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1 SPH4U: Lecture 7 Today s Agenda rcton What s t? Systeatc catagores of forces How do we characterze t? Model of frcton Statc & Knetc frcton (knetc = dynac n soe languages) Soe probles nvolvng frcton ew Topc: rcton What does t do? It opposes relatve oton of two obects that touch! How do we characterze ths n ters we have learned (forces)? rcton results n a force n the drecton opposte to the drecton of relatve oton (knetc frcton, statc pendng ot) APPLIED a RICTIO soe roughness here Surface rcton... rcton s caused by the croscopc nteractons between the two surfaces: Surface rcton... orce of frcton acts to oppose relatve oton: Parallel to surface. Perpendcular to oral force. a Page 1

2 Model for Sldng (knetc) rcton Model... The drecton of the frctonal force vector s perpendcular to the noral force vector. Dynacs: : K = a The agntude of the frctonal force vector s proportonal to the agntude of the noral force. = K ( = K n the prevous exaple) : so = K = a (ths works as long as s bgger than frcton,.e. the left hand sde s postve) The heaver soethng s, the greater the frcton wll be...akes sense! The constant K s called the coeffcent of knetc frcton. a These relatons are all useful APPROXIMATIOS to essy realty. K Lecture 7, Act 1 orces and Moton A box of ass 1 = 1.5 kg s beng pulled by a horzontal strng havng tenson T = 90. It sldes wth frcton ( k = 0.51) on top of a second box havng ass 2 = 3 kg, whch n turn sldes on a frctonless floor. (T s bgger than rcton, too.) What s the acceleraton of the second box? Lecture 7, Act 1 Soluton rst draw BD of the top box: 1 (a) a = 0 /s 2 (b) a = 2.5 /s 2 (c) a = 3.0 /s 2 T 1 f = K 1 = K 1 g Hnt: draw BDs of both blocks that s 2 dagras T 1 sldes wth frcton ( k =0.51 ) 1 g a =? 2 sldes wthout frcton Page 2

3 Lecture 7, Act 1 Soluton ewtons 3rd law says the force box 2 exerts on box 1 s equal and opposte to the force box 1 exerts on box 2. As we ust saw, ths force s due to frcton: Lecture 7, Act 1 Soluton ow consder the BD of box 2: 2 (contact fro ) 1 f 1,2 = K 1 g (frcton fro ) f 2,1 2 f 2,1 = k 1 g 2 (gravty fro ) 2 g 1 g (contact fro ) Lecture 7, Act 1 Soluton nally, solve = a n the horzontal drecton: Inclned Plane wth rcton: Draw free-body dagra: K 1 g = 2 a a = 2.5 /s 2 a 1 k kg g s 3kg 2 K a f 2,1 = K 1 g 2 Page 3

4 Inclned plane... Consder and coponents o ET = a : sn K = a K = cos a sn K cos = a a / g = sn K cos cos sn Statc rcton... So far we have consdered frcton actng when the two surfaces ove relatve to each other- I.e. when they slde.. We also know that t acts n when they ove together: the statc case. In these cases, the force provded by frcton wll depend on the OTHER forces on the parts of the syste. Statc rcton (wth one surface statonary) Just lke n the sldng case except a = 0. : = 0 : = Whle the block s statc: Statc rcton The axu possble force that the frcton between two obects can provde s f MAX = S, where s s the coeffcent of statc frcton. So S. As one ncreases, gets bgger untl = S and the obect starts to ove. If an obect doesn t ove, t s statc frcton If an obect does ove, t s dynac frcton Page 4

5 Statc rcton... S s dscovered by ncreasng untl the block starts to slde: : MAX S = 0 : = S MAX / Lecture 7, Act 2 orces and Moton A box of ass =10.21 kg s at rest on a floor. The coeffcent of statc frcton between the floor and the box s s = 0.4. A rope s attached to the box and pulled at an angle of = 30 o above horzontal wth tenson T = 40. Does the box ove? (a) yes (b) no (c) too close to call MAX T S statc frcton ( s = 0.4 ) Lecture 7, Act 2 Soluton y Lecture 7, Act 2 Soluton y Pck axes & draw BD of box: x y: = 80 x Apply ET = a x: T cos - R = a X y: + T sn - = a Y = 0 = - T sn = 80 The box wll ove f T cos - R > 0 x: T cos - R = a X The box wll ove f T cos - R > 0 R T T cos = 34.6 f MAX = s = (.4)(80) = 32 f MAX = s T So T cos > f MAX and the box does ove ow use dynac frcton: a x = Tcos - K Page 5

6 Statc rcton: We can also consder S on an nclned plane. Statc rcton... The force provded by frcton,, depends on. In ths case, the force provded by frcton wll depend on the angle of the plane. a = 0 (block s not ovng) sn f f (ewton s 2nd Law along x-axs) Statc rcton... We can fnd s by ncreasng the rap angle untl the block sldes: S sn f f In ths case, when t starts to slde: f f S S cos M sn M S cos M Addtonal coents on rcton: Snce =, knetc frcton does not depend on the area of the surfaces n contact. (Ths s a surprsngly good rule of thub, but not an exact relaton. Do you see why??) By defnton, t ust be true that S K for any syste (thnk about t...). M S tan M Page 6

7 Model for Surface rcton The drecton of the frctonal force vector s perpendcular to the noral force vector, n the drecton opposng relatve oton of the two surfaces. Asde: Graph orctonal force vs Appled force: Knetc (sldng): The agntude of the frctonal force vector s proportonal to the agntude of the noral force. = K It oves, but t heats up the surface t oves on! = S = K Statc: The frctonal force balances the net appled forces such that the obect doesn t ove. The axu possble statc frctonal force s proportonal to. S and as long as ths s true, then = f A n opposte drecton It doesn t ove! = A A Exaple Deterne the acceleraton of the 2.0kg Block n the Dagra below (gven that the coeffcent of frcton s 0.5). Soluton Deterne the acceleraton of the 2.0kg Block n the Dagra below (gven that the coeffcent of frcton s 0.5). rst we note that snce all three blocks are connected, they all have the sae acceleraton agntude (ust dfferent drectons). Obvously the 3.0 kg ass beats the 1.0 kg ass so, t wll dctate the drecton of oveent. Let s draw a free body dagra, treatng the three obects as one (snce connected), and usng the rght drecton as postve. g left 1kg g frcton left 2kg 6 kg g rght 3kg Page 7

8 Soluton Proble: Box on Truck a rght left left frcton 6kg A box wth ass sts n the back of a truck. The coeffcent of statc frcton between the box and the truck s S. What s the axu acceleraton a that the truck can have wthout the box slppng? 3.0kg kg kg kg a kg kg kg kg a kg a a S a s Therefore the acceleraton of the 2.0 kg block s about 1.6 /s 2 to the rght. Proble: Box on Truck Draw ree Body Dagra for box: Consder case where s ax... (.e. f the acceleraton were any larger, the box would slp). Proble: Box on Truck Use ET = a for both and coponents S = a MAX = a MAX = S g a MAX = S = S Page 8

9 Lecture 7, Act 3 orces and Moton An nclned plane s acceleratng wth constant acceleraton a. A box restng on the plane s held n place by statc frcton. What s the drecton of the statc frctonal force? Lecture 7, Act 3 Soluton rst consder the case where the nclned plane s not acceleratng. S a f All the forces add up to zero! (a) (b) (c) Lecture 7, Act 3 Soluton If the nclned plane s acceleratng, the noral force decreases and the frctonal force ncreases, but the frctonal force stll ponts along the plane: Puttng on the brakes Ant-lock brakes work by akng sure the wheels roll wthout slppng. Ths axzes the frctonal force slowng the car snce S > K. a All the forces add up to a! = a The answer s (a) a Page 9

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