Halliday/Resnick/Walker 7e Chapter 6

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1 HRW 7e Chapter 6 Page of Halliday/Renick/Walker 7e Chapter 6 3. We do not conider the poibility that the bureau might tip, and treat thi a a purely horizontal motion problem (with the peron puh F in the +x direction). Applying ewton econd law to the x and y axe, we obtain F f,max = ma F mg = 0 repectively. The econd equation yield the normal force F = mg, whereupon the maximum tatic friction i found to be (from Eq. 6-) f,max = µ mg. Thu, the firt equation become F µ mg = ma = 0 where we have et a = 0 to be conitent with the fact that the tatic friction i till (jut barely) able to prevent the bureau from moving. (a) With µ = 045. and m = 45 kg, the equation above lead to F = 98. To bring the bureau into a tate of motion, the peron hould puh with any force greater than thi value. Rounding to two ignificant figure, we can therefore ay the minimum required puh i F =.0 0. (b) Replacing m = 45 kg with m = 8 kg, the reaoning above lead to roughly F = The greatet deceleration (of magnitude a) i provided by the maximum friction force (Eq. 6-, with F = mg in thi cae). Uing ewton econd law, we find a = f,max /m = µ g. Eq. -6 then give the hortet ditance to top: x = v /a = 36 m. In thi calculation, it i important to firt convert v to 3 m/. 7. We chooe +x horizontally rightward and +y upward and oberve that the 5 force ha component F x = F co θ and F y = F in θ. (a) We apply ewton econd law to the y axi: F F in θ mg = 0 F = (5) in 40 + (3.5)(9.8) = 44. With µ k = 0.5, Eq. 6- lead to f k =. (b) We apply ewton econd law to the x axi: ( 5) co 40 Fco θ fk = ma a = = 0.4 m/. 3.5

2 HRW 7e Chapter 6 Page of Since the reult i poitive-valued, then the block i accelerating in the +x (rightward) direction. 9. Applying ewton econd law to the horizontal motion, we have F µ k m g = ma, where we have ued Eq. 6-, auming that F = mg (which i equivalent to auming that the vertical force from the broom i negligible). Eq. -6 relate the ditance traveled and the final peed to the acceleration: v = a x. Thi give a =.4 m/. Returning to the force equation, we find (with F = 5 and m = 3.5 kg) that µ k = (a) The free-body diagram for the crate i hown below. T i the tenion force of the rope on the crate, F i the normal force of the floor on the crate, mg i the force of gravity, and f i the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. We aume the crate i motionle. The equation for the x and the y component of the force according to ewton econd law are: T co θ f = 0 Tinθ + F mg = 0 where θ = 5 i the angle between the rope and the horizontal. The firt equation give f = T co θ and the econd give F = mg T in θ. If the crate i to remain at ret, f mut be le than µ F, or T co θ < µ (mg T inθ). When the tenion force i ufficient to jut tart the crate moving, we mut have We olve for the tenion: T = µ mg co θ + µ in θ ( 0.50) ( 68) ( 9.8) = co in = T co θ = µ (mg T in θ). (b) The econd law equation for the moving crate are T co θ f = ma F + T in θ mg = 0. ow f =µ k F, and the econd equation give F = mg Tinθ, which yield f = µ ( mg T inθ ). Thi expreion i ubtituted for f in the firt equation to obtain k T co θ µ k (mg T in θ) = ma,

3 HRW 7e Chapter 6 Page 3 of o the acceleration i T ( co θ + µ k in θ) a = µ k g. m umerically, it i given by a = b gb g b gc h 304 co in m/ =.. 3. m/. 68 kg 5. (a) The free-body diagram for the block i hown below. F i the applied force, F i the normal force of the wall on the block, f i the force of friction, and mg i the force of gravity. To determine if the block fall, we find the magnitude f of the force of friction required to hold it without accelerating and alo find the normal force of the wall on the block. We compare f and µ F. If f < µ F, the block doe not lide on the wall but if f > µ F, the block doe lide. The horizontal component of ewton econd law i F F = 0, o F = F = and µ F = (0.60)( ) = 7.. The vertical component i f mg = 0, o f = mg = 5.0. Since f < µ F the block doe not lide. (b) Since the block doe not move f = 5.0 and F =. The force of the wall on the block i F = F ˆi + f ˆj = ˆi ˆ j where the axe are a hown on Fig. 6-5 of the text. w ( ) ( ) 6. We find the acceleration from the lope of the graph (recall Eq. -): a = 4.5 m/. The force are imilar to what i dicued in Sample Problem 6- but with the angle φ equal to 0 (the applied force i horizontal), and in thi problem the horizontal acceleration i not zero. Thu, ewton econd law lead to F µ k mg = ma, where F = 40.0 i the contant horizontal force applied. With m = 4. kg, we arrive at µ k = Fig. 6-4 in the textbook how a imilar ituation (uing φ for the unknown angle) along with a free-body diagram. We ue the ame coordinate ytem a in that figure. (a) Thu, ewton econd law lead to 3

4 HRW 7e Chapter 6 Page 4 of Tco φ f = ma along x axi Tinφ + F mg = 0 along y axi Setting a = 0 and f = f,max = µ F, we olve for the ma of the box-and-and (a a function of angle): F HG m T coφ = inφ + g µ which we will olve with calculu technique (to find the angle φ m correponding to the maximum ma that can be pulled). dm dt F HG T inφ m = coφ m g µ I KJ I = KJ Thi lead to tan φ m = µ which (for µ = 035. ) yield φ m = 9. (b) Plugging our value for φ m into the equation we found for the ma of the box-and-and yield m = 340 kg. Thi correpond to a weight of mg = The free-body diagram for block B and for the knot jut above block A are hown next. T i the tenion force of the rope pulling on block B or pulling on the knot (a the cae may be), T i the tenion force exerted by the econd rope (at angle θ = 30 ) on the knot, f i the force of tatic friction exerted by the horizontal urface on block B, F i normal force exerted by the urface on block B, W A i the weight of block A (W A i the magnitude of m Ag), and WB i the weight of block B (W B = 7 i the magnitude of mg B ). For each object we take +x horizontally rightward and +y upward. Applying ewton econd law in the x and y direction for block B and then doing the ame for the knot reult in four equation: 4

5 HRW 7e Chapter 6 Page 5 of T f,max = 0 F WB = 0 T co θ T = 0 T in θ W = 0 A where we aume the tatic friction to be at it maximum value (permitting u to ue Eq. 6-). Solving thee equation with µ = 0.5, we obtain W = Treating the two boxe a a ingle ytem of total ma m C + m W = = 4.0 kg, ubject to a total (leftward) friction of magnitude = 6.0, we apply ewton econd law (with +x rightward): F f = m a total = (4.0) a which yield the acceleration a =.5 m/. We have treated F a if it were known to the nearet tenth of a ewton o that our acceleration i good to two ignificant figure. Turning our attention to the larger box (the Wheatie box of ma m W = 3.0 kg) we apply ewton econd law to find the contact force F' exerted by the Cheerio box on it. W W A F f = m a total F 4.0 = (3.0)(.5) Thi yield the contact force F' = (a) Free-body diagram for the block A and C, conidered a a ingle object, and for the block B are hown below. T i the magnitude of the tenion force of the rope, F i the magnitude of the normal force of the table on block A, f i the magnitude of the force of friction, W AC i the combined weight of block A and C (the magnitude of force F hown in the figure), and WB i the weight of block B (the magnitude of force F gb hown). Aume the block are not moving. For the block on the table we take the x axi to be to the right and the y axi to be upward. From ewton econd law, we have x component: T f = 0 y component: F W AC = 0. For block B take the downward direction to be poitive. Then ewton econd law for that block i W B T = 0. The third equation give T = W B and the firt give f = T = W B. The econd equation give F = W AC. If liding i not to occur, f mut be le than µ F, or W B < µ W AC. The mallet that W AC can be with the block till at ret i W AC = W B /µ = ( )/(0.0) = 0. gac 5

6 HRW 7e Chapter 6 Page 6 of Since the weight of block A i 44, the leat weight for C i (0 44) = 66. (b) The econd law equation become T f = (W A /g)a F W A = 0 W B T = (W B /g)a. In addition, f = µ k F. The econd equation give F = W A, o f = µ k W A. The third give T = W B (W B /g)a. Subtituting thee two expreion into the firt equation, we obtain Therefore, W B (W B /g)a µ k W A = (W A /g)a. a ( ) ( µ W) (9.8 m/ ) ( 0.5)( 44 ) gwb k A = = = W + W 44 + A B.3 m/. 3. Uing Eq. 6-6, we olve for the area mg A Cρ v t which illutrate the invere proportionality between the area and the peed-quared. Thu, when we et up a ratio of area of the lower cae to the fater cae we obtain A A low fat 30 km / h 60 km / h = F H G I K J = The magnitude of the acceleration of the car a it round the curve i given by v /R, where v i the peed of the car and R i the radiu of the curve. Since the road i horizontal, only the frictional force of the road on the tire make thi acceleration poible. The horizontal component of ewton econd law i f = mv /R. If F i the normal force of the road on the car 6

7 HRW 7e Chapter 6 Page 7 of and m i the ma of the car, the vertical component of ewton econd law lead to F = mg. Thu, uing Eq. 6-, the maximum value of tatic friction i f,max = µ F = µ mg. If the car doe not lip, f µ mg. Thi mean v g v R g. R µ µ Conequently, the maximum peed with which the car can round the curve without lipping i vmax = µ Rg = (0.60)(30.5)(9.8) = 3 m/ 48 km/h. 37. The magnitude of the acceleration of the cyclit a it round the curve i given by v /R, where v i the peed of the cyclit and R i the radiu of the curve. Since the road i horizontal, only the frictional force of the road on the tire make thi acceleration poible. The horizontal component of ewton econd law i f = mv /R. If F i the normal force of the road on the bicycle and m i the ma of the bicycle and rider, the vertical component of ewton econd law lead to F = mg. Thu, uing Eq. 6-, the maximum value of tatic friction i f,max = µ F = µ mg. If the bicycle doe not lip, f µ mg. Thi mean v R µ g R v µ g. Conequently, the minimum radiu with which a cyclit moving at 9 km/h = 8. m/ can round the curve without lipping i R v (8. m/) = = = m. (0.3)(9.8 m/ ) min µ g 39. Perhap urpriingly, the equation pertaining to thi ituation are exactly thoe in Sample Problem 6-9, although the logic i a little different. In the Sample Problem, the car move along a (tationary) road, wherea in thi problem the cat i tationary relative to the merry-go-around platform. But the tatic friction play the ame role in both cae ince the bottom-mot point of the car tire i intantaneouly at ret with repect to the race track, jut a tatic friction applie to the contact urface between cat and platform. Uing Eq. 6-3 with Eq. 4-35, we find µ = (πr/t ) /gr = 4π R/gT. With T = 6.0 and R = 5.4 m, we obtain µ = At the top of the hill, the ituation i imilar to that of Sample Problem 6-7 but with the normal force direction revered. Adapting Eq. 6-9, we find 7

8 HRW 7e Chapter 6 Page 8 of F = m(g v /R). Since F = 0 there (a tated in the problem) then v = gr. Later, at the bottom of the valley, we revere both the normal force direction and the acceleration direction (from what i hown in Sample Problem 6-7) and adapt Eq. 6-9 accordingly. Thu we obtain F = m(g + v /R) = mg = (a) We note that the peed 80.0 km/h in SI unit i roughly. m/. The horizontal force that keep her from liding mut equal the centripetal force (Eq. 6-8), and the upward force on her mut equal mg. Thu, F net = (mg) + (mv /R) = 547. (b) The angle i tan [(mv /R)/(mg)] = tan (v /gr) = 9.53º (a meaured from a vertical axi). 45. (a) At the top (the highet point in the circular motion) the eat puhe up on the tudent with a force of magnitude F = 556. Earth pull down with a force of magnitude W = 667. The eat i puhing up with a force that i maller than the tudent weight, and we ay the tudent experience a decreae in hi apparent weight at the highet point. Thu, he feel light. (b) ow F i the magnitude of the upward force exerted by the eat when the tudent i at the lowet point. The net force toward the center of the circle i F b W = mv /R (note that we are now chooing upward a the poitive direction). The Ferri wheel i teadily rotating o the value mv R i the ame a in part (a). Thu, F mv = + W = = 778. R (c) If the peed i doubled, mv point we have W F = mv R, which lead to R increae by a factor of 4, to 444. Therefore, at the highet F = = 3. (d) Similarly, the normal force at the lowet point i now found to be F = k. 49. For the puck to remain at ret the magnitude of the tenion force T of the cord mut equal the gravitational force Mg on the cylinder. The tenion force upplie the centripetal force that keep the puck in it circular orbit, o T = mv /r. Thu Mg = mv /r. We olve for the peed: Mgr (.50)(9.80)(0.00) v = = =.8 m/. m.50 8

9 HRW 7e Chapter 6 Page 9 of 55. We apply ewton econd law (a F puh f = ma). If we find F puh < f max, we conclude no, the cabinet doe not move (which mean a i actually 0 and f = F puh ), and if we obtain a > 0 then it i move (o f = f k ). For f max and f k we ue Eq. 6- and Eq. 6- (repectively), and in thoe formula we et the magnitude of the normal force equal to 556. Thu, f max = 378 and f k = 3. (a) Here we find F puh < f max which lead to f = F puh =. (b) Again we find F puh < f max which lead to f = F puh = 334. (c) ow we have F puh > f max which mean it move and f = f k = 3. (d) Again we have F puh > f max which mean it move and f = f k = 3. (e) The cabinet move in (c) and (d). 56. Sample Problem 6-3 treat the cae of being in danger of liding down the θ ( = 35.0º in thi problem) incline: tanθ = µ = (Eq. 6-3). Thi value repreent a 3.4% decreae from the given 0.75 value. 58. (a) The x component of F trie to move the crate while it y component indirectly contribute to the inhibiting effect of friction (by increaing the normal force). ewton econd law implie x direction: Fcoθ f = 0 y direction: F Finθ mg = 0. To be on the verge of liding mean f = f,max = µ F (Eq. 6-). Solving thee equation for F (actually, for the ratio of F to mg) yield Thi i plotted below (θ in degree). F mg µ = coθ µ inθ. 9

10 HRW 7e Chapter 6 Page 0 of (b) The denominator of our expreion (for F/mg) vanihe when For µ = 0.70, we obtain θ = inf = µ coθ µ inθ 0 θ tan inf = = tan 55 µ. (c) Reducing the coefficient mean increaing the angle by the condition in part (b). (d) For µ = 0.60 we have θ inf = = µ tan (a) The tenion will be the greatet at the lowet point of the wing. ote that there i no ubtantive difference between the tenion T in thi problem and the normal force F in Sample Problem 6-7. Eq. 6-9 of that Sample Problem examine the ituation at the top of the circular path (where F i the leat), and rewriting that for the bottom of the path lead to where F i at it greatet value.. T = mg + mv /r (b) At the breaking point T = 33 = m(g + v /r) where m = 0.6 kg and r = 0.65 m. Solving for the peed, we find that the cord hould break when the peed (at the lowet point) reache 8.73 m/. 70. (a) The coefficient of tatic friction i µ = tan(θ lip ) = (b) Uing mg inθ f = ma f = f k = µ k F = µ k mg coθ and a = d/t (with d =.5 m and t = 4.0 ), we obtain µ k = Replace f with f k in Fig. 6-5(b) to produce the appropriate force diagram for the firt part of thi problem (when it i liding downhill with zero acceleration). Thi amount to replacing the tatic coefficient with the kinetic coefficient in Eq. 6-3: µ k = tanθ. ow (for the econd part of the problem, with the block projected uphill) the friction direction i revered from what i hown in Fig. 6-5(b). ewton econd law for the uphill motion (and Eq. 6-) lead to m g inθ µ k m g coθ = m a. 0

11 HRW 7e Chapter 6 Page of Canceling the ma and ubtituting what we found earlier for the coefficient, we have g inθ tanθ g coθ = a. Thi implifie to g inθ = a. Eq. -6 then give the ditance to top: x = v o /a. (a) Thu, the ditance up the incline traveled by the block i x = v o /(4ginθ ). (b) We uually expect µ > µ k (ee the dicuion in ection 6-). Sample Problem 6-3 treat the angle of repoe (the minimum angle neceary for a tationary block to tart liding downhill): µ = tan(θ repoe ). Therefore, we expect θ repoe > θ found in part (a). Conequently, when the block come to ret, the incline i not teep enough to caue it to tart lipping down the incline again. 95. Except for replacing f with f k, Fig 6-5 in the textbook i appropriate. With that figure in mind, we chooe uphill a the +x direction. Applying ewton econd law to the x axi, we have W fk Win θ = ma where m=, g and where W = 40, a = m/ and θ = 5. Thu, we find f k = 0. Along the y axi, we have o that µ k = f k / F = F = 0 F = Wcoθ y 0. (a) The free-body diagram for the peron (hown a an L-haped block) i hown below. The force that he exert on the rock lab i not directly hown (ince the diagram hould only how force exerted on her), but it i related by ewton third law) to the normal force F and F exerted horizontally by the lab onto her hoe and back, repectively. We will how in part (b) that F = F o that we there i no ambiguity in aying that the magnitude of her puh i F. The total upward force due to (maximum) tatic friction i f = f + f where f = µ F and f = µ F. The problem give the value µ =. and µ = 0.8.

12 HRW 7e Chapter 6 Page of (b) We apply ewton econd law to the x and y axe (with +x rightward and +y upward and there i no acceleration in either direction). F F = f + f mg = 0 0 The firt equation tell u that the normal force are equal F = F = F. Conequently, from Eq. 6-, f f = µ F = µ F we conclude that Therefore, f + f mg = 0 lead to = f µ f µ. µ µ + f = mg which (with m = 49 kg) yield f = 9. From thi we find F = f /µ = 40. Thi i equal to the magnitude of the puh exerted by the rock climber. (c) From the above calculation, we find f = µ F = 88 which amount to a fraction or 60% of her weight. f 88 = = W b g b g 060.

a = f s,max /m = s g. 4. We first analyze the forces on the pig of mass m. The incline angle is.

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