1) caused by the interaction of 2 + objects. 2) opposite (opposes) motion. 3) Types Kinetic, static, sliding, rolling

Transcription

1 Friction: 1) caused by the interaction of 2 + objects 2) opposite (opposes) motion 3) Types Kinetic, static, sliding, rolling 4) size determined by: nature of surfaces force pushing surfaces together frictional characteristic of the surfaces "coefficient of friction" force of friction = μ Normal Force force pushing surfaces together T to surface 1

2 "μ" is the "coefficient of friction" and is the frictional property of the two surfaces in contact. Each pair of surfaces has its own unique value of µ. Note that µ is a ratio of forces, and therefore a pure number. μ = / is parallel to surface is perpendicular to surface force of 125 N. What is 1) data? 2

3 force of 125 N. What is Plane of motion: Motion is along the horizontal With the question of Type Motion and Type force you have to consider planes parallel and perpendicular to motion. 2) Type motion? 3) Type force? force of 125 N. What is horizontal (x) vertical (y) as the box slides across the floor in is not moving up or down, therefore it is at rest in the "y" (perpendicular) axis 1 st Law rest 1 st Law 4) picture? (diagram) 3

4 force of 125 N. What is horizontal (x) vertical (y) 1 st Law rest 1 st Law 5) force diagram force of 125 N. What is horizontal (x) vertical (y) 1 st Law rest 1 st Law F h = 125 N the box is sliding right, there is left!!! 6) Determine any forces you don't know 4

5 force of 125 N. What is mbox = 25 kg fh = 125 N + F h = 0 = F h vc horizontal (x) vertical (y) vc 1 st Law rest horizontal (x) vertical (y) + = 0 = 125 N 245 N 245 N F h = 125 N = mg = 25 kg ( 9.8 m/s 2 ) = 245 N = (125 N) = 125 N = (mg) = ( 245 N) = (245 N) 7) Find unknown force of 125 N. What is + F h = 0 = F h mbox = 25 kg fh = 125 N vc horizontal (x) vertical (y) vc 1 st Law rest horizontal (x) = (125 N) = 125 N vertical (y) + = 0 = = (mg) = ( 245 N) = (245 N) 125 N 245 N 245 N F h = 125 N µ = / = mg = 25 kg ( 9.8 m/s 2 ) = 245 N µ = 125 N/245 N µ = The " " in µ is meaningless because it is a vector direction and µ is not a vector it is a pure number, therefore it is dropped 7) Find unknown 5

6 You pull a 25.0 kg box across a waxed floor with an acceleration of 2.50 m/s 2 with a horizontal force of 125 N. What is You pull a 25.0 kg box across a waxed floor with an acceleration of 2.50 m/s 2 with a horizontal force of 125 N. What is a = 2.50 m/s 2 horizontal vertical a 2 nd Law rest ΣF = ma + F h = ma = ma F h = 25.0 kg(2.50 m/s 2 ) (125 N) = 62.5 N (125 N) = 62.5 N + = 0 = ( ) = ( 245 N) = 245 N 245 N 245 N F h = 125 N 6

7 You pull a 25.0 kg box across a waxed floor with an acceleration of 2.50 m/s 2 with a horizontal force of 125 N. What is a = 2.50 m/s 2 horizontal a 2 nd Law rest ΣF = ma µ = / µ = 62.5 N/245 N µ = N vertical = 62.5 N 245 N F h = 125 N \ You pull a 25.0 kg box across a floor at a consant velocity with a force of 125 N directed above the horizontal. What is 7

8 You pull a 25.0 kg box across a floor at a consant velocity with a force of 125 N directed above the horizontal. What is f a = 125 N at horizontal vertical 1 st Law rest parallel to motion µ = / F a = 125 N F a isn't parallel or perpendicular to motion, it's both, so break it down into its parallel and perpendicular components. perpendicular to motion break down (resolve) all forces into their parallel and perpendicular components!!!!! perpendicular plane of motion parallel F a = 125 N plane of motion perpendicular 8

9 parallel perpendicular F ay F a = 125 N F ax perpendicular F ay = sin (125 N) = N Fax = cos (125 N) = +108 N parallel perpendicular Now that you've broken the angled 125 N force into its "x" and "y" (horizontal and vertical) components you don't use it (the 125 N force), you just use its components!!! You pull a 25.0 kg box across a floor at a consant velocity with a force of 125 N directed above the horizontal. What is f a = 125 N at = mg = 25 kg ( 9.8 m/s 2 ) = 245 N horizontal vertical 1 st Law rest + F ax = 0 = F ax = (108N) = 108 N F w + + F ay = 0 = F ay = 108 N = ( 245 N) (62.5N) = N µ = / 245 N µ = 108 N/182.5 N µ = F a = 125 N F ax = +108 N F ay = N 9

10 You pull a 25.0 kg box across a waxed floor with an acceleration of 2.00 m/s 2 with a force of 125 N directed above the horizontal. What is You pull a 25.0 kg box across a waxed floor with an acceleration of 2.00 m/s 2 with a force of 125 N directed above the horizontal. What is f a = 125 N at = mg acceleration = 25 kg ( 9.8 m/s 2 ) = 245 N horizontal vertical a 2 nd Law rest Σ F = 0 ΣF = ma F f + F ax = ma + + F ay = 0 = ma F = F ay ax = 108 N F ax = +108 N = ( 245 N) (62.5N) = N = 25.0 kg(2.00 m/s 2 ) (108N) = 58 N µ = / 245 N µ = 58 N/182.5 N µ = 0.32 F a = 125 N F ay = N 10

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