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Friction & Applying Newton’s 2 nd Law Chapter 6.2 System.

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Presentation on theme: "Friction & Applying Newton’s 2 nd Law Chapter 6.2 System."— Presentation transcript:

1 Friction & Applying Newton’s 2 nd Law Chapter 6.2 System

2 What is Friction? Friction is a force that is parallel to the surfaces of two objects that are in contact with one another that resists the relative motion of the two objects. FfFf

3 What causes friction? Friction is caused by the temporary electrostatic bonds created between two objects in contact with one another.

4 Friction How does friction affect the motion of objects? –It can slow an object down like the friction between the tires and the road. –It is responsible for increasing the speed of an object like a car. –It is also responsible for objects being able to change direction.

5 Static Friction (Stationary Objects) System F applied F friction F ground-on-crate F gravity F applied F friction F net = F applied – F friction Since the crate is not accelerating, F net = 0 F applied = F friction Note: As long as the crate does not move, F applied = F friction Static Friction: –The resistive force that keeps an object from moving. –Since v = 0, a = 0.

6 Kinetic Friction (Objects in Motion) System F applied F friction F ground-on-crate F gravity Kinetic Friction: –The resistive force that opposes the relative motion of two contacting surfaces that are moving past one another. –F net may or may not be 0. F applied F friction F net = F applied – F friction F net Note: If the crate moves at a constant velocity, then F applied = F friction and F net = 0.

7 Determining the Frictional Force (The Coefficient of Friction,  ) The force of friction (F f ) is proportional to the normal force (F N ) and a proportionality constant (  - pronounced mu) called the coefficient of friction. For static friction: –0 < F f, static <  s F N For kinetic friction: –F f, kinetic =  k F N As per the formula, the greater , the greater the frictional force. Note: F N = the force normal (perpendicular) to the frictional force on the object.  is dimensionless (NO UNITS) F f, static > F f, kinetic FfFf FNFN

8 The Normal Force The normal force is a force that most often opposes the Earth’s gravitational attraction and is perpendicular to the surface that an object rests or is moving on. –For a horizontal surface, F N = F g = mg. –For a surface that is not horizontal, as seen in the figure below, F N = F g cos  FNFN 

9 The Normal Force FNFN FgFg FNFN  FgFg F N = F g = mg cos  = adj/hyp F N = F g cos  = mg cos 

10 Example 2: Determining Friction (Balanced Forces) Assume that the man in the figure is pushing a 25 kg wooden crate across a wooden floor at a constant speed of 1 m/s. –How much force is exerted on the crate? System F applied FfFf FNFN FgFg

11 Diagram the Problem System F applied FfFf FNFN FgFg FfFf FNFN FgFg y-direction: F N = F g x-direction: F net = F applied - F f Since the crate is moving with constant speed, a = 0, F net = 0, and F applied = F f +y +x

12 State the Known and Unknowns What is known? oMass (m) = 25 kg oSpeed = 1 m/s oAcceleration (a) = 0 m/s 2 o  k = 0.3 (wood on wood from reference table) What is not known? oF applied = ?

13 Perform Calculations y-direction: oF g = F N = mg(Note: horizontal surface) x-direction: Since a = 0, F net = 0 oF net = F applied – F f oF applied = F f oF applied =  k F N F applied =  k mg oF applied = (0.3)(25 kg)(9.8 m/s 2 ) oF applied = 73.5 N 0

14 Example 3: Determining Friction (Unbalanced Forces) Assume that the man in the figure is pushing a 25 kg wooden crate across a wooden floor at a speed of 1 m/s with a force of 73.5 N. –If he doubled the force on the crate, what would the acceleration be? System F applied FfFf FNFN FgFg

15 Diagram the Problem System F applied FfFf FNFN FgFg FfFf FNFN FgFg y-direction: F N = F g x-direction: Since a > 0, F net = F applied - F f +y +x

16 State the Known and Unknowns What is known? oForce = 147 N oMass (m) = 25 kg oSpeed = 1 m/s o  k = 0.3 (wood on wood) What is not known? oa ?

17 Perform Calculations y-direction: oF g = F N = mg x-direction: a > 0 oF net = F applied – F f oma = F applied – F f oa = 147N – 73.5N 25kg oa = 2.9 m/s 2

18 Key Ideas Friction is an opposing force that exists between two bodies. Friction is proportional to the normal force and the coefficient of friction; static or kinetic. The force required to overcome static friction is greater than that required to overcome kinetic friction.


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