s shown bove, = 0 nd we find Σ F = n mgcosθ = m = 0 n = mgcosθ Also, when sttic conditions still previl, but the coin is on the verge of slipping, we hve m = 0, θ= θc, nd f = fs = μsn = μsmgcosθc. hen, Newton s second lw becomes Σ F = mgsinθ μ mgcosθ = m c s nd μ s mg cosθ c = mg sinθ c ielding μ c = 0 s sinθc = = cosθ c tnθ c (d) Once the coin strts to slide, kinetic conditions previl nd the friction fce is At θ = θc < c f = f = μ n= μ mgcosθ k k k θ, the coin slides with constnt velocit, nd = 0 gin. Under these conditions, Newton s second lw gives Σ F = mgsin θc μkmgcosθ c = m = 0 nd μ k mg cos θ c = mg sinθ c ielding sin θc μk = = tnθ c cosθ c Pge 4.7
Chpter 4 4.7 () When the pole eerts fce downwrd nd towrd the rer on the lkebed, the lkebed eerts n oppositel directed fce of equl mgnitude, F, on the end of the pole. As the bot flots on the surfce of the lke, its verticl ccelertion is = 0. hus, Newton s second lw gives the mgnitude of the buont fce, F B, s Σ F = F + F cosθ mg = 0 B nd, with θ = 35.0, F = mg F cosθ = 370 kg 9.80 m s 40 N cos 35.0 B F = = B 3.43 03 N 3.43 kn Appling Newton s second lw to the hizontl motion of the bot gives ur Σ F = Fsinθ = m Fdrg ( 40 N) sin 35.0 47.5 N = = 370 kg 0.44 m s After n elpsed time t = 0.450 s, v = v0 + t gives the velocit of the bot s v = 0.857 m s + 0.44 m s 0.450 s = 0.967 m s (c) If ngle θ incresed while the mgnitude of F ur remined constnt, the verticl component of this fce would decrese. he buont fce would hve to increse to suppt me of the weight of the bot nd its contents. At the sme time, the hizontl component of F ur would increse, which would increse the ccelertion of the bot. 4.7 () Appling Newton s second lw to the rope ields Σ F = m F = m = F m r r hen, ppling Newton s second lw to the block, we find Σ F = m = m F m = m b r b [] Pge 4.73
Chpter 4 which gives = m b F + m r (c) Substituting the ccelertion found bove bck into eqution [] gives the tension t the left end of the rope s F m = F mr = F mr = F mb + m r b + m b m r + m r m r mb = F m + m b r (d) From the result of (c) bove, we see tht s m r pproches zero, pproches F. hus, the tension in cd of negligible mss is constnt long its length. 4.73 Choose the positive is to be down the incline nd the is perpendiculr to this s shown in the free-bod digrm of the to. he ccelertion of the to then hs components of m Δ v + 30.0 m s 0, nd 5.00 m s = = = = + Δt 6.00 s q mg q Appling the second lw to the to gives: () Σ F = mg sin θ = m sin θ = m mg = g, nd 5.00 m s θ = sin = sin = 30.7 g 9.80 m s Σ F = mg cosθ = m = 0, = mgcosθ = 0.00 kg 9.80 m s cos 30.7 = 0.843 N Pge 4.74
Chpter 4 4.74 he sketch t the right gives the free-bod digrm of the person. he scle simpl reds the mgnitude of the nml fce eerted on the student b the set. From Newton s second lw, we obtin Σ F = m =0 n mgcos 30.0 = 0 n = mgcosθ = 00 lb cos 30.0 = 73 lb 4.75 he ccelertion the cr hs s it is coming to stop is ( Δ) v 0 0 35 m s v = = = 0.6 m s 000 m hus, the mgnitude of the totl retrding fce cting on the cr is w 8 80 N F = m = g = = 9.80 m s 0.6 m s 5.5 0 N 4.766 () In the verticl direction, we hve Σ F = 8 000 N sin 65.0 w = m = 0 8 000 N 65.0 so w = 3 ( 8 000 N) sin 65.0 = 7.5 0 N Along the hizontl, Newton s second lw ields w Σ F = ( 8 000 N) cos 65.0 = m = g g 8 000 N cos 65.0 9.80 m s 8 000 N cos 65.0 = = = w 7.5 03 N w mg Pge 4.75
Chpter 4 4.77 Since the bd is in equilibrium, Σ = 0 nd we see tht F the nml fces must hve the sme mgnitudes on both sides of the bd. Also, if the minimum nml fces (compression fces) re being pplied, the bd is on the verge of slipping nd the friction fce on ech side is f = f = μ n s m s n f f n he bd is lso in equilibrium in the verticl direction, so Σ F = f w = 0, f = w w 95.5 N he minimum compression fce needed is then n = f w 95.5 μ = μ = N 0.663 s s = 7.0 N 4.78 Consider the two free-bod digrms, one of the penguin lone nd one of the combined sstem consisting of penguin plus sled. n n he nml fce eerted on the f f penguin b the sled is n = w = m g w 70.0 N w totl 30 N F nd the nml fce eerted on the combined sstem b the ground is n = wtotl = mtotlg = 30 N he penguin is ccelerted fwrd b the sttic friction fce eerted on it b the sled. When the penguin is on the verge of slipping, this ccelertion is m ( f ) n μs m m μs = = = m m ( g) m = μ s g = 0.700 9.80 m s = 6.86 m s Pge 4.76
Chpter 4 Since the penguin does not slip on the sled, the combined sstem must hve the sme ccelertion s the penguin. Appling Newton s second lw to this sstem gives which ields wtotl Σ F = F f = mtotl m F = f + mtotl m = μk ( wtotl ) + g m 30 N F = ( 0.00)( 30 N) + ( 6.86 m s ) = 04 N 9.80 m s 4.79 First, we will compute the needed ccelertions: () Befe it strts to move: = 0 () During the first 0.80 s: v v. m s 0 = = =.5 m s t 0.80 s 0 (3) While moving t constnt velocit: = 0 (4) During the lst.5 s: v v 0. m s = = = 0.80 m s t.5 s 0 he spring scle reds the nml fce the scle eerts on the mn. Appling Newton s second lw to the verticl motion of the mn gives: Σ F = n mg = m n = m( g + ) () When = 0, n = + = 7 kg 9.80 m s 0 7. 0N When =.5 m s, n = 7 kg 9.80 m s +.5 m s = 8. 0N (c) When = 0, n = + = 7 kg 9.80 m s 0 7. 0N Pge 4.77
Chpter 4 (d) When = 0.80 m s, n = 7 kg 9.80 m s 0.80 m s = 6.5 0N 4.80 he friction fce eerted on the mug b the moving tblecloth is the onl hizontl fce the mug eperiences during this process. hus, the hizontl ccelertion of the mug will be mug fk 0.00 N = = = m 0.00 kg mug 0.500 m s he cloth nd the mug both strt from rest ( v 0 = 0 ) t time t = 0. hen, t time t > 0, the hizontl 0 displcements of the mug nd cloth re given b Δ = v t + t s: nd Δ mug = 0 + 0.500 m s t = 0.50 m s t (.50 m ) Δ = 0 + 3.00 m s t = s t cloth In der f the edge of the cloth to slip under the mug, it is necessr tht t.50 m s = 0.50 m s t + 0.300 m he elpsed time when this occurs is Δ = Δ + cloth mug 0.300 m, 0.300 m t = = (.50 0.50 ) m s 0.490s At this time, the mug hs moved distnce of Δ 0.50 m s 0.490 s 6.00 0 mug = = m = 6.00 cm Pge 4.78
Chpter 4 4. 8 () Consider the first free-bod digrm in which Chris nd the chir treted s combined sstem. he weight of this sstem is w totl = 480 N, nd its mss is 50 N n m totl wtotl = = g 49.0 kg king upwrd s positive, the ccelertion of this sstem is found 50 N 50 N from Newton s second lw s Σ F = w = m totl totl hus 50 N 480 N = = + 0.408 m s 49.0 kg w totl 30 N 60 N w Chris 30 N 0.408 m s upwrd he downwrd fce tht Chris eerts on the chir hs the sme mgnitude s the upwrd nml fce eerted on Chris b the chir. his is found from the free-bod digrm of Chris lone s Σ F = + n w = m Chris Chris so n = mchris + wchris Hence, n 30 N = ( 0.408 m ) 9.80 m s s + 30 N 50 N = 83.3 N 4.8 Let R ur represent the hizontl fce of ir resistnce. Since the helicopter nd bucket move t constnt velocit, = = 0. he second lw then gives: Also, Σ F = cos 40.0 mg = 0 mg = cos40.0 Σ F = sin 40.0 R= 0 R= sin 40.0 R 40.0 w m g Pge 4.79
Chpter 4 hus, mg R = = = cos 40.0 3 sin 40.0 60 kg 9.80 m s tn 40.0 5.0 0 N Pge 4.80