Introduction to Newton’s Laws Newton’s First Law.

Introduction to Newton’s Laws Newton’s First Law.
*Slides adapted from Peggy Bertrand 1

Isaac Newton Arguably the greatest scientific genius ever.
Came up with 3 Laws of Motion to explain the observations and analyses of Galileo and Johannes Kepler. Discovered that white light was composed of many colors all mixed together. Invented new mathematical techniques such as calculus and binomial expansion theorem in his study of physics. Published his Laws in 1687 in the book Mathematical Principles of Natural Philosophy. 2

What is Force? A force is a push or pull on an object.
Forces cause an object to accelerate… To speed up To slow down To change direction 3

Newton’s First Law A body in motion stays in motion at constant velocity and a body at rest stays at rest unless acted upon by a net external force. This law is commonly referred to as the Law of Inertia. 4

The First Law is Counterintuitive
Aristotle firmly believed this. But Physics B students know better! 5

Implications of Newton’s 1st Law
If there is zero net force on a body, it cannot accelerate, and therefore must move at constant velocity, which means it cannot turn, it cannot speed up, it cannot slow down. 6

What is Zero Net Force? SF = 0 The table pushes up on the book. FT
A book rests on a table. Physics Gravity pulls down on the book. FG Even though there are forces on the book, they are balanced. Therefore, there is no net force on the book. SF = 0 7

Diagrams Draw a force diagram and a free body diagram for a book sitting on a table. Physics N W Force Diagram N W Free Body Diagram 8

Sample Problem A monkey hangs by its tail from a tree branch. Draw a force diagram representing all forces on the monkey FT FG 9

Sample Problem Now the monkey hangs by both hands from two vines. Each of the monkey’s arms are at a 45o from the vertical. Draw a force diagram representing all forces on the monkey. Fa1 Fa2 FG 10

Mass and Inertia Chemists like to define mass as the amount of “stuff” or “matter” a substance has. Physicists define mass as inertia, which is the ability of a body to resist acceleration by a net force. 11

Sample Problem A heavy block hangs from a string attached to a rod. An identical string hangs down from the bottom of the block. Which string breaks when the lower string is pulled with a slowly increasing force? when the lower string is pulled with a quick jerk? Top string breaks due to its greater force. Bottom string breaks because block has lots of inertia and resists acceleration. Pulling force doesn’t reach top string. 12

Sample problem A 5.0-kg bag of potatoes sits on the bottom of a stationary shopping cart. Sketch a free-body diagram for the bag of potatoes. Now suppose the cart moves with a constant velocity. How does this affect the free-body diagram? 13

Newton’s Second Law 14

Newton’s Second Law A body accelerates when acted upon by a net external force. The acceleration is proportional to the net force and is in the direction which the net force acts. 15

Newton’s Second Law ∑F = ma
where ∑F is the net force measured in Newtons (N) m is mass (kg) a is acceleration (m/s2) 16

Units of force Newton (SI system)
1 N = 1 kg m /s2 1 N is the force required to accelerate a 1 kg mass at a rate of 1 m/s2 Pound (British system) 1 lb = 1 slug ft /s2 17

Working 2nd Law Problems
Draw a force or free body diagram. Set up 2nd Law equations in each dimension. SFx = max and/or SFy = may Identify numerical data. x-problem and/or y-problem Solve the equations. Substitute numbers into equations. “plug-n-chug” 18

Sample Problem In a grocery store, you push a 14.5-kg cart with a force of 12.0 N. If the cart starts at rest, how far does it move in 3.00 seconds? 19

Sample Problem A catcher stops a 92 mph pitch in his glove, bringing it to rest in 0.15 m. If the force exerted by the catcher is 803 N, what is the mass of the ball? 20

Sample Problem A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50 x 105 kg, its speed is 27.0 m/s, and the net braking force is 4.30 x 105 N a) what is its speed 7.50 s later? b) How far has it traveled in this time? 21

Newton’s Third Law 22

Newton’s Third Law For every action there exists an equal and opposite reaction. If A exerts a force F on B, then B exerts a force of -F on A. 23

Examples of Newton’s 3rd Law
Copyright James Walker, “Physics”, 1st edition 24

Sample Problem You rest an empty glass on a table.
a) How many forces act upon the glass? b) Identify these forces with a free body diagram. c) Are these forces equal and opposite? d) Are these forces an action-reaction pair? 25

Tension Tension is a pulling force that arises when a rope, string, or other long thin material resists being pulled apart without stretching significantly. Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string. 26

A physical picture of tension
Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string. Copyright James Walker, “Physics”, 1st ed. 27

Tension examples Note that the pulleys shown are magic! They affect the tension in any way, and serve only to bend the line of action of the force. Copyright James Walker, “Physics”, 1st ed. 28

Sample problem A 1,500 kg crate hangs motionless from a crane cable. What is the tension in the cable? Ignore the mass of the cable. Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now? 29

Systems of Objects (Forces Day 2)
30

Sample Problem A force of magnitude 7.50 N pushes three boxes with masses m1 = 1.30 kg, m2 = 3.20 kg, and m3 = 4.90 kg as shown. Find the contact force between (a) boxes 1 and 2 and (b) between boxes 2 and 3. Copyright James Walker, “Physics”, 1st edition 32

Gravity A very common accelerating force is gravity. Here is gravity in action. The acceleration is g. 33

Slowing gravity down The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g. 34

Magic pulleys on a flat table
Magic pulleys bend the line of action of the force without affecting tension. T m2g N m1g SF = ma m2g + T – T = (m1 + m2)a a = m2g/(m1+m2) -x x m1 Frictionless table m2 35

Sample problem Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find the acceleration of each block. the tension in the connecting string. m1 m2 36

Mass and Weight 37

Mass and Weight Many people think mass and weight are the same thing. They are not. Mass is inertia, or resistance to acceleration. Weight can be defined as the force due to gravitation attraction. W = mg 38

Normal Force 40

Normal force The normal force is a force that keeps one object from penetrating into another object. The normal force is always perpendicular a surface. The normal exactly cancels out the components of all applied forces that are perpendicular to a surface. 41

Normal force on flat surface
The normal force is equal to the weight of an object for objects resting on horizontal surfaces. N = W = mg N mg 42

Normal force not associated with weight.
A normal force can exist that is totally unrelated to the weight of an object. applied force friction weight normal N = applied force 43

The normal force most often equals the weight of an object…
44

…but this is by no means always the case!
45

Newton’s Laws in 2D 46

Newton’s 2nd Law in 2-D The situation is more complicated when forces act in more than one dimension. You must still identify all forces and draw your force diagram. You then resolve your problem into an x-problem and a y-problem (remember projectile motion????). 47

Forces in 2-D Copyright James Walker, “Physics”, 1st edition 48

Sample problem A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 320 with the surface of the lawn. If a 249 N force is applied along the handle of the 21-kg mower, what is the normal force exerted by the lawn on the mower? 54

Sample problem Larry pushes a 200 kg block on a frictionless floor at a 45o angle below the horizontal with a force of 150 N while Moe pulls the same block horizontally with a force of 120 N. a) Draw a free body diagram. b) What is the acceleration of the block? c) What is the normal force exerted on the block? 55

Apparent weight If an object subject to gravity is not in free fall, then there must be a reaction force to act in opposition to gravity. We sometimes refer to this reaction force as apparent weight. 56

Elevator rides When you are in an elevator, your actual weight (mg) never changes. You feel lighter or heavier during the ride because your apparent weight increases when you are accelerating up, decreases when you are accelerating down, and is equal to your weight when you are not accelerating at all. 57

Going Up? v = 0 a = 0 v > 0 a > 0 v > 0 a = 0 v > 0
W Wapp Ground floor Normal feeling v = 0 a = 0 W Wapp Just starting up Heavy feeling v > 0 a > 0 W Wapp Between floors Normal feeling v > 0 a = 0 W Wapp Arriving at top floor Light feeling v > 0 a < 0 Going Up? 58

Going Down? v = 0 a = 0 v < 0 a < 0 v < 0 a = 0 v < 0
Wapp Top floor Normal feeling v = 0 a = 0 W Wapp Beginning descent Light feeling v < 0 a < 0 W Wapp Between floors Normal feeling v < 0 a = 0 W Wapp Arriving at Ground floor Heavy feeling v < 0 a > 0 Going Down? 59

Sample Problem An 85-kg person is standing on a bathroom scale in an elevator. What is the person’s apparent weight a) when the elevator accelerates upward at 2.0 m/s2? b) when the elevator is moving at constant velocity between floors? c) when the elevator begins to slow at the top floor at 2.0 m/s2? 60

Introduction to Friction
62

Friction Friction is the force that opposes a sliding motion.
Friction is due to microscopic irregularities in even the smoothest of surfaces. Friction is highly useful. It enables us to walk and drive a car, among other things. Friction is also dissipative. That means it causes mechanical energy to be converted to heat. We’ll learn more about that later. 63

Microscopic View N Fpush f W (friction) Big view:
Surfaces look perfectly smooth. Fpush f (friction) Small view: Microscopic irregularities resist movement. Friction may or may not exist between two surfaces. The direction of friction, if it exists, is opposite to the direction object will slide when subjected to a horizontal force. It is always parallel to the surface. 64

Friction depends on the normal force.
The friction that exists between two surfaces is directly proportional to the normal force. Increasing the normal force increases friction; decreasing the normal force decreases friction. This has several implications, such as… Friction on a sloping surface is less than friction on a flat surface (since the normal force is less on a slope). Increasing weight of an object increases the friction between the object and the surface it is resting on. Weighting down a car over the drive wheels increases the friction between the drive wheels and the road (which increases the car’s ability to accelerate). 65

Static Friction This type of friction occurs between two surfaces that are not slipping relative to each other. fs  sN fs : static frictional force (N) s: coefficient of static friction N: normal force (N) 66

fs < msN is an inequality!
The fact that the static friction equation is an inequality has important implications. Static friction between two surfaces is zero unless there is a force trying to make the surfaces slide on one another. Static friction can increase as the force trying to push an object increases until it reaches its maximum allowed value as defined by ms. Once the maximum value of static friction has been exceeded by an applied force, the surfaces begin to slide and the friction is no longer static friction. 67

Static friction and applied horizontal force
Physics N W Force Diagram surface fs = 0 There is no static friction since there is no applied horizontal force trying to slide the book on the surface. 68

Physics N W Force Diagram F fs surface 0 < fs < msN and fs = F Static friction is equal to the applied horizontal force, and there is no movement of the book since SF = 0. 69

Physics N W Force Diagram F fs surface fs = msN and fs = F Static friction is at its maximum value! It is still equal to F, but if F increases any more, the book will slide. 70

Physics N W Force Diagram fk F surface fs = msN and fs < F Static friction cannot increase any more! The book accelerates to the right. Friction becomes kinetic friction, which is usually a smaller force. 71

Kinetic Friction This type of friction occurs between surfaces that are slipping past each other. fk = kN fk : kinetic frictional force (N) k: coefficient of kinetic friction N: normal force (N) Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces. 72

Sample Problem A 10-kg box rests on a ramp that is laying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the maximum horizontal force that can be applied to the box before it begins to slide? What force is necessary to keep the box sliding at constant velocity? 73

Sample Problem A 10-kg wooden box rests on a ramp that is lying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is What is the friction force between the box and ramp if no force horizontal force is applied to the box? a 20 N horizontal force is applied to the box? a 60 N horizontal force is applied to the box? 74

More on the Normal Force
75

Normal force on ramp N mg N = mgcos  
The normal force is perpendicular to angled ramps as well. It’s always equal to the component of weight perpendicular to the surface. N = mgcos N  mgcos mgsin mg  76

Draw a free body diagram for the skier.
77

Sample problem Find the normal force exerted on a 2.5-kg book resting on a surface inclined at 28o above the horizontal. If the angle of the incline is reduced, do you expect the normal force to increase, decrease, or stay the same? 80

Static friction on a ramp
surface fs N Without friction, the book will slide down the ramp. If it stays in place, there is sufficient static friction holding it there. Physics W = mg q Wx = mgsinq and N = mgcosq At maximum angle before the book slides, we can prove that ms = tanq. 81

Static friction on a ramp
x surface fs Assume q is maximum angle for which book stays in place. N SF = 0 Wx = fs mgsinq = msmgcosq ms = sinq/cosq = tanq Physics q Wx W = mg q fs = mgsinq and N = mgcosq At maximum angle before the book slides, we can prove that ms = tanq. 82

Problem A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is What is the friction force between the box and ramp if the ramp is at a 25o angle? the ramp is at a 45o angle? what is the acceleration of the box when the ramp is at 45o? 83

Springs (Hooke’s Law) The magnitude of the force exerted by a spring is proportional to the amount it is stretched. F = kx F: force exerted by the spring (N) k: force constant of the spring (N/m or N/cm) x: displacement from equilibrium (unstretched and uncompressed) position (m or cm) The direction of the force is back toward the equilibrium (or unstretched) position. 86

Sample problem A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length? 87

Sample problem A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60. Copyright James Walker, “Physics”, 1st ed. 88

Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary. m1 m2 90

Sample problem - solution
SF = 0 m2g - T + T – fs = 0 fs = m2g msN = m2g msm1g = m2g ms = m2/m1 = 0.50 T m2g N m1g fs m1 m2 91

Note: we know from previous problem that the static friction is not enough to hold the blocks in place! Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If ms = 0.30 and mk = 0.20, what is the acceleration of each block? the tension in the connecting string? m1 m2 92

Sample problem – solution (a)
SF = ma m2g - T + T – fk = ma m2g - mkm1g = (m1 + m2)a a = (m2- mkm1) g/(m1 + m2) a = 2.0 m/s2 T m2g N m1g fk m1 m2 93

Sample problem – solution (b)
Using that the acceleration is 2.0 m/s2 from part a) Sample problem – solution (b) Using block 2 SF = ma m2g - T = m2a T = m2(g – a) T = 40 N Using block 1 SF = ma T - fk = m1a T = m1(a + mkg) T = 40 N T m2g N m1g fk m1 m2 94

Pulleys and Ramps - together
Ramps and Pulleys – together! 96

Magic pulleys on a ramp q
It’s a little more complicated when a magic pulley is installed on a ramp. m1g N T m2g m1gsinq m1gcosq SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) m1 m2 q 97

Sample problem Two blocks are connected by a string as shown in the figure. What is the acceleration, assuming there is no friction? 10 kg 5 kg 45o 98

m1g N T m2g m1gsinq m1gcosq SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = m/s2 10 kg 5 kg 45o 99

m1g N T m2g m1gsinq m1gcosq SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = m/s2 10 kg 5 kg 45o How would this change if there is friction on the ramp? 100

Newton’s Laws Applications

Coefficients of Friction Laboratory

Problem A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is What is the friction force between the box and ramp if the ramp is at a 25o angle? the ramp is at a 45o angle? what is the acceleration of the box when the ramp is at 45o?

Laboratory Determine the coefficients of static and kinetic friction between the wooden block (felt side) and the cart track. The only additional equipment you may use is a meter stick, a clamp, and a pole. For homework, calculate the coefficients of friction. Include diagrams (free-body), calculations, and results for each kind of friction.

More experiments to determine coefficients of friction

Laboratory Determine the coefficients of friction (static and kinetic) between the felt-covered wood block and the cart track using spring scales. Tomorrow you will turn in the following for each coefficient of friction: Free body diagram Calculation of coefficient of friction. Balances are available for measurement of mass.

Strings and Springs

Tension Tension is a pulling force that arises when a rope, string, or other long thin material resists being pulled apart without stretching significantly. Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string.

A physical picture of tension
Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string. Copyright James Walker, “Physics”, 1st ed.

Tension examples Note that the pulleys shown are magic! They affect the tension in any way, and serve only to bend the line of action of the force. Copyright James Walker, “Physics”, 1st ed.

Sample problem A 1,500 kg crate hangs motionless from a crane cable. What is the tension in the cable? Ignore the mass of the cable. Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now?

Springs (Hooke’s Law) The magnitude of the force exerted by a spring is proportional to the amount it is stretched. F = kx F: force exerted by the spring (N) k: force constant of the spring (N/m or N/cm) x: displacement from equilibrium (unstretched and uncompressed) position (m or cm) The direction of the force is back toward the equilibrium (or unstretched) position.

Sample problem A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?

Sample problem A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60. Copyright James Walker, “Physics”, 1st ed.

Laboratory Using the ramp at an angle, determine the coefficient of kinetic friction between the felt side of the block and the ramp by allowing the block to accelerated down the ramp. Tomorrow, you will turn in the appropriate free body diagrams and calculations.

Connected Objects

Sample problem A 5.0 kg object (m1) is connected to a 10.0 kg object (m2) by a string. If a pulling force F of 20 N is applied to the 5.0 kg object as shown, A) what is the acceleration of the system? B) what is the tension in the string connecting the objects? (Assume a frictionless surface.) Copyright James Walker, “Physics”, 1st ed.

Gravity A very common accelerating force is gravity. Here is gravity in action. The acceleration is g.

Slowing gravity down The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g.

Magic pulleys on a flat table
Magic pulleys bend the line of action of the force without affecting tension. T m2g N m1g SF = ma m2g + T – T = (m1 + m2)a a = m2g/(m1+m2) -x x m1 Frictionless table m2

Sample problem Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find the acceleration of each block. the tension in the connecting string. m1 m2

Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary. m1 m2

SF = 0 m2g - T + T – fs = 0 fs = m2g msN = m2g msm1g = m2g ms = m2/m1 = 0.50 T m2g N m1g fs m1 m2

Note: we know from previous problem that the static friction is not enough to hold the blocks in place! Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If ms = 0.30 and mk = 0.20, what is the acceleration of each block? the tension in the connecting string? m1 m2

Sample problem – solution (a)
SF = ma m2g - T + T – fk = ma m2g - mkm1g = (m1 + m2)a a = (m2- mkm1) g/(m1 + m2) a = 2.0 m/s2 T m2g N m1g fk m1 m2

Sample problem – solution (b)
Using that the acceleration is 2.0 m/s2 from part a) Sample problem – solution (b) Using block 2 SF = ma m2g - T = m2a T = m2(g – a) T = 40 N Using block 1 SF = ma T - fk = m1a T = m1(a + mkg) T = 40 N T m2g N m1g fk m1 m2

Laboratory Using the pulley, masses, friction block, cart (for added weight), and cart track, determine the coefficient of static and kinetic friction for the felt side of the wood block and the track. For static friction, assume SF = 0 and find the maximum value of static friction. For kinetic friction, you must base your results on an accelerating system. Photogate timers may be used to increase timing accuracy.

Pulleys and Ramps - together
Ramps and Pulleys – together!

Magic pulleys on a ramp q
It’s a little more complicated when a magic pulley is installed on a ramp. m1g N T m2g m1gsinq m1gcosq SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) m1 m2 q

Sample problem Two blocks are connected by a string as shown in the figure. What is the acceleration, assuming there is no friction? 10 kg 5 kg 45o

m1g N T m2g m1gsinq m1gcosq SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = m/s2 10 kg 5 kg 45o

m1g N T m2g m1gsinq m1gcosq SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = m/s2 10 kg 5 kg 45o How would this change if there is friction on the ramp?

Laboratory Determine the coefficients of static and of kinetic friction using a RAMP and PULLEY SYSTEM combined. For kinetic friction, timing may be done with photo-gate timers or with stopwatches. Do 4 trials: Static friction; tendency of the block to slide up the ramp. Static friction; tendency of the block to slide down the ramp. Kinetic friction; block accelerating up the ramp. Kinetic friction; block accelerating down the ramp. Full lab report will be collected Friday. All free body diagrams and calculations may be done by hand; all other information must be typed.

Introduction to Newton’s Laws Newton’s First Law.

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