PHYSICS 151 Notes for Online Lecture 2.3

Transcription

1 PHYSICS 151 Note for Online Lecture.3 riction: The baic fact of acrocopic (everda) friction are: 1) rictional force depend on the two aterial that are liding pat each other. bo liding over a waed floor eperience le friction than a bo liding over an unwaed floor. ) There are two tpe of friction for ot urface: tatic friction and kinetic (or liding) friction. 3) The frictional force, fr alwa act in the direction oppoite the direction of otion. 4) The frictional force doe not depend on the urface area of the object in contact. Static friction i the reitance an object ha to tarting to ove. or an interface, we ut pull on an object with oe oderate force before the object begin to ove. or the kinetic cae fr ( kinetic) =µ N with direction oppoite to the direction of otion k Kinetic friction i the friction eperienced b a oving object. Once otion begin, the frictional force decreae and we can eert a aller force on the object (to balance the kinetic frictional force) to keep the object oving at contant velocit. or an interface the dr liding frictional force i roughl independent of the relative peed of the object over the urface. or urface lubricated with liquid or for vicou friction, (a for a boat peeding acro a lake) there i oe peed dependence. or the tatic cae µ ( ) N fr tatic Everthing we know about friction i uarized in a paraeter called the coefficient of friction. The coefficient of friction i different for the tatic and kinetic cae -- µ i the coefficient of tatic friction. µ k i the coefficient of kinetic friction. Both of thee nuber are alwa le than 1. or eaple, teel oving along teel ha a kinetic frictional coefficient around 0.6. If ou lubricate the urface, ou can reduce the coefficient of kinetic friction to around 0.. lo, ou ight notice that it i often harder to get the object oving than to keep it oving at a contant velocit. Thi ean that the aiu value of the tatic coefficient of friction i alwa greater than the kinetic coefficient of friction. Notice that in the equation for the tatic frictional force, there i a le than or equal to ign. µ fr ( tatic) N The tatic frictional force i a trangel adaptable force. If I puh lightl on a heav bo itting on the floor, the bo doe not ove. I can puh a little harder and till there i no otion. Since there i no acceleration in either of thee cae, the tatic frictional force ut adapt itelf to jut balance whatever force I appl to the bo o that the net force on the bo i zero.

2 n invetigation of friction naturall lead to wondering what happen at the interface at the atoic level. We know that friction i different for different aterial; o the particular tructure of each aterial ut be ignificant. However, we will entirel focu on the acrocopic, where block of aterial can be characterized b a coefficient of friction. E. 1: bo of a 10.0 kg i pulled at an acceleration of.0 / with the force applied trictl horizontall. If the coefficient of friction i 0.0, what force i now needed to aintain thi acceleration? = 10 kg N + + f W -direction = a µ fr(kinetic) k N = a = a -direction = a 0 = N W We now need to find the agnitude of the Noral orce. We do thi fro the -equation 0 = N W N = W = g Plu thi reult into the -equation = µ N + a = µ g + a = ( µ g + a ) = 39.6 N = 40 N ( 0.0(9.8 ) +.0 ) = 10.0kg In the frictionle cae we would onl need to appl a force of 0 N (let µ = 0 to ee thi), o we have to appl an additional 0 N to overcoe friction. E. : Let return to the bo of a 10.0 kg pulled at an acceleration of.0 / but now with the force applied at an angle of 30 degree with the horizontal. a) If the coefficient of friction i 0.0, what force i now needed to aintain thi acceleration? b) What i the agnitude of the noral force?

3 bo of a 10 kg i pulled at an acceleration of /. If the force i applied at an angle of 30 with repect to the horizontal and the coefficient of friction i 0., what force i necear to aintain thi acceleration? = 10 kg = 10 kg N f 30 -coponent -coponent Break the applied force into coponent = co (30) = in(30) Write Newton Second Law for each direction = a = a = a N W + = 0 f co( 30 ) µ N = a co(30) µ N = a To get the applied force, we need the noral force N = W co(30) µ ( g in (30)) = ( co(30) + µ in (30)) = a + µ g ( co(30) + µ in (30)) = ( a + µ g) ( a + µ g) = ( co(30) + µ in (30)) (10 kg) = N = g Plug thi reult back into the -equation a (.0 ( 0.0)( ) ( co (30) in (30)) 39.6 N = = 41N ( 0.966) in(30) Copare thi to the reult if there i no friction (3 N). Now ue thi to find the agnitude of the noral force N = W N = + g N = (10kg) in(30) ( 9.8 ) 41N in (30) N = 78 N Note that the noral force i reduced due to the applied force being upward. W

4 E. 3: book i on an inclined plane. I the noral force: a) greater than g b) le than g c) equal to g + hown in the free bod diagra below, the noral force will be balanced b the coponent of the weight parallel to the noral direction. or inclined plane proble, it often eaier to change the coordinate ae, 30 a hown. The coponent of the weight in the direction will be proportional to the coine of the angle arked θ. Regardle of what θ i, the in or co will be le than W 30 one, o the reulting coponent will be le than the value of the weight. So the noral force will be le than g. In our lat eaple, there were coponent in both direction, and. Reeber fro our anali of otion in D that the and the coponent act independentl. In the free bod diagra I ve hown above, r what ut be happening to the book? nwer i: it ut be accelerating, a we have = in θ = a Now ou know wh undertanding vector i o iportant! + c 90 d 90 b e a We are given that the plane i at an angle of θ (angle a). Looking at the right triangle fored b the inclined plane and the weight vector, angle b ut be 90-θ. B the alternate interior angle theore (I think that the theore), the angle directl oppoite thi angle (c) ut alo be 90-θ. Then angle d ut be 90 (90-θ) = θ. Uing the ae theore, the angle that we re intereted in angle e ut be θ.

5 E. 4: kier lide down a hill of lope 30 degree. She tart at ret. How fat i he going b the tie he traveled 30. ue no friction. kier lide down a hill of lope 35 degree. How fat i he going after he traveled a ditance of 30? ue no friction W 30 With thi choice of coordinate ae, all of the otion i taking place in the -direction. v = v + a 0 v = a But we have to know a! = a The Noral force i entirel along the -ai; however, the weight i not entirel along either ai, o we have to decopoe the weight. = a W = a ro the picture, W = +W in θ W = -W co θ Winθ = a W inθ = a g inθ = a ginθ = a ( 9.8 ) in 30 = a Now put thi back in the equation for velocit: 4.9 = a v = a v = (4.9 )(30 ) v = 94 v = 17.1

6 Now let include friction. ue the coefficient of friction i 0.15 and find the velocit after 30. What change? ll we have to do i to add the frictional force to the free bod diagra. + f + w w But now we have to find the noral force. g inθ µ g coθ = a g( inθ µ coθ) = a ( 9.8 )( in co 30) 3.63 = a W = a f g in θ µ N = a = a = a = a W + N = 0 Wcoθ + N = 0 Wcoθ = N v v = a = (3.63 )(30 ) v = 18 v = 14.8

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