Introduction to Mechanics Applying Newton s Laws Friction

Transcription

1 Introduction to Mechanics Applying Newton s Laws Friction Lana heridan De Anza College Nov 9, 2017

2 Last time kinds of forces and problem solving objects accelerated together the Atwood machine and variants

3 Overview friction friction examples

4 Question You push on a heavy crate and moves it across the floor. However, even as you push it does not accelerate and if you stop pushing, the box stops moving. Why?

5 ome Types of Forces: Friction Friction is a resistive force. It opposes motion. Tiny defects in the surface of the floor and the crate catch on one another as the crate is pushed. Air resistance is another resistive force. 1 Figure from boundless.com

6 Friction There are actually two types of friction: kinetic (moving) static (stationary) Kinetic Friction fk n mg Motion F kinetic friction normal force f k = µ k n µ k is the coefficient of kinetic friction

7 ome types of forces Kinetic Friction The kinetic friction force always acts to oppose motion of the surfaces relative to each other. That means the kinetic friction F kf always points opposite to the velocity vector.

8 Friction tatic Friction n max. static friction normal force fs F f s µ s n µ s is the coefficient of static friction mg

9 Friction magnitude of the applied force. can breaks free and accelerates to the right. n n Motion fs F fk F mg mg a f b f s,max f s F f k m k n O tatic region Kinetic region F c

10 Friction Example 1 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull horizontally on a sled of mass 10 kg that is at rest initially. You exert a force of 5 N on the sled. What is the magnitude of the static frictional force that acts on the sled?

11 Friction Example 1 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull horizontally on a sled of mass 10 kg that is at rest initially. You exert a force of 5 N on the sled. What is the magnitude of the static frictional force that acts on the sled? F s,max = µ s n = µ s mg = (0.14)(10 kg)g = 13.7 N 13.7 N > 5 N, so the sled will remain at rest. If the sled is at rest and remains at rest, it does not accelerate. F net = 0.

12 Friction Example 1 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull horizontally on a sled of mass 10 kg that is at rest initially. You exert a force of 5 N on the sled. What is the magnitude of the static frictional force that acts on the sled? F s,max = µ s n = µ s mg = (0.14)(10 kg)g = 13.7 N 13.7 N > 5 N, so the sled will remain at rest. If the sled is at rest and remains at rest, it does not accelerate. F net = 0. F net,x = 0 = 5 N F s F s = 5 N (directed opposite to the pulling force)

13 Friction Example 2 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull horizontally on a sled of mass 10 kg that is at rest initially. How much force do you need to apply to get the sled moving? If you continue to apply that force, what will the magnitude of sled s acceleration be once it is moving? ketch. Hypothesis: 13.7 N, should be the max static friction force we just worked out; 1 m/s 2

14 Friction Example 2 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull horizontally on a sled of mass 10 kg that is at rest initially. How much force do you need to apply to get the sled moving? If you continue to apply that force, what will the magnitude of sled s acceleration be once it is moving? ketch. Hypothesis: 13.7 N, should be the max static friction force we just worked out; 1 m/s 2 To get the sled moving F app F sf F sf = µ s n = (0.14)(10 kg)g = 13.7 N

15 Friction Example 2 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull on a sled of mass 10 kg that is at rest initially. How much force do you need to apply to get the sled moving? If you continue to apply that force, what will the magnitude of sled s acceleration be once it is moving?

16 Friction Example 2 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull on a sled of mass 10 kg that is at rest initially. How much force do you need to apply to get the sled moving? If you continue to apply that force, what will the magnitude of sled s acceleration be once it is moving? F net,x = ma x F app F kf = µ k n = 13.7 (0.1)(10 kg)g = 3.92 N a = F m = 3.92 N 10 kg = 0.39 ms 2

17 Friction Example 2 For waxed wood on wet snow, µ s = 0.14 and µ k = 0.1. You pull on a sled of mass 10 kg that is at rest initially. How much force do you need to apply to get the sled moving? If you continue to apply that force, what will the magnitude of sled s acceleration be once it is moving? F net,x = ma x F app F kf = µ k n = 13.7 (0.1)(10 kg)g = 3.92 N a = F m = 3.92 N 10 kg = 0.39 ms 2 Reasonable?: Yes for the force. The acceleration was a bit less than my guess, but same order of magnitude.

18 Friction Example 6-2 A trained sea lion slides from rest with constant acceleration down a 3.0-m-long ramp into a pool of water. If the ramp is inclined at an angle of 23 above the horizontal and the coefficient of kinetic friction between the sea lion and the ramp is 0.26, how long does it take for the sea lion to make a splash in the pool? 0 Walker, Physics.

19 force vectors in such a case, we begin by resolving each vector into its x and y components. Friction Example 6-2 A trained sea lion slides from rest with constant acceleration down EXAMPLE 6 2 Making a Big plash a 3.0-m-long ramp into a pool of water. If the ramp is inclined at an angle of 23 above the horizontal and the coefficient of kinetic does it take for the sea lion to make a splash in the pool? friction between the sea lion and the ramp is 0.26, how long does Picture the Problem it take for the sea lion to make a splash in the pool? ketch: On an incline, align one axis the surface, and the other axi dicular to the surface. That wa is in the x direction. ince no m in the y direction, we know A trained sea lion slides from rest with constant acceleration down a 3.0-m-long ramp into a pool of water. If the ramp at an angle of 23 above the horizontal and the coefficient of kinetic friction between the sea lion and the ramp is 0.26, As is usual with inclined surfaces, we choose one axis to be parallel to the surface and the other to be perpendicular to sketch, the sea lion accelerates in the positive x direction 1a x 7 02, having started from rest, v 0x = 0. We are free to c initial position of the sea lion to be There is no motion in the y direction, and therefore Finally, we note free-body diagram that N! = NyN, f! x 0 = 0. and W! a y = 0. k = -m k NxN, = 1mg sin u2xn + 1-mg cos u2yn. +y f k N W +x Physical picture y x N f k mg sin mg cos W = mg Free-b diagra trategy We can use the kinematic equation relating position to time, x = x 0 + v 0x t a xt 2, to find the time of the sea lion s sli be necessary, however, to first determine the acceleration of the sea lion in the x direction, a x. To find a x we apply Newton s second law to the sea lion. First, we can find N by setting F y = ma y equal to zero (sinc 0 Walker, Physics.

20 force vectors in such a case, we begin by resolving each vector into its x and y components. Friction Example 6-2 king a Big plash A trained sea lion slides from rest with constant acceleration down EXAMPLE 6 2 Making a Big plash a 3.0-m-long ramp into a pool of water. If the ramp is inclined at an angle of 23 above the horizontal and the coefficient of kinetic does it take for the sea lion to make a splash in the pool? 1a friction between the sea x 7 02, v lion and the ramp is 0.26, 0x = 0. how long does Picture x the Problem 0 = 0. a y = 0. it take for the sea lion to make a splash in the pool? ketch: On an incline, align one axis the surface, and the other axi dicular to the surface. That wa is in the x direction. ince no m s from rest with constant acceleration down a 3.0-m-long ramp into a pool of water. If the ramp is inclin in the y direction, we know e the horizontal and the coefficient of kinetic friction between the sea lion and the ramp is 0.26, how lon lion to make a splash in the pool? A trained sea lion slides from rest with constant acceleration down a 3.0-m-long ramp into a pool of water. If the ramp ed surfaces, at we an angle choose of 23 one above axis the to horizontal be parallel and the to coefficient the surface of kinetic and the friction other between to be the perpendicular sea lion and the ramp to it. is In 0.26, o elerates in the positive x direction having started from rest, We are free to choose t ea lion to be There is no motion in the y direction, and therefore Finally, we note from t As is usual with inclined surfaces, we choose one axis to be parallel to the surface and the other to be perpendicular to t N! = NyN, f! sketch, the sea lion and accelerates W! k = -m k NxN, = 1mg in the sin positive u2xn + x direction 1-mg cos 1a x 7u2yN. 02, having started from rest, v 0x = 0. We are free to c initial position of the sea lion to be There is no motion in the y direction, and therefore Finally, we note free-body diagram that N! = NyN, f! x 0 = 0. and W! a y = 0. k = -m k NxN, = 1mg sin u2xn + 1-mg cos u2yn. +y +y W f k N +x Physical picture W +x N f N k f mg sin k mg sin y y Physical picture Free-bodyFree-b mg cos mg cos diagra W diagram = mg W = mg x x trategy We can use the kinematic equation relating position to time, x = x 0 + v 0x t a xt 2, to find the time of the sea lion s sli be necessary, however, to first determine the acceleration of the sea lion in the x direction, a x. tic equation 0 To Walker, relating find a x we position Physics. apply Newton s time, second x = law x 0 to + the vsea 0x t lion. + 1 First, a x t 2, we to can find find the N by time setting of the F y sea = ma lion s y equal slide. to zero It (sinc w

21 Friction Example 6-2 Hypothesis: About 5 seconds.

22 Friction Example 6-2 Hypothesis: About 5 seconds. trategy: Use Newton s 2nd law, find acceleration, use kinematics equation.

23 Friction Example 6-2 Hypothesis: About 5 seconds. trategy: Use Newton s 2nd law, find acceleration, use kinematics equation. F net,y = n mg cos θ = 0 n = mg cos θ F net,x = mg sin θ f k = ma

24 Friction Example 6-2 Hypothesis: About 5 seconds. trategy: Use Newton s 2nd law, find acceleration, use kinematics equation. F net,y = n mg cos θ = 0 n = mg cos θ F net,x = mg sin θ f k = ma mg sin θ µ k n = ma mg sin θ µ k (mg cos θ) = ma a = g(sin θ µ k cos θ)

25 Friction Example 6-2 Hypothesis: About 5 seconds. trategy: Use Newton s 2nd law, find acceleration, use kinematics equation. F net,y = n mg cos θ = 0 n = mg cos θ F net,x = mg sin θ f k = ma mg sin θ µ k n = ma mg sin θ µ k (mg cos θ) = ma a = g(sin θ µ k cos θ) a = 1.5 m/s 2

26 Friction Example 6-2 Given: x = 3 m, a = 1.5 m/s 2, v i = 0 m/s. Asked for: t.

27 Friction Example 6-2 Given: x = 3 m, a = 1.5 m/s 2, v i = 0 m/s. Asked for: t. x = v i t at2 t = 2.0 s

28 ummary friction 2nd Test due Thursday, Nov 16. Homework Walker Physics: Ch 6, onwards from page 177. Questions: 3, 15; Problems: 1, 3, 7, 11, 13, 15, 87