Phyic 100 Homewor 5 Chapter 6 Contact Force Introduced ) When two object lide againt one another, the magnitude of the frictional force i alway equal to μ B) When two object are in contact with no relative motion, the magnitude of the frictional force may be either le or equal to μ C) When a board with a bo on it i lowly tilted to a larger and larger angle, common eperience how that the bo will at ome point "brea looe" and tart to accelerate down the board. he bo begin to lide once the component of it weight parallel to the board, W, equal the maimum force of tatic friction. he bo accelerate down the board after it begin to lide becaue the coefficient of inetic friction i le than the coefficient of tatic friction. Comparing Friction Force with Gravity )he crate i at ret on the incline becaue the frictional force point up the incline. B) When the phyicit attempt to move the create by puhing up the incline, he find that the force of tatic friction point down the incline. Contant velocity C) When the two phyicit puh the crate up the incline with contant velocity, their combined force i the ame a the um of the force of gravity parallel to the incline plu the force of inetic friction. Both force pointing down the incline. Kinetic Friction in a Bloc-and-Pullet Sytem. f B W W B Conider the ytem hown in the figure. Bloc ha weight 4.91 and bloc B ha weight.94. Once bloc B i et into downward motion, it decend at a contant peed. ume that the ma and friction of the pulley are negligible. Copyright 010 Pearon Education, Inc. ll right reerved. hi material i protected under all copyright law a they currently eit. o 6 1
Chapter 6: pplication of ewton Law Jame S. Waler, Phyic, 4 th Edition Bloc (no acceleration) f f W = μ = μ W > = μ W Bloc B (no acceleration) = M B g Combining both epreion for μ W = W B WB.94 μ = =.6 W 4.91 Puhing a Chair long the Floor chair of weight 80.0 lie atop a horizontal floor; the floor i not frictionle. You puh on the chair with a force of F= 4.0 directed at an angle of 41.0 below the horizontal and the chair lide along the floor. F in 41 F co 41 41 f W Vertical direction F in( 41) W = W + F in( 41) = 80 + 4 in(41) = 108 6.3 baeball player lide into third bae with an initial peed of 4.0 m/. If the coefficient of inetic friction between the player and the ground i 0.46, how far doe the player lide before coming to ret? a f Horizontal direction. he only force i that of friction. f μmg f = ma a = = = μ g m m Copyright 010 Pearon Education, Inc. ll right reerved. hi material i protected under all copyright law a they currently eit. o 6
Chapter 6: pplication of ewton Law Jame S. Waler, Phyic, 4 th Edition 3. Picture the Problem: baeball player lide in a traight line and come to ret due to the frictional force. Strategy: Find the acceleration of the player uing ewton Second Law, and inert the reult into equation -1 to find the lide ditance. Let the initial velocity v r 0 point in the poitive direction. Solution: 1. Find the acceleration of the player uing ewton Second Law:. Ue equation -1 to find the lide ditance: v v 0 v ( 4.0 m/) 0 0 Δ = = = = 1.8 m a μ g 0.46 9.81 m/ Inight: If the player were running fater, ay, 8.0 m/, a µ of 0.46 would reult in a lide ditance of 7.1 m, a quarter of the 7 m between the bae! He might very well overhoot the bae and be tagged out. 6.1 45-g crate i placed on an inclined ramp. When the angle the ramp mae with the horizontal i increaed to 6, the crate begin to lide downward. 1. Picture the Problem: he crate lide down the incline when the angle i jut 6. free-body diagram of the ituation i depicted at right. Strategy: Chooe the -ai to be parallel to the ramp, pointing uphill, and the y ai to point perpendicular to the ramp. Write ewton Second Law in the y direction to find the normal force. hen write ewton Second Law in the direction with a (the bo jut begin to lide) to find the coefficient of tatic friction μ. Solution: 1. (a) Write ewton Second Law in the y direction: F = mgcoθ y = mgcoθ. Write ewton Second Law in the F = μ mg inθ = ma direction, etting a (the bo jut μ ( mg coθ) mg inθ begin to lide) and olving for μ : μ = tanθ = tan 6 = 0.49 3. (b) Since θ only depend upon μ, changing the ma will not change θ and the angle remain 6. Inight: Increaing the ma doe increae the normal force and hence the friction force, but the component of the crate weight that pull it down the ramp i alo increaed. he two effect cancel and θ depend upon μ only. Lifting a Bucet 6-g bucet of water i being pulled traight up by a tring at a contant peed. ) he tenion in the rope W = mg = ( 6)(9.8) = 59, about 60 a=0 contant peed W Copyright 010 Pearon Education, Inc. ll right reerved. hi material i protected under all copyright law a they currently eit. o 6 3
Chapter 6: pplication of ewton Law Jame S. Waler, Phyic, 4 th Edition B) t a certain point the peed of the bucet begin to change. he bucet now ha an upward contant acceleration of magnitude 3m/. What i the tenion in the rope now? W = ma = mg + ma = m( g + a) = (6)(9.8 + 3) = 76. 8, about 76 C) If the acceleration i downward. W = ma = mg ma = m( g a) = (6)(9.8 3) = 40. 8, about 4 6.31) picture hang on the wall upended by two tring, a hown in the figure. he tenion in tring 1 i 1.7. 31. Picture the Problem: he free-body diagram for the contact point between the two tring i depicted at right. Strategy: he horizontal component of the tring tenion mut be equal becaue the picture i not accelerating. he ame i true of the vertical component of the force. Ue ewton Second Law in the horizontal direction to find the tenion in tring, and in the vertical direction to find the weight of the picture. Solution: 1. (a) he tenion in tring i le than the tenion in tring 1, becaue it provide motly a ideway component of force that i balanced by the horizontal component of tring 1. hat mean tring 1 mut upport mot of the weight of the picture plu balance tring horizontal component, giving it a larger tenion than tring.. (b) Write ewton Second Law in the horizontal direction in order to find the tenion in tring : 3. (c) Write ewton Second Law in the vertical direction in order to find the picture weight: F = 1coθ1+ coθ coθ co 65 1.7 0.85 1 = 1 = = coθ co 3 F = inθ + inθ W y 1 1 W = 1.7 in 65 + 0.85 in 3 =.0 Inight: the angle of tring 1 approache 90 and the angle of tring approache 0, the tenion in tring drop to zero and the entire.0 weight of the picture i upported by tring 1. 6.37) wo bloc are connected by a tring, a hown in the figure. he mooth inclined urface mae an angle of 4 with the horizontal, and the bloc on the incline ha a ma of 6.7g. 37. Picture the Problem: he force eerted on the left and right bloc are depicted at right. Strategy: Becaue the pulley i ideal, the tenion in the tring i equal to the weight of the hanging bloc. hi can be verified by ewton Second Law in the vertical direction for the hanging bloc: Fy = mg. Write ewton Second Law along the direction parallel to the incline for the 6.7 g bloc and ubtitute = mginto the reulting equation to find m. Solution: 1. Write ewton Second Law along the direction parallel to the incline:. Subtitute = mginto the reulting equation to find m. F = Mginθ = mg = Mginθ m= M inθ = 6.7 g in 4 = 4.5 g Copyright 010 Pearon Education, Inc. ll right reerved. hi material i protected under all copyright law a they currently eit. o 6 4
Chapter 6: pplication of ewton Law Jame S. Waler, Phyic, 4 th Edition Inight: larger m i required if the angle θ i increaed. If it i increaed all the way to θ =90, the large ma will be hanging traight down and the ma m required to maintain equilibrium would be 6.7 g.. (b) he tenion in the tring wa determined in Eample 6-6 to be m m g ( m m ) doubled the tenion would be = m g ( m + m ) =. tenion would increae. new 1 1 = + If both mae were 1 1. m We conclude that if both mae were doubled the Inight: he tenion mut increae in the econd cae becaue there i twice a much ma that mut be accelerated at the ame rate a before. 6.44) Find the acceleration of the mae hown in the figure, given that m1 = 1.0g, m =. 0g, and m3 = 3. 0g. ume the table i frictionle and the mae move freely. 44. Picture the Problem: he phyical apparatu i hown at right. Strategy: Write ewton Second Law for each of the three bloc and add the equation to eliminate the unnown 1 and. Solve the reulting equation for the acceleration a. Let be poitive in the direction of each ma motion. Solution: 1. Write ewton Second Law for each of the three bloc and add the equation:. Solve the reulting bloc 1 bloc bloc 3 F = = ma 1 1 F = + = m a 1 F = + m g = m a 3 3 mg= m+ m + m a 3 1 3 m 3.0 g = = 9.81 m/ = 4.9 m/ m1+ m + m3 6.0 g 3 equation for a: a g Inight: ote that the bloc move a if they were a ingle bloc of ma 6.0 g under the influence of a force equal to mg= 3 9. Inight: comet that travel along an elliptical path eperience the larget gravitational force when it i cloet to the Sun. In uch a cae the required centripetal force i larget both becaue v i greatet and r i leat at that point. In orbit terminology the point cloet to the Sun i called perihelion. Ma in a urntable mall metal cylinder ret on a circular turntable that i rotating at a contant rate, a illutrated in the diagram. Copyright 010 Pearon Education, Inc. ll right reerved. hi material i protected under all copyright law a they currently eit. o 6 5
Chapter 6: pplication of ewton Law Jame S. Waler, Phyic, 4 th Edition ) In circular motion with contant peed the acceleration and force point radially inward and the velocity i tangential. (e) B) he tangential peed i v = rω, and the centripetal acceleration i a cp = rω, where the angular velocity ω i contant. If the radiu i halved, the peed and the acceleration are halved. 6.56) Find the linear peed of the bottom of a tet tube in a centrifuge if the centripetal acceleration there i 5. 10 4 time the acceleration of gravity. he ditance from the ai of rotation to the bottom of the tet tube i 7.5 cm. 56. Picture the Problem: he tet tube travel along a circular path at contant peed. Strategy: Solve equation 6-15 for the peed required to attain the deired acceleration. Solution: Solve equation 6-15 for the peed: v = racp = r( 5,000g) = ( 0.075 m)( 5,000)( 9.81 m/ ) 0 m/.0 m/ Inight: hi peed correpond to 5,000 revolution per minute for the centrifuge, or 415 revolution per econd. Inight: he tenion doe not quadruple in part (a) becaue the acceleration i the um of gravitational and centripetal acceleration. he centripetal acceleration quadruple, but the um doe not becaue gravity remain contant. o calculate the number of revolution above: v 00 ω,666 ω = = =,666 rad/ ω = πf f = = = 44 rev/ r 0.075 6.6 Driving in your car with a contant peed of 1m/, you encounter a bump in the road that ha a circular cro-ection, a indicated in the figure. If the radiu of curvature of the bump i 35m, find the apparent weight of a 67-g peron in your car a you pa over the top of the bump. π π 6. Picture the Problem: he car follow a circular path at contant peed a it pae over the bump. Strategy: he centripetal acceleration i downward, toward the center of the circle, a the car pae over the bump. Write ewton Second Law in the vertical direction and olve for the normal force, which i alo the apparent weight of the paenger. Solution: 1. Write ewton y Second Law for the paenger and olve for = m( g v r) :. Inert numerical value: F = mg = macp = mv r 1 m/ = 67 g 9.81 m/ = 380.38 35 m Inight: hi apparent weight i 4% le than the normal 0.66- weight of the paenger. Copyright 010 Pearon Education, Inc. ll right reerved. hi material i protected under all copyright law a they currently eit. o 6 6