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FORCE. Newton’s Laws Three Laws of Motion Aristotle’s Motion Natural Motion is up or down Natural Motion is up or down  Down for falling objects  Up.

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Presentation on theme: "FORCE. Newton’s Laws Three Laws of Motion Aristotle’s Motion Natural Motion is up or down Natural Motion is up or down  Down for falling objects  Up."— Presentation transcript:

1 FORCE

2 Newton’s Laws Three Laws of Motion

3 Aristotle’s Motion Natural Motion is up or down Natural Motion is up or down  Down for falling objects  Up for smoke  Circular for heavenly bodies since without end Violent Motion Violent Motion  Due to imposed forces such as wind pushing a ship or someone pulling a cart Natural state of motion is rest Natural state of motion is rest  A force is needed to keep something moving

4 Aristotle’s Basic Error Friction not understood as a force Friction not understood as a force

5 Galileo’s Motion Force is a push or a pull Force is a push or a pull Friction is a force that occurs when objects move past each other Friction is a force that occurs when objects move past each other Friction due to tiny irregularities Friction due to tiny irregularities Only when friction is present is a force required to keep something moving Only when friction is present is a force required to keep something moving

6 Galileo’s Inclined Planes Ball rolling downhill speeds up Ball rolling downhill speeds up Ball rolling uphill slows down Ball rolling uphill slows down He asked about ball on smooth level surface He asked about ball on smooth level surface Concluded it would roll forever in absence of friction Concluded it would roll forever in absence of friction

7 Inertia Resistance to change in state of motion Resistance to change in state of motion Galileo concluded all objects have inertia Galileo concluded all objects have inertia Contradicted Aristotle’s theory of motion Contradicted Aristotle’s theory of motion No force required to keep Earth in motion around sun because no friction No force required to keep Earth in motion around sun because no friction

8 Newton Born 1665 Born 1665 Built on Galileo’s ideas Built on Galileo’s ideas Proposed three laws of motion at age of 23 Proposed three laws of motion at age of 23

9 Newton’s First Law Every object continues in its state of rest, or of motion in a straight line at constant speed, unless compelled to change that state by forces exerted on it. Every object continues in its state of rest, or of motion in a straight line at constant speed, unless compelled to change that state by forces exerted on it. Also called Law of Inertia: things move according to their own inertia Also called Law of Inertia: things move according to their own inertia Things keep on doing what they are doing Things keep on doing what they are doing Examples: Hockey puck on ice, rolling ball, ball in space, person sitting on couch Examples: Hockey puck on ice, rolling ball, ball in space, person sitting on couch Ourtesy hockey.gif

10 Mass Amount of inertia depends on amount of mass…or amount of material (number and kind of atoms) Amount of inertia depends on amount of mass…or amount of material (number and kind of atoms) Measured in kilograms Measured in kilograms Question: Which has more mass, a kilogram of lead or a kilogram of feathers? Question: Which has more mass, a kilogram of lead or a kilogram of feathers? Mass vs. Volume: volume is how much space something occupies Mass vs. Volume: volume is how much space something occupies

11 Experiencing Inertia Inertia is resistance to shaking Inertia is resistance to shaking Which is easier to shake, a pen or a person? Which is easier to shake, a pen or a person? Why is it so hard to stop a heavy boat? Why is it so hard to stop a heavy boat?

12 Inertia in a Car Discuss three examples of inertia in a car Discuss three examples of inertia in a car Car hitting a wall Car hit from behind by a truck Car going around a corner

13 Newton’s Second Law Law of Acceleration Law of Acceleration The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force, and is inversely proportional to the mass of the body. The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force, and is inversely proportional to the mass of the body. Acceleration = net force ÷mass Acceleration = net force ÷mass F =ma F =ma Acceleration is in direction of net force Acceleration is in direction of net force

14 Units F = ma F = ma Unit of force is the Newton (N) Unit of force is the Newton (N) 1 N = 1 kg m/s 2 1 N = 1 kg m/s 2

15 Net Force Net Force means sum of all forces acting Net Force means sum of all forces acting Sum is Vector sum Sum is Vector sum F1F1 F2F2 Resultant force

16 Understanding the Second Law The cause of acceleration is… The cause of acceleration is… _________ resists acceleration _________ resists acceleration The greater the force, the ________ the The greater the force, the ________ the ______________ ______________ The greater the mass, the _________ the acceleration. The greater the mass, the _________ the acceleration. Force Mass or inertia acceleration less greater

17 F = ma is Three Equations F and a are vectors F and a are vectors So F = ma equation is really three So F = ma equation is really three  F x = ma x  F y = ma y  F z = ma z  F x = ma x  F y = ma y  F z = ma z

18 Examples What force is required to accelerate a 1000 kg car at 2.0 m/s 2 ? What force is required to accelerate a 1000 kg car at 2.0 m/s 2 ? Answer: F = ma = 1000 kg x 2.0 m/s 2 = 2000 N. Answer: F = ma = 1000 kg x 2.0 m/s 2 = 2000 N. What is the acceleration of a 145 g baseball thrown with a force of 20.0 N? What is the acceleration of a 145 g baseball thrown with a force of 20.0 N? a = F/m = 20.0 N/0.145kg = 138 m/s 2 a = F/m = 20.0 N/0.145kg = 138 m/s 2

19 F = ma Example; m unknown An astronaut puts a N force on an object of unknown mass producing an accelerations of m/s 2. What was the mass? An astronaut puts a N force on an object of unknown mass producing an accelerations of m/s 2. What was the mass? M = F/a = 500.0N/0.462 m/s 2 = 1082 Kg = 1.08 x 10 3 Kg M = F/a = 500.0N/0.462 m/s 2 = 1082 Kg = 1.08 x 10 3 Kg

20 Net force example If four teams are playing tug of war (imagine a rope that looks like a cross, with the flag tied in the middle). Each team is 90 ⁰ from each other. Team A pulls with an overall force of 350 N to the North, Team B pulls with an overall force of 270 N to the South, Team C pulls with an overall force of 150 N to the East and Team D pulls with an overall force of 250 N to the West. If the flag in the middle has a mass of.25 kg, what is the magnitude and direction of its acceleration?

21 Putting it all together……. Calculate the change in force of a car that has a mass of 2500 kg if it goes from 45 m/s to rest in 7 seconds at a stop sign, then accelerates up to 65 m/s in 5 seconds.

22 a= v f -v i /t or a = F/m a 1 = 0-45/7 = m/s 2 a 2 = 65-0/5 = 13 m/s 2 The difference between them is m/s 2. F = m x a = 2500 kg x m/s 2 = N difference between the two accelerations

23 Newton’s Third Law Forces always come in pairs Forces always come in pairs Two forces on different objects Two forces on different objects Every action has an equal and opposite reaction Every action has an equal and opposite reaction Whenever one object exerts a force on a second object, the second exerts an equal and opposite force on the first Whenever one object exerts a force on a second object, the second exerts an equal and opposite force on the first Example: hammer hits nail Example: hammer hits nail

24 Example: pushing on wall What are the forces when you push on a wall? What are the forces when you push on a wall? You exert force on wall You exert force on wall You accelerate in the opposite direction You accelerate in the opposite direction Wall must have exerted a force on you in the direction you accelerated (by 2 nd Law) Wall must have exerted a force on you in the direction you accelerated (by 2 nd Law)

25 Example: person walking Foot exerts force backward on ground Foot exerts force backward on ground Ground exerts force forward on foot Ground exerts force forward on foot

26 Example: Throwing ball Pitcher exerts force on ball Pitcher exerts force on ball Ball exerts equal and opposite force on pitcher Ball exerts equal and opposite force on pitcher Why doesn’t pitcher move? Why doesn’t pitcher move?

27 Example: Rocket Rocket engine exerts rearward force on gas molecules Rocket engine exerts rearward force on gas molecules Molecules exert forward force on rocket. Molecules exert forward force on rocket.

28 Book on Table The mass of the book is one kg. What is the force (magnitude and direction) on the book? The mass of the book is one kg. What is the force (magnitude and direction) on the book? 9.8 N upward 9.8 N upward

29 Really putting it all together…… Calculate the Force necessary to launch a cannonball with a mass of 15 kg if it is fired at an angle of 43 ⁰ if it hits a target 210 m away in 6.3 seconds? What can we solve in this problem? What equations do we need to solve this problem?

30 What we need to solve the force Vx = dx/t = 210/6.3 = 33.3 m/s Vf 2 = Vi a(d) Vi = 0 for this problem a = Vf 2 /2d = / 2(210) = 2.64 m/s 2 Force of the cannon: F = m(a) F = 15 kg (2.64 m/s 2 ) = 39.6 N

31 The Horse and the Cart Problem If there is always an equal an opposite reaction, how does anything move? For example, if you have a horse and a cart, how does the horse pull the cart? If there is always an equal an opposite reaction, how does anything move? For example, if you have a horse and a cart, how does the horse pull the cart?

32 The Horse and Cart Problem. A= - B B= - C C= -D A=B=C=D no acc! These appear to be the equalizing forces. These appear to be the equalizing forces.

33 The Horse and Cart Problem. Because it is accelerating, the force the horse exerts on the cart has increased. By Newton's third law, the force of the cart on the horse has increased by the same amount. But the horse is also accelerating, so the friction of the ground on its hooves must be larger than the force the cart exerts on the horse. The friction between hooves and ground is static (not sliding or rolling) friction, and can increase as necessary (up to a limit, when slipping might occur, as on a slippery mud surface or loose gravel). So, when accelerating, we still have B = -C, by Newton's third law, but D>C and B>A, so D>A.

34 More Examples Can you think of some more examples of Newton’s Third Law in Action? Can you think of some more examples of Newton’s Third Law in Action? Imagine an astronaut floating in deep space, with only his spacesuit. Is there any way for him to move himself back to earth? Imagine an astronaut floating in deep space, with only his spacesuit. Is there any way for him to move himself back to earth?

35 Mass vs. Weight Mass is intrinsic property of any object Mass is intrinsic property of any object Weight measures gravitational force on an object, usually due to a planet Weight measures gravitational force on an object, usually due to a planet Weight depends on location of object Weight depends on location of object Question 1: How does mass of a rock compare when on Earth and on moon? Question 1: How does mass of a rock compare when on Earth and on moon? Question 2: How does its weight compare? Question 2: How does its weight compare?

36 Review Mass vs. Weight What is mass? What is mass? Answer: quantity of matter in something or a measure of its inertia Answer: quantity of matter in something or a measure of its inertia What is weight? What is weight? Answer: Force on a body due to gravity Answer: Force on a body due to gravity

37 Weight of 1 Kilogram 9.8 Newtons 9.8 Newtons About 2.2 pounds About 2.2 pounds Compare the weight of 1 kg nails with 1 kg styrofoam Compare the weight of 1 kg nails with 1 kg styrofoam Answer: Same Answer: Same

38 Weight Examples What does a 70 kg person weigh? What does a 70 kg person weigh? Weight = mass x g(acceleration due to gravity) Weight = mass x g(acceleration due to gravity) W = mg = 70 kg x 9.80 N/m 2 = 686 N W = mg = 70 kg x 9.80 N/m 2 = 686 N An object weighs 9800 N on Earth. What is its mass? An object weighs 9800 N on Earth. What is its mass? m = W/g = 9800 / 9.8 m/s 2 = 1000 kg m = W/g = 9800 / 9.8 m/s 2 = 1000 kg

39 Atwoods Lab You have 25 washers on your lab setup, if you have a unbalanced force, you will have acceleration. You will be using the stopwatch function of your data collector. You have 25 washers on your lab setup, if you have a unbalanced force, you will have acceleration. You will be using the stopwatch function of your data collector. Make a chart to record mass, time, acceleration and force. Make a chart to record mass, time, acceleration and force. Put all washers on one side, raise that side to the top, then release it timing how long it takes to reach the bottom. Record this time. Put all washers on one side, raise that side to the top, then release it timing how long it takes to reach the bottom. Record this time. The mass of one washer is 16 g. It is the difference in mass that causes the acceleration. Calculate the difference in mass and record in table. 1 st mass is 25 x 16, 2 nd mass is 23 x 16, 3 rd mass is 21 x 16 etc. The mass of one washer is 16 g. It is the difference in mass that causes the acceleration. Calculate the difference in mass and record in table. 1 st mass is 25 x 16, 2 nd mass is 23 x 16, 3 rd mass is 21 x 16 etc. Calculate the Acceleration = 2d/t 2 (d = 1 m for the fall) so a = 2/ t 2 Calculate the Acceleration = 2d/t 2 (d = 1 m for the fall) so a = 2/ t 2 Calculate the Net force of the fall and record. (F= ma) Calculate the Net force of the fall and record. (F= ma) Move one washer at a time over to the other side and repeat. Move one washer at a time over to the other side and repeat. Continue until the machine no longer turns (12 or 13 trials) Continue until the machine no longer turns (12 or 13 trials)

40 FRICTION Sliding (motion) & Static (stationary)

41 Sliding Friction Often called kinetic friction Often called kinetic friction A force opposite to direction of motion A force opposite to direction of motion Due to bumps in surfaces and electric forces Due to bumps in surfaces and electric forces FfFf Surface under microscope

42 Kinetic Friction is… Dependent on nature of the two surfaces Dependent on nature of the two surfaces Directly proportional to the normal force between the surfaces Directly proportional to the normal force between the surfaces Normal Force is perpendicular to the surface. If it is on a flat surface, it is equal to the weight of the object. Normal Force is perpendicular to the surface. If it is on a flat surface, it is equal to the weight of the object. Independent of velocity Independent of velocity

43 Reducing Friction In order to reduce friction we can: In order to reduce friction we can: A. Reduce surface area A. Reduce surface area B. Reduce weight of object B. Reduce weight of object C. Change type of friction C. Change type of friction  - sliding(the greatest amount)  - rolling (use wheels to ease friction)  - fluid ( Eliminate contact by using liquids or gases)

44 Coefficient of friction  k Generally between zero and one Generally between zero and one Based on comparing Friction Force to Normal Force Based on comparing Friction Force to Normal Force Normal Force is always perpendicular to surface Normal Force is always perpendicular to surface Calculate from F f / F N = µ k Calculate from F f / F N = µ k Can be more than one for special rubber Can be more than one for special rubber Very low for ice, Teflon, lubricated surfaces, ball bearings Very low for ice, Teflon, lubricated surfaces, ball bearings

45 Friction: Good or Bad Mostly undesirable since reduces useful force and wastes energy Mostly undesirable since reduces useful force and wastes energy Friction produces heat Friction produces heat Necessary for walking! Necessary for walking! Necessary for braking Necessary for braking

46 Static Friction Force to start something moving Force to start something moving Usually larger than kinetic friction for same surfaces Usually larger than kinetic friction for same surfaces Requires force to be exerted Requires force to be exerted Before sliding begins, is equal and opposite to applied force Before sliding begins, is equal and opposite to applied force

47 Where are all the forces? Block on an inclined plane Block on an inclined plane

48 Free Body Diagram Example 1 If the box below accelerates to the right at 1 m/s 2 Solve all of the following:

49 Solution 1 Fgrav = m x g = 5 x 9.8 = 49 N Fgrav = m x g = 5 x 9.8 = 49 N Using the angle and the F applied, we can calculate the X and Y component of that force. Using the angle and the F applied, we can calculate the X and Y component of that force. Fx= 15 sin 45 Fy = 15 cos 45 Fx= 15 sin 45 Fy = 15 cos 45 Fx = 10.6 N Fy = 10.6 N Fx = 10.6 N Fy = 10.6 N If the force of gravity is 49 N down and the applied force is 10.6 N up, then the normal force applied is the difference between the two. F norm= = 38.4 N If the force of gravity is 49 N down and the applied force is 10.6 N up, then the normal force applied is the difference between the two. F norm= = 38.4 N

50 Solution 1 cont. If the object has an a of 1 m/s 2 and a mass of 5 kg, then it has a net force of 5 N in the X direction. If the object has an a of 1 m/s 2 and a mass of 5 kg, then it has a net force of 5 N in the X direction. If the applied force in the X is 10.6 and the net is 5, then the force of friction is the difference between the two. If the applied force in the X is 10.6 and the net is 5, then the force of friction is the difference between the two. Ffric= = 5.6 N Ffric= = 5.6 N To solve the coefficient of friction we use this equation: F f =  k F N To solve the coefficient of friction we use this equation: F f =  k F N  k = F f /F N = 5.6/ 38.4 =.145

51 Flat pull If you pull a 2505 g box with a force of 15 N at an angle of 53 ⁰ to the horizon and the box accelerates at 2.0 m/s 2 to the right, calculate the following: If you pull a 2505 g box with a force of 15 N at an angle of 53 ⁰ to the horizon and the box accelerates at 2.0 m/s 2 to the right, calculate the following: Fn, Fg, Ff, Fnet, Fapp, Fx, Fy and µ Fn, Fg, Ff, Fnet, Fapp, Fx, Fy and µ

52 Friction Lab Put a ramp flat in your lab space. Place two photogates relatively close together. Put a ramp flat in your lab space. Place two photogates relatively close together. If the mass of the sled is.040 kg calculate the Fnormal (Fn=Fg if on flat surface) If the mass of the sled is.040 kg calculate the Fnormal (Fn=Fg if on flat surface) Now, using your sled car (no wheels) launch the car with your rubber band. Make sure that it goes through both photo gates (you may have to adjust photo gates). Use our acceleration procedure from lab and calculate the rate of deceleration. Now, using your sled car (no wheels) launch the car with your rubber band. Make sure that it goes through both photo gates (you may have to adjust photo gates). Use our acceleration procedure from lab and calculate the rate of deceleration. Calculate Ffric= mass of sled x deceleration Calculate Ffric= mass of sled x deceleration Calculate µ = Ff/Fn Calculate µ = Ff/Fn

53 Free body diagram example 2 Say a box is sitting on 30 ⁰ slope and is frictionless, so the only forces are the normal force and gravity. What is the block's acceleration down the slope if the mass is 3.0 kg? What is the normal force? Say a box is sitting on 30 ⁰ slope and is frictionless, so the only forces are the normal force and gravity. What is the block's acceleration down the slope if the mass is 3.0 kg? What is the normal force?

54 Free Body Diagram example 3 A box is sitting on a 35 ⁰ inclined plane. It is being pulled up the ramp by you with an acceleration of 2.5 m/s 2. If the box has a mass of 25 kg and the force of friction is 3.5 N, solve all of the following: Fnet, Fnormal, Fgravity, Fapplied, and µ. A box is sitting on a 35 ⁰ inclined plane. It is being pulled up the ramp by you with an acceleration of 2.5 m/s 2. If the box has a mass of 25 kg and the force of friction is 3.5 N, solve all of the following: Fnet, Fnormal, Fgravity, Fapplied, and µ.

55 A 50-N applied force (30 degrees to the horizontal) accelerates a box across a horizontal sheet of ice (see diagram). Glen Brook, Olive N. Glenveau, and Warren Peace are discussing the problem. Glen suggests that the normal force is 50 N; Olive suggests that the normal force in the diagram is 75 N; and Warren suggests that the normal force is 100 N. While all three answers may seem reasonable, only one is correct. Indicate which two answers are wrong and explain why they are wrong. A 50-N applied force (30 degrees to the horizontal) accelerates a box across a horizontal sheet of ice (see diagram). Glen Brook, Olive N. Glenveau, and Warren Peace are discussing the problem. Glen suggests that the normal force is 50 N; Olive suggests that the normal force in the diagram is 75 N; and Warren suggests that the normal force is 100 N. While all three answers may seem reasonable, only one is correct. Indicate which two answers are wrong and explain why they are wrong.

56 Review: Newton’s Laws of Motion Newton’s First Law: Newton’s First Law: Every object continues in its state of rest, or of motion in a straight line at constant speed, unless compelled to change that state by forces exerted on it. Every object continues in its state of rest, or of motion in a straight line at constant speed, unless compelled to change that state by forces exerted on it. Newton’s Second Law: Newton’s Second Law: The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force, and is inversely proportional to the mass of the body. The acceleration produced by a net force on an object is directly proportional to the magnitude of the net force, and is inversely proportional to the mass of the body. Newton’s Third Law: Newton’s Third Law: Whenever one object exerts a force on a second object, the second exerts an equal & opposite force on the first Whenever one object exerts a force on a second object, the second exerts an equal & opposite force on the first

57 Action- Reaction Lab Adjust the smart track (or lab table) to be as level as possible(may have to put lab book under) put rubber band around one car. Adjust the smart track (or lab table) to be as level as possible(may have to put lab book under) put rubber band around one car. Squeeze two cars together and attach with the car link. Squeeze two cars together and attach with the car link. Position car in middle of track, making sure all wheels are on track. Position car in middle of track, making sure all wheels are on track. With a quick upward motion, pull the link straight up and out from the cars. With a quick upward motion, pull the link straight up and out from the cars. Describe how the cars move in a data table. Describe how the cars move in a data table. Start adding marbles to cars and repeat procedures above Start adding marbles to cars and repeat procedures above Make all these combinations of marbles in cars Make all these combinations of marbles in cars 0,0 0,1 0,2 0,3 1,1 1,2 1,3 2,2 2,3 3,3 0,0 0,1 0,2 0,3 1,1 1,2 1,3 2,2 2,3 3,3 Sum up the action reaction effect on cars and marbles. Sum up the action reaction effect on cars and marbles.

58 Draw the free body diagram

59 Draw the free body diagram, if a =.1 m/s 2 and the force you push on the lawnmower is 25 N, solve for every force you know.

60 Say a box is sitting on 40 ⁰ slope ramp. If the mass is 3.0 kg? What are all the forces acting on the box and what is µ? Say a box is sitting on 40 ⁰ slope ramp. If the mass is 3.0 kg? What are all the forces acting on the box and what is µ?

61 Renee is on Spring Break and pulling her 21-kg suitcase through the airport at a constant speed of 0.47 m/s. She pulls on the strap with 120 N of force at an angle of 38° above the horizontal. Determine the normal force and the total resistance force (friction and air resistance) experienced by the suitcase. Renee is on Spring Break and pulling her 21-kg suitcase through the airport at a constant speed of 0.47 m/s. She pulls on the strap with 120 N of force at an angle of 38° above the horizontal. Determine the normal force and the total resistance force (friction and air resistance) experienced by the suitcase.

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64 For each collection of listed forces, determine the vector sum or the net force. For each collection of listed forces, determine the vector sum or the net force. Set A 58 N, right 42 N, left 98 N, up 98 N, down Set A 58 N, right 42 N, left 98 N, up 98 N, down

65 Hector is walking his dog (Fido) around the neighborhood. Upon arriving at Fidella's house (a friend of Fido's), Fido turns part mule and refuses to continue on the walk. Hector yanks on the chain with a 67.0 N force at an angle of 30.0° above the horizontal. Determine the horizontal and vertical components of the tension force. Hector is walking his dog (Fido) around the neighborhood. Upon arriving at Fidella's house (a friend of Fido's), Fido turns part mule and refuses to continue on the walk. Hector yanks on the chain with a 67.0 N force at an angle of 30.0° above the horizontal. Determine the horizontal and vertical components of the tension force.

66 Helen is parasailing. She sits in a seat harness which is attached by a tow rope to a speedboat. The rope makes an angle of 51° with the horizontal and has a tension of 350 N. Determine the horizontal and vertical components of the tension force. Helen is parasailing. She sits in a seat harness which is attached by a tow rope to a speedboat. The rope makes an angle of 51° with the horizontal and has a tension of 350 N. Determine the horizontal and vertical components of the tension force.

67 Jerome and Michael, linebackers for South’s varsity football team, delivered a big hit to the halfback in last weekend’s game. Striking the halfback simultaneously from different directions with the following forces: Jerome and Michael, linebackers for South’s varsity football team, delivered a big hit to the halfback in last weekend’s game. Striking the halfback simultaneously from different directions with the following forces: F Jerome = 1230 N at 53° F Michael = 1450 at 107° F Jerome = 1230 N at 53° F Michael = 1450 at 107° Determine the resultant force applied by Jerome and Michael to the halfback. (The directions of the two forces are stated as counter-clockwise angles of rotation with East.) Determine the resultant force applied by Jerome and Michael to the halfback. (The directions of the two forces are stated as counter-clockwise angles of rotation with East.)

68 2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below to determine the values for the blanks. 2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below to determine the values for the blanks.

69 Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = ma; F frict = μF norm ; F grav = mg) Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = ma; F frict = μF norm ; F grav = mg)

70 Friday Problem 1 A 5-kg mass below is moving with a an acceleration of 4 m/s 2 to the right. The coefficient of friction for this surface is.2. Use your understanding of force relationships and vector components to determine all your forces. The 5-kg mass below is moving with a constant speed of 4 m/s to the right. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = ma; F frict = μF norm ; F grav = mg)

71 Friday Problem 2 You are pushing a 200 kg block up a 20 ⁰ hill with a force of 200 N. If the box moves up the hill with a constant speed of 2 m/s, calculate all the forces involved and calculate µ. You are pushing a 200 kg block up a 20 ⁰ hill with a force of 200 N. If the box moves up the hill with a constant speed of 2 m/s, calculate all the forces involved and calculate µ.

72 Tuesday Problem 1 5. The following object is being pulled at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = ma; F frict = μF norm ; F grav = mg) 5. The following object is being pulled at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = ma; F frict = μF norm ; F grav = mg)

73 At one moment during a walk around the block, there are four forces exerted upon Fido - a 10.0 kg dog. The forces are: At one moment during a walk around the block, there are four forces exerted upon Fido - a 10.0 kg dog. The forces are: F app = 67.0 N at 30.0° above the horizontal (rightward and upward) F norm = 64.5 N, up F frict = 27.6 N, left F grav = 98 N, down F app = 67.0 N at 30.0° above the horizontal (rightward and upward) F norm = 64.5 N, up F frict = 27.6 N, left F grav = 98 N, down Resolve the applied force (F app ) into horizontal and vertical components, then add the forces up as vectors to determine the net force and calculate the acceleration. Resolve the applied force (F app ) into horizontal and vertical components, then add the forces up as vectors to determine the net force and calculate the acceleration.

74 A box is sliding down a ramp at an angle of 47 ⁰ to the horizontal. If it is accelerating at 2.5 m/s 2 and has a mass of 150 kg, what is the Fnormal, Fnet, Fgravity, Ffric and µ? A box is sliding down a ramp at an angle of 47 ⁰ to the horizontal. If it is accelerating at 2.5 m/s 2 and has a mass of 150 kg, what is the Fnormal, Fnet, Fgravity, Ffric and µ?

75 Ramp Problem #1 Say a box is sitting on 30 ⁰ slope and is frictionless, so the only forces are the normal force and gravity. What is the block's acceleration down the slope if the mass is 3.0 kg? What is the normal force? Say a box is sitting on 30 ⁰ slope and is frictionless, so the only forces are the normal force and gravity. What is the block's acceleration down the slope if the mass is 3.0 kg? What is the normal force?

76 Rotation & Centripetal Force How to Keep it Straight Without Getting Dizzy Coutesy Space.com

77 Rotation In addition to side to side (linear) motion, rotation plays an important role in physics, engineering, and life. In addition to side to side (linear) motion, rotation plays an important role in physics, engineering, and life. Name some common phenomena or devices that show rotation Name some common phenomena or devices that show rotation Tops, planets, bicycle, car wheels, gears, pulleys, fans etc

78 Speed on a Wheel Which horses on a carousel move the fastest, inner or outer? Which horses on a carousel move the fastest, inner or outer? Outer v = radius x angular speed v = r 

79 Mass at the End of a String What force must the What force must the string exert on the mass? What is the direction of this force? A force toward the center of the circle

80 Centripetal Force Any force directed toward the center of a circle is called centripetal. Any force directed toward the center of a circle is called centripetal. Centripetal forces have clear causes such as tension in a string, gravity, friction etc. Centripetal forces have clear causes such as tension in a string, gravity, friction etc. Some people call centripetal force a “pseudoforce.” (not real) Some people call centripetal force a “pseudoforce.” (not real) They say “a real force such as friction provides centripetal force.” They say “a real force such as friction provides centripetal force.”

81 How Big is Centripetal Force? F c = mv 2 /r F c = mv 2 /r The faster the speed the more the force The faster the speed the more the force The tighter (smaller) the radius the more the force The tighter (smaller) the radius the more the force v 2 /r is called centripetal acceleration v 2 /r is called centripetal acceleration

82 Is a mass moving at steady speed in a circle accelerating? Yes. The direction is changing Yes. The direction is changing What is the direction of this acceleration? What is the direction of this acceleration? Toward the center of the circle

83 Car on a Curve When auto rounds corner, sideways acting friction between tires and road provides centripetal force that holds car on road When auto rounds corner, sideways acting friction between tires and road provides centripetal force that holds car on road

84 Don’t Confuse Inertia With Force Tub’s inner wall exerts centripetal force on clothes, forcing them into circular path Tub’s inner wall exerts centripetal force on clothes, forcing them into circular path Water escapes through Water escapes through holes because it tends to move by inertia in a straight line path holes because it tends to move by inertia in a straight line path Clothes Washer Photo courtesy HowStuffWorks.com

85 How Can Water Stay In The Bucket? Bucket swung in a Bucket swung in a vertical circle vertical circle What force pushes on the What force pushes on the water? water? You have to swing the bucket fast enough for the bucket to fall as fast as the water There must be a “normal” force exerted by the bottom of the bucket on the water, in addition to gravity Weight and normal force down

86 Centrifugal Force The force ON THE PAIL is inward (centripetal) The force ON THE PAIL is inward (centripetal) The force ON THE STRING is outward (centrifugal) The force ON THE STRING is outward (centrifugal) If the string broke, which way would the can go? If the string broke, which way would the can go? Tangent to the circle

87 Change Your Point of View In rest frame of the can there appears to be a centrifugal force. This pseudoforce(or fictitious force) is a result of rotation In rest frame of the can there appears to be a centrifugal force. This pseudoforce(or fictitious force) is a result of rotation Unlike real forces, centrifugal force is not part of an interaction

88 Book on a Car Seat When a car goes around a curve to the left, a book slides When a car goes around a curve to the left, a book slides Which way does it slide? Which way does it slide? Why doesn’t it keep moving with the car? Why doesn’t it keep moving with the car? There is not enough static friction force to keep it going in a circle. This friction must provide the necessary centripetal force. The explanation in the rotating rest frame is different. How?

89 Roller Coaster Lab- Centripetal Force You are dropping the ball from 45 ⁰, practice dropping the steel ball and the plastic ball to observe when it gets around the track. You are dropping the ball from 45 ⁰, practice dropping the steel ball and the plastic ball to observe when it gets around the track. Attach the photogate and calculate the speed and centripetal force of the marble at the top of the loop from various distances for both marbles. Width of ball=.019 m Attach the photogate and calculate the speed and centripetal force of the marble at the top of the loop from various distances for both marbles. Width of ball=.019 m Complete the table for both marbles. (as many trials as necessary) Complete the table for both marbles. (as many trials as necessary) steel =.028 kg plastic =.004 kg Fc= mv 2 /r radius of loop =.05 m Draw a free body force diagram when the ball is at the top of the loop, label all forces. Do the following lab to solve for the minimum force needed to keep the ball (steel and plastic) on the loop. Draw a free body force diagram when the ball is at the top of the loop, label all forces. Do the following lab to solve for the minimum force needed to keep the ball (steel and plastic) on the loop. Mass(kg)Weight(N)Photogate Time (sec) Speed (m/s)Centripetal Force (N) Did the marble stay on track?


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