Chapter 4 Newton s Laws. September 24 and September 29, 2009

Transcription

1 Chapter 4 Newton s Laws September 24 and September 29, 2009

2 First exam for Physics 201: Tuesday, September 29, :45 7:00 Bascom 272 Review sessions: Saturday 9/26, 10am-12pm, 2241 Chamberlin (Jialu) Saturday 9/26, 2pm-4pm, 2241 Chamberlin (Jared) Saturday 9/26, 6pm-8pm, 1121 Humanities (Andrew) Sunday 9/27, 2pm-4pm, 2241 Chamberlin (Eunsong) Monday 9/28, 8pm - 10pm, 2241 Chamberlin (Daniel) practice questions are available on course web page: follow link from

3 AcceleraJon in Uniform Circular MoJon Magnitude of accelerajon in uniform circular mojon is a = v2 R Derivation I: Say it takes time T to go around a circle of radius R. The travel distance in one revolution is 2πR. Magnitude of velocity v=2πr/t. (Note: this implies T=2πR/v.) Velocity sweeps around one revolution in same time T, so total change in velocity in one revolution is 2πv. So magnitude of acceleration a is: a = 2πv T = 2πv ( 2πR / v) = v2 R direction is always inward along radius

4 AcceleraJon in Uniform Circular MoJon Magnitude of accelerajon in uniform circular mojon is a = v2 R Derivation II: r(t) δθ r(t + δt) v(t) v(t + δt) v(t + δt) v(t) is perpendicular to r(t). v(t + δt) is perpendicular to r(t + δt). v(t) So the angle between r(t) and r(t + δt) is the same as the angle between v(t) and v(t + δt). δθ

5 AcceleraJon in Uniform Circular MoJon Magnitude of accelerajon in uniform circular mojon is a = v2 R Derivation II (continued): v(t + δt) v(t) δ v = vδθ a δv δt = v δθ δt v dθ dt = vω = v2 R.

6 QuesJon An object apached to a string is rotajng around an axis with a constant speed in a plane, horizontal with respect to the Earth's surface. Its accelerajon is: A. zero B. constant, poinjng to the axis of rotajon C. constant, poinjng away from the axis of rotajon E. conjnuously varying

7 QuesJon An object apached to a string is rotajng around an axis with a constant speed in a plane, horizontal with respect to the Earth's surface. Its accelerajon is: A. zero B. constant, poinjng to the axis of rotajon correct C. constant, poinjng away from the axis of rotajon E. conjnuously varying

8 Newton s first law: if the net force on an object is zero, its velocity is constant. First Law. An object at rest stays at rest unless acted on by an external force. An object in motion continues to travel with constant speed in a straight line unless acted on by an external force. NEWTON S FIRST LAW Another way to say the same thing: No net force velocity is constant acceleration is zero

9 Newton s first law

10 Is Madison a good inerjal reference frame? Is Madison accelerating? YES! Madison is on the earth. The earth is rotating. T = 1 day = 8.64 x 10 4 sec, R ~ R E = 6.4 x 10 6 meters. What is the centripetal acceleration of Madison? a M = v2 R = ω2 R = 2π T Plug this in: a M = 0.34 m/s 2 ( ~ 1/300 g) This is close enough to zero that we will ignore it. Madison is a pretty good inertial reference frame. 2 R

11 A force is defined as an external influence or action on an object that causes the object to change velocity relative to an inertial reference frame. Contact and Field (AcJon At A Distance) Forces

12 The Four Fundamental Forces Types Strong nuclear force Electromagnetic force Weak nuclear force Gravity Characteristics All field forces Listed in order of decreasing strength Only gravity and electromagnetic forces are relevant in classical mechanics, which deals with motion of macroscopic objects.

13 Combining Forces Force is a vector. If two or more forces simultaneously act on an object, the net force is equal to the vector sum of the individual forces.

14 QuesJon 1 lift drag thrust weight

15 QuesJon 1 lift correct drag thrust weight Velocity is constant implies that there is no net force acting on it. Therefore, all components of the net force are zero. Forces are balanced in all three dimensions.

16 Mass or InerJa Mass (m) is the property of an object that measures how hard it is to change its velocity. Units: [M] = kg Importance of mass Braking car versus train Car versus bus going around curve Push on light or heavy object Mass determines how much force is exerted or is needed to achieve desired motion.

17 Newton s Second Law This law tells us how the motion of an object is affected when a force is applied. In words. acceleration = (net force)/mass in symbols: F = M a net

18 Newton s Second Law Units: [F] = [M] [a] [F] = kg-m/s 2 1 Newton (N) 1 kg-m/s 2 A vector equation:» F net,x = Ma x» F net,y = Ma y

19 Newton s Second Law F net = F i i = m a

20 Force due to gravity: weight Near the earth s surface, the magnitude of g, the acceleration due to the earth s gravitational field, is and is directed towards the earth s center. The acceleration due to gravity is the same for all objects.

21 Weight: gravitajonal force on object Mass (or inerja): intrinsic property of object If you stand on the moon, your weight is 1/6 th of the value that it has on earth, but your mass is the same as on earth (or in space, or anywhere).

22 Contact Forces: Solids, Springs, and Strings Contact forces exerted by solids: Normal force Frictional force Normal force

23 The Normal Force When holding the bag above the table, the person must exert a force on the bag. When the bag is placed on the table, the table supplies the force that holds the bag on it That force is perpendicular or normal to the surface of table

24 Springs Hooke s Law: F x = -kx

25 Strings A string can pull but not push. The magnitude of the force that one segment of the string exerts on the adjacent segment is called tension (usually denoted by T). Forces on gymnast: T left T right Mg Since gymnast is not accelerating, T left +T right = Mg

26 Tension Tension is a force along the length of a medium Tension can be transmitted around corners If there is no friction in the pulleys, T remains the same

27 Free Body Diagrams Free-body diagrams are diagrams of the forces on an object. First, isolate the object in question. Then, identify the forces on it. Forces acting on sled: 1. The gravitational force on the sledrope 2. The contact force exerted by the ice on the runners. (Without friction, the contact force is directed normal to the ice.) 3. The contact force exerted by the dog on the rope. (Since the sled remains on the ice, the y-components of the force sum to zero.)

28 Example: Hanging a Picture Free-body diagram for the picture Picture is not accelerating Forces on picture sum to zero. cos(30 ) x component : T 1 cos(30 ) - T 2 cos(60 ) = 0 T 2 = T 1 cos(60 ) y component : T 1 sin(30 ) + T 2 sin(60 ) - mg = 0 T 1 = mg sin30 + cos30 cos60 sin60 = mg / 2 1/ = mg 2 ; T 2 = T 1 cos(30 ) cos(60 ) = 3T 1

29 Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is a) larger than b) identical to c) less than QuesJon 2 the downward weight W of the person.

30 Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is a) larger than b) identical to c) less than QuesJon 2 the downward weight W of the person. N mg Person is accelerating upwards - net upwards force is non zero

31 Newton s Third Law For every acjon, there is an equal and opposite reacjon. F finger box Finger pushes on box F finger box = force exerted on box by finger F box finger Box pushes on finger F box finger = force exerted on finger by box Third Law: F box finger = - F finger box

32 Newton s Third Law Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the one it exerts.

33 System of Interest Newton s second law applies to every system of interest. System 1: Acceleration of the professor and the cart System 2: Force the professor exerts on the cart f - All forces opposing the motion

34 QuesJon 3 Consider a car at rest. We can conclude that the downward gravitational pull of Earth on the car and the upward contact force of Earth on it are equal and opposite because A. The two forces form an action-reaction pair B. The net force on the car is zero C. Neither of the above

35 QuesJon 3 Consider a car at rest. We can conclude that the downward gravitational pull of Earth on the car and the upward contact force of Earth on it are equal and opposite because A. The two forces form an action-reaction pair B. The net force on the car is zero C. Neither of the above The two forces cannot be an action-reaction pair because they act on the same object (car). Car is at rest - therefore, it has no net forces acting on it. The forces acting on it add up to zero

36 QuesJon 4: Pulley I What is the tension in the string? A) T<mg B) T=mg C) mg<t<2mg m m g

37 QuesJon 4: Pulley I What is the tension in the string? A) T<mg B) T=mg g C) mg<t<2mg m m Forces on weight on left: T Forces on weight on right: T mg mg accelerations are equal and opposite T-mg = mg-t T=mg

38 QuesJon 5: Pulley II What is the tension in the string (assume the pulley is fricjonless)? A) T<mg B) T=mg C) mg<t<2mg 2m m

39 QuesJon 5: Pulley II What is the tension in the string (assume the pulley is fricjonless)? A) T<mg B) T=mg C) mg<t<2mg acceleration a is same for both blocks a 2m m a 2ma = T 2mg ma = mg T (note tricky sign!) a = mg / 3 and T = 4mg/3.

40 Solving Problems IdenJfy force using Free Body Diagram This is the most important step! Set up axes and origin x and y Write F net =ma for each axis (components of forces) Calculate accelerajon components Setup kinemajc equajons Solve! Strong suggesjon:» work problem algebraically (using symbols)» plug in numbers only at the end

41 Example 1 µ k = 0.2 v 0 =8 m/s Find stopping distance

42 Example 1 µ k = 0.2 v 0 =8 m/s µ k N N Mg Find stopping distance Normal force is balanced by gravity because there is no vertical motion: N = Mg, if M is the mass of the object Kinetic frictional force that decelerates the block is, f = µ k N = µ k Mg Therefore, deceleration (direction opposite of v 0 ), a = -f/m = -µ k g Given deceleration, use kinematics equation to obtain the answer. v 2 = v aΔx Δx = v 2 v 0 2 2a = v 2 v 0 2 2µ k g = m =16.3 m Answer: 16.3 m - is independent of the mass

43 Example 2 µ k = 0.2 T=50 N M = 5 kg Find acceleration of block

44 Example 2 µ k = 0.2 T=50 N M = 5 kg Find acceleration of block Frictional force: f = µ k Mg opposes the tension T Net force: F net = T - f = T - µ k Mg Acceleration: a = F net / M = (( x 5 x 9.8) / 5) m/s 2 Answer: 8.04 m/s 2

45 Example 3 T=50 N µ k = 0.2 θ=50 0 M = 5 kg Find acceleration of block

46 Example 3 T=50 N µ k = 0.2 θ=50 0 M = 5 kg Find acceleration of block Hints: Resolve T in x and y components: T x =T sinθ, T y =T cosθ Draw free body diagram Solve for y-component of force, and note that y-acceleration is zero (Obtain relationship between T and N) Solve for x-component of force, then use a x =F x /m a = F net M = T x µ k N M = T x µ k ( Mg T y ) M = T cosθ µ k Mg T sinθ ( ) M Answer: acceleration of block is 6.0 m/s 2 in +x direction