Newton’s Laws

Tuesday, September 18, 2007 Re-introduction to Newton’s 3 Laws

What is Force? A push or pull on an object. A push or pull on an object. Unbalanced forces cause an object to accelerate… Unbalanced forces cause an object to accelerate… To speed up To speed up To slow down To slow down To change direction To change direction Force is a vector! Force is a vector!

Types of Forces Contact forces: involve contact between bodies. Contact forces: involve contact between bodies. Normal, Friction Normal, Friction Field forces: act without necessity of contact. Field forces: act without necessity of contact. Gravity, Electromagnetic, Strong, Weak Gravity, Electromagnetic, Strong, Weak Question: Is there really any such thing as a contact force? Question: Is there really any such thing as a contact force?

Forces and Equilibrium If the net force on a body is zero, it is in equilibrium. If the net force on a body is zero, it is in equilibrium. An object in equilibrium may be moving relative to us (dynamic equilibrium). An object in equilibrium may be moving relative to us (dynamic equilibrium). An object in equilibrium may appear to be at rest (static equilibrium). An object in equilibrium may appear to be at rest (static equilibrium).

Galileo’s Thought Experiment © The Physics Classroom, Tom Henderson

This thought experiment lead to Newton’s First Law. Galileo’s Thought Experiment © The Physics Classroom, Tom Henderson

Newton’s First Law The Law of Inertia. The Law of Inertia. A body in motion stays in motion in a straight line unless acted upon by an external force. A body in motion stays in motion in a straight line unless acted upon by an external force. This law is commonly applied to the horizontal component of velocity, which is assumed not to change during the flight of a projectile. This law is commonly applied to the horizontal component of velocity, which is assumed not to change during the flight of a projectile.

Newton’s Second Law A body accelerates when acted upon by a net external force A body accelerates when acted upon by a net external force The acceleration is proportional to the net (or resultant) force and is in the direction which the net force acts. The acceleration is proportional to the net (or resultant) force and is in the direction which the net force acts. This law is commonly applied to the vertical component of velocity. This law is commonly applied to the vertical component of velocity.  F = ma  F = ma

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

Commonly Confused Terms Inertia: or the resistance of an object to being accelerated Inertia: or the resistance of an object to being accelerated Mass: the same thing as inertia (to a physicist). Mass: the same thing as inertia (to a physicist). Weight: gravitational attraction Weight: gravitational attraction inertia = mass  weight inertia = mass  weight

Sample Problem – 1 st Law Two forces, F 1 = (4i – 6j + k) N and F 2 = (i – 2j - 8k) N, act upon a body of mass 3.0 kg as it is moving at constant speed. What do you know must be true?

Sample Problem – 2 nd Law Two forces, F 1 = (4i – 6j + k) N and F 2 = (i – 2j - 8k) N, act upon a body of mass 3.0 kg. No other forces act upon the body at this time. What do you know must be true?

Sample Problem – 3 rd Law A tug-of-war team ties a rope to a tree and pulls hard horizontally to create a tension of 30,000 N in the rope. Suppose the team pulls equally hard when, instead of a tree, the other end of the rope is being pulled by another tug-of-war team such that no movement occurs. What is the tension in the rope in the second case?

Working a 2 nd Law Problem Working Newton’s 2 nd Law Problems is best accomplished in a systematic fashion. Working Newton’s 2 nd Law Problems is best accomplished in a systematic fashion. The more complicated the problem, the more important it is to have a general procedure to follow in working it. The more complicated the problem, the more important it is to have a general procedure to follow in working it.

2 nd Law Procedure 1. Identify the body to be analyzed. 2. Select a reference frame, stationary or moving, but not accelerating 3. Draw a force or free body diagram. 4. Set up ΣF = ma equations for each dimension. 5. Use kinematics or calculus where necessary to obtain acceleration. 6. Substitute known quantities. 7. Calculate the unknown quantities.

Wednesday, September 19, 2007 Types of Forces Commonly Found in Newton’s Law Problems

Sample Problem A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. The bullet is accelerated by expanding gases while it travels down the m long barrel. Assuming constant acceleration and negligible friction, what is the force on the bullet? A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. The bullet is accelerated by expanding gases while it travels down the m long barrel. Assuming constant acceleration and negligible friction, what is the force on the bullet?

Sample Problem A 3.00 kg mass undergoes an acceleration given by a = (2.50i j) m/s 2. Find the resultant force F and its magnitude. A 3.00 kg mass undergoes an acceleration given by a = (2.50i j) m/s 2. Find the resultant force F and its magnitude.

Normal force The force that keeps one object from invading another object. The force that keeps one object from invading another object. Our weight is the force of attraction of our body for the center of the planet. Our weight is the force of attraction of our body for the center of the planet. We don’t fall to the center of the planet. We don’t fall to the center of the planet. The normal force keeps us up. The normal force keeps us up.

Normal Force on Flat surface

N = mg for objects resting on horizontal surfaces. mg N

Normal Force on Ramp 

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

Normal Force on Ramp mgmg What will acceleration be in this situation?  F= ma mgsin  = ma gsin  = a N   mgcos  mgsin  N = mgcos 

Normal Force on Ramp mgmg How could you keep the block from accelerating? N   mgcos  mgsin  N = mgcos  T

Tension A pulling force. A pulling force. Generally exists in a rope, string, or cable. Generally exists in a rope, string, or cable. Arises at the molecular level, when a rope, string, or cable resists being pulled apart. Arises at the molecular level, when a rope, string, or cable resists being pulled apart.

Tension (static 2D) The horizontal and vertical components of the tension are equal to zero if the system is not accelerating. 15 kg 30 o 45 o 3 2 1

Tension (static 2D) The horizontal and vertical components of the tension are equal to zero if the system is not accelerating. 15 kg 30 o 45 o T1T1 T2T2 T3T3  F x = 0  F y = 0 T3T3 mgmg 3 2 1

Tension (elevator) What about when an elevator is accelerating upward? M

Tension (elevator) What about when an elevator is accelerating upward? M T MgMg

Tension (elevator) What about when the elevator is moving at constant velocity between floors? M

Tension (elevator) What about when the elevator is moving at constant velocity between floors? M T MgMg

Tension (elevator) What about when the elevator is slowing at the top floor? M

Tension (elevator) What about when the elevator is slowing at the top floor? M T MgMg

Tension (elevator) What about if the elevator cable breaks? M

Tension (elevator) What about if the elevator cable breaks? M MgMg

Magic pulleys simply bend the coordinate system. Pulley problems m1m1 m2m2

-x x Magic pulleys simply bend the coordinate system. Pulley problems m1m1 m2m2 m1gm1g N T T m2gm2g  F = ma m2gm2g = (m 1 +m 2 )a

All problems should be started from a force diagram. Pulley problems m1m1 m2m2 

Tension is determined by examining one block or the other Pulley problems m1m1 m2m2 m1gm1g N T T m2gm2g  F  m 1 +m 2 )a m 2 g – T+T – m 1 gsin  = (m 1 +m 2 )a 

Tension is determined by examining one block or the other Pulley problems m1m1 m2m2 m1gm1g N T T m2gm2g  F 2  m 2 a m 2 g - T = m 2 a   F 1  m 1 a  m 1 gsin  = m 1 a

Atwood machine A device for measuring g. A device for measuring g. If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured. If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured. Then, g can be measured. Then, g can be measured. m1m1 m2m2

Atwood machine A device for measuring g. A device for measuring g. If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured. If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured. Then, g can be measured. Then, g can be measured. m1m1 m2m2 T m1gm1g T m2gm2g  F  ma m 2 g-m 1 g = (m 2 +m 1 )a

Thursday, September 20, 2007 Post-Test Review

Friday, September 21, 2007 Workday Exam Corrections Homework Review

Atwood Machine Mini-lab Draw diagram of apparatus in lab book. Draw diagram of apparatus in lab book. Record all data. Record all data. Calculate g. Calculate g.

Easy Problem How fast will the block be sliding at the bottom of the frictionless ramp? 20 o L = 12 m 5.0 kg

Easy Problem How high up the frictionless ramp will the block slide? 5.0 kg 20 o v = 12.0 m/s

Tuesday, September 25, 2007 Friction

Moderate Problem Describe acceleration of the 5 kg block. Table and pulley are magic and frictionless. 20 o 5.0 kg 1.0 kg

Friction Friction opposes a sliding motion. Friction opposes a sliding motion. Static friction exists before sliding occurs Static friction exists before sliding occurs (f s   s N). (f s   s N). Kinetic friction exists after sliding occurs Kinetic friction exists after sliding occurs f k =  k N f k =  k N

Friction on flat surfaces x y Draw a free body diagram for a braking car. x y Draw a free body diagram for a car accelerating from rest.

But we don’t want cars to skid! Why don’t we? Why don’t we? Let’s use DataStudio to see if we can detect the difference in magnitude between static and kinetic friction. Let’s use DataStudio to see if we can detect the difference in magnitude between static and kinetic friction.

Wednesday, September 26, 2007 Friction Lab

Thursday, September 27, 2007

Friction on a ramp   Sliding down Sliding up

Friction is always parallel to surfaces…. A 1.00 kg book is held against a wall by pressing it against the wall with a force of N. What must be the minimum coefficient of friction between the book and the wall, such that the book does not slide down the wall? FW f N (0.20)

Problem #1 Assume a coefficient of static friction of 1.0 between tires and road. What is the minimum length of time it would take to accelerate a car from 0 to 60 mph?

Problem #2 Assume a coefficient of static friction of 1.0 between tires and road and a coefficient of kinetic friction of 0.80 between tires and road. How far would a car travel after the driver applies the brakes if it skids to a stop?

Centripetal Force Inwardly directed force which causes a body to turn; perpendicular to velocity. Inwardly directed force which causes a body to turn; perpendicular to velocity. Centripetal force always arises from other forces, and is not a unique kind of force. Centripetal force always arises from other forces, and is not a unique kind of force. Sources include gravity, friction, tension, electromagnetic, normal. Sources include gravity, friction, tension, electromagnetic, normal. ΣF = ma ΣF = ma a = v 2 /r a = v 2 /r ΣF = m v 2 /r ΣF = m v 2 /r

Highway Curves R r z Friction turns the vehicle Normal force turns the vehicle

Sample problem Find the minimum safe turning radius for a car traveling at 60 mph on a flat roadway, assuming a coefficient of static friction of 0.70.

Sample problem Derive the expression for the period best banking angle of a roadway given the radius of curvature and the likely speed of the vehicles.

Friday, September 28, 2007 More on Circular Motion

Conical Pendulum For conical pendulums, centripetal force is provided by a component of the tension. z r T mgmg L T = 2  L cos  g

Sample problem Derive the expression for the period of a conical pendulum with respect to the string length and radius of rotation.

Non-uniform Circular Motion Consider circular motion in which either speed of the rotating object is changing, or the forces on the rotating object are changing. Consider circular motion in which either speed of the rotating object is changing, or the forces on the rotating object are changing. If the speed changes, there is a tangential as well as a centripetal component to the force. If the speed changes, there is a tangential as well as a centripetal component to the force. In some cases, the magnitude of the centripetal force changes as the circular motion occurs. In some cases, the magnitude of the centripetal force changes as the circular motion occurs.

Sample problem You swing a 0.25-kg rock in a vertical circle on a 0.80 m long rope at 2.0 Hz. What is the tension in the rope a) at the top and b) at the bottom of your swing?

Monday, October 1 Time-dependent forces

Sample problem A 40.0 kg child sits in a swing supported by 3.00 m long chains. If the tension in each chain at the lowest point is 350 N, find a) the child’s speed at the lowest point and b) the force exerted by the seat on the child at the lowest point.

Sample problem A 900-kg automobile is traveling along a hilly road. If it is to remain with its wheels on the road, what is the maximum speed it can have as it tops a hill with a radius of curvature of 20.0 m?

Non-constant Forces Forces can vary with time. Forces can vary with time. Forces can vary with velocity. Forces can vary with velocity. Forces can vary with position. Forces can vary with position.

Calculus Concepts for Forces that Vary with Time Differentiation Differentiation the tangent (or slope) of a function the tangent (or slope) of a function position -> velocity -> acceleration position -> velocity -> acceleration Integration Integration the area under a curve the area under a curve acceleration -> velocity -> position acceleration -> velocity -> position

Evaluating Integrals If a(t) = t n then t n dt = t n+1 / (n+1) + C t n dt = t n+1 / (n+1) 0 t 0 t

Sample problem Consider a force that is a function of time: Consider a force that is a function of time: F(t) = (3.0 t – 0.5 t 2 )N If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the resulting velocity and position of the particle? If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the resulting velocity and position of the particle?

Sample problem Consider a force that is a function of time: Consider a force that is a function of time: F(t) = (16 t 2 – 8 t + 4)N If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the resulting change in velocity of the particle? If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the resulting change in velocity of the particle?

Drag Forces Drag forces Drag forces slow an object down as it passes through a fluid. slow an object down as it passes through a fluid. act in opposite direction to velocity. act in opposite direction to velocity. are functions of velocity. are functions of velocity. impose terminal velocity. impose terminal velocity.

Drag as a Function of Velocity f D = bv + cv 2 f D = bv + cv 2 b and c depend upon b and c depend upon shape and size of object shape and size of object properties of fluid properties of fluid b is important at low velocity b is important at low velocity c is important at high velocity c is important at high velocity

Drag Force in Free Fall mgmg mgmg fDfD mgmg fDfD mgmg fDfD when f D equals mg, terminal velocity has been reached

Drag Force in Free Fall for fast moving objects F D = c v 2 c = 1/2 D  A Where D = drag coefficient  = density of fluid A = cross-sectional area F D = bv + c v 2 for slow moving objects F D = b v mg FDFD

Sample Problem: Slow moving objects Show that v T = mg/b Show that v T = mg/b mg F D = bv

Sample Problem: Slow Moving Objects Show that v(t) = (mg/b)(1 – e –bt/m ) Show that v(t) = (mg/b)(1 – e –bt/m ) mg F D = bv

Sample Problem: Fast moving objects Show that v T = (2mg / D  A) 1/2 Show that v T = (2mg / D  A) 1/2 mg F D = 1/2D  Av 2

Sample Problem: Fast moving objects Derive an expression for the velocity of the object as a function of time Derive an expression for the velocity of the object as a function of time mg F D = 1/2D  Av 2