Use Newton’s 1st law for bodies in equilibrium (statics) Goals for Chapter 6 Use Newton’s 1st law for bodies in equilibrium (statics) Use Newton’s 2nd law for accelerating bodies (dynamics) Study types of friction & fluid resistance Solve circular motion problems

Using Newton’s First Law when forces are in equilibrium VISUALIZE (create a coordinate system; decide what is happening?) SKETCH FREE-BODY DIAGRAM Isolate one point/body/object Show all forces in that coordinate system ON that body (not acting by that body!) LABEL all forces clearly, consistently Normal forces from surfaces Friction forces from surfaces Tension Forces from ropes Contact forces from other objects Weight from gravity

Using Newton’s First Law when forces are in equilibrium BREAK ALL applied forces into components based on your coordinate system. Apply Newton’s Laws to like components ONLY SFx = max; SFy = may

One-dimensional equilibrium: Tension in a massless rope A gymnast hangs from the end of a massless rope. Example (mg = 50 kg; what is weight & force on rope?)

One-dimensional equilibrium: Tension in a massless rope A gymnast hangs from the end of a massless rope. Example (mg = 50 kg; what is weight & force on rope?) Tension of Rope on Gymnast

One-dimensional equilibrium: Tension in a massless rope A gymnast hangs from the end of a massless rope. Example (mg = 50 kg; what is weight & force on rope?) Tension of Ceiling on Rope!

One-dimensional equilibrium: Tension in a rope with mass What is the tension in the previous example if the rope has mass? (Say weight of rope = 120 N)

Two-dimensional equilibrium A car engine hangs from several chains. Weight of car engine = w; ignore chain weights

A car on an inclined plane An car rests on a slanted ramp (car of weight w)

A car on an inclined plane Coordinate system choice #1: (y) parallel to slope & (x) perpendicular to slope y x a b

A car on an inclined plane Coordinate system choice #1: (y) parallel to slope & (x) perpendicular to slope y T (in x) x N (in y) W (in x & y) a b

A car on an inclined plane Coordinate system choice #2: (y) perpendicular to ground (x) parallel to ground N in both x & y! T (in x & y!) y a W (in y only) a x

A car on an inclined plane Coordinate system choice #2: (y) perpendicular to ground (x) parallel to ground N T y Ny Tx a Tx b Nx a W a x

Example: Bodies connected by a cable and pulley Cart connected to bucket by cable passing over pulley. Initially, assume pulley is massless and frictionless! Pulleys REDIRECT force – they don’t amplify or reduce. Tension in rope pulls upwards along slope SAME tension in rope pulls upwards on bucket

Example 5.5: Bodies connected by a cable and pulley Draw separate free-body diagrams for the bucket and the cart.

A note on free-body diagrams Only the force of gravity acts on the falling apple. ma does not belong in a free-body diagram! It is the SUM of all the forces you find! 

Ex: Straight-line motion with constant force Wind exerts a constant horizontal force on the boat. 4.0 s after release, v = 6.0 m/s; mass = 200 kg.

Ex: Straight-line motion with constant force Wind exerts a constant horizontal force on the boat. 4.0 s after release, v = 6.0 m/s; mass = 200 kg. Find W, the force of the wind!

Example: Straight-line motion with friction For the ice boat in the previous example, a constant horizontal friction force of 100 N opposes its motion What constant force needed by wind to create the same acceleration (a = + 1.5 m/s/s)?

Example: Straight-line motion with friction For the ice boat in the previous example, a constant horizontal friction force now opposes its motion (100N); what constant force needed by wind to create the same acceleration (a = + 1.5 m/s/s)?

Example: Tension in an elevator cable Elevator (800 kg) is moving downward @ 10 m/s but slowing to a stop over 25.0 m. What is the tension in the supporting cable?

Example: Tension in an elevator cable Elevator (800 kg) is moving downward @ 10 m/s but slowing to a stop over 25.0 m. What is the tension in the supporting cable?

Example: Tension in an elevator cable Compare Tension to Weight while elevator slows?

Example 5.8: Tension in an elevator cable Compare Tension to Weight while elevator slows? What if elevator was accelerating upwards at same rate?

Example 5.8: Tension in an elevator cable What if elevator was accelerating upwards at same rate? Same Free Body Diagram! Same result!

Ex 5.9 Apparent weight in an accelerating elevator A woman inside the elevator of the previous example is standing on a scale. How will the acceleration of the elevator affect the scale reading?

Ex 5.9 Apparent weight in an accelerating elevator A woman inside the elevator of the previous example is standing on a scale. How will the acceleration of the elevator affect the scale reading?

Ex 5.9 Apparent weight in an accelerating elevator What if she was accelerating downward, rather than slowing? Increasing speed down?

Acceleration down a hill What is the acceleration of a toboggan sliding down a friction-free slope?

Acceleration down a hill What is the acceleration of a toboggan sliding down a friction-free slope?

Two common free-body diagram errors  The normal force must be perpendicular to the surface. There is no separate “ma force.”

Two bodies with the same magnitude of acceleration The glider on the air track and the falling weight move in different directions, but their accelerations have the same magnitude and relative direction (both increasing, or both decreasing)

Two bodies with the same magnitude of acceleration What is the tension T, and the acceleration a, of the system?

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Friction forces are essential: Picking things up Walking, biking, driving anywhere Writing with a pencil Building with nails, weaving cloth © 2014 John Wiley & Sons, Inc. All rights reserved. 34

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction But overcoming friction forces is also important: Efficiency in engines (20% of the gasoline used in an automobile goes to counteract friction in the drive train) Roller skates, fans Anything that we want to remain in motion © 2014 John Wiley & Sons, Inc. All rights reserved. 35

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Three experiments: Slide a book across a counter. The book slows and stops, so there must be an acceleration Direction? parallel to the surface and opposite the direction of motion. © 2014 John Wiley & Sons, Inc. All rights reserved. 36

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Three experiments: Push a book at a constant speed across the counter. There must be an equal and opposite force opposing you, otherwise the book would accelerate. Again the force is parallel to the surface and opposite the direction of motion. © 2014 John Wiley & Sons, Inc. All rights reserved. 37

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Three experiments: Push a crate or other heavy object that does not move. To keep the crate stationary, an equal and opposite force must oppose you. If you push harder, the opposing force must also increase to keep the crate stationary. Keep pushing harder… Eventually the opposing force will reach a maximum, and the crate will slide. © 2014 John Wiley & Sons, Inc. All rights reserved. 38

© 2014 John Wiley & Sons, Inc. All rights reserved. When a body rests or slides on a surface, the friction force is parallel to the surface. Friction between two surfaces arises from interactions between molecules on the surfaces. © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Microscopic picture: surfaces are bumpy Friction occurs as contact points slide over each other Two specially prepared metal surfaces can cold-weld together and become impossible to slide, because there is so much contact between the surfaces Greater force normal to the contact plane increases the friction because the surfaces are pressed together and make more contact Sliding that is jerky, due to the ridges on the surface, produces squeaking/squealing/sound Figure 6-2 © 2014 John Wiley & Sons, Inc. All rights reserved. 40

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 TWO types of friction The static frictional force: The opposing force that prevents an object from moving Can have any magnitude from 0 N up to a maximum Once the maximum is reached, forces are no longer in equilibrium and the object slides The kinetic frictional force: The opposing force that acts on an object in motion Has only one value Generally smaller than the maximum static frictional force © 2014 John Wiley & Sons, Inc. All rights reserved. 41

© 2014 John Wiley & Sons, Inc. All rights reserved. Kinetic friction acts when a body slides over a surface. The kinetic friction force is fk = µkn. Static friction acts when there is no relative motion between bodies. The static friction force can vary between zero and its maximum value: fs ≤ µsn. © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. Before the box slides, static friction acts. But once it starts to slide, kinetic friction acts. © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. Before the box slides, static friction acts. But once it starts to slide, kinetic friction acts. © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 STATIC Friction The properties of static friction If the body does not move, then the applied force and frictional force balance along the direction parallel to the surface: equal in magnitude, opposite in direction The magnitude of fs has a maximum fs,max given by: where μs is the coefficient of static friction. If the applied force increases past fs,max, sliding begins. Eq. (6-1) © 2014 John Wiley & Sons, Inc. All rights reserved. 45

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Kinetic Friction The properties of kinetic friction Once sliding begins, the frictional force decreases to fk given by: where μk is the coefficient of kinetic friction. Magnitude FN of the normal force measures how strongly the surfaces are pushed together The values of the friction coefficients are unitless and must be determined experimentally Eq. (6-2) © 2014 John Wiley & Sons, Inc. All rights reserved. 46

© 2014 John Wiley & Sons, Inc. All rights reserved. Kinetic friction acts when a body slides over a surface. The kinetic friction force is fk = µkn. Static friction acts when there is no relative motion between bodies. The static friction force can vary between zero and its maximum value: fs ≤ µsn. Assume that μk does not depend on velocity Note that these equations are not vector equations © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction © 2014 John Wiley & Sons, Inc. All rights reserved. 48

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Answer: (a) 0 (b) 5 N (c) no (d) yes (e) 8 N © 2014 John Wiley & Sons, Inc. All rights reserved. 49

Some approximate coefficients of friction

Friction in horizontal motion Move a 500-N crate across a floor with friction by pulling with a force of 230 N. Initially, pull harder to get it going; later pull easier (at 200N once it is going). What are ms? mk?

Friction in horizontal motion Before the crate moves, static friction acts on it.

Friction in horizontal motion After it starts to move, kinetic friction acts.

Static friction can be less than the maximum Static friction only has its maximum value just before the box “breaks loose” and starts to slide. Force Force builds in time to maximum value, then object starts moving & slipping Time

Pulling a crate at an angle The angle of the pull affects the normal force, which in turn affects the friction force.

Motion on a slope having friction – ex 5.16 Consider a toboggan going down a slope at constant speed. What is m? Now consider same toboggan on steeper hill, so it is now accelerating. What is a?

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Example For a force applied at an angle: STEP 1: COORDINATES STEP 2: COMPONENTS! Decompose the force into x and y components STEP 3: No Fruit Salad! Balance the vertical components (FN, Fg, Fy) Balance the horizontal components (f, Fx) STEP 4: You’ll need Normal to get Friction! Note FN and f are related © 2014 John Wiley & Sons, Inc. All rights reserved. 57

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-1 Friction Example For a force applied at an angle: STEP 1: COORDINATES STEP 2: COMPONENTS! Decompose the force into x and y components STEP 3: No Fruit Salad! Balance the vertical components (FN, Fg, Fy) Balance the horizontal components (f, Fx) STEP 4: You’ll need Normal to get Friction! Note FN and f are related © 2014 John Wiley & Sons, Inc. All rights reserved. 58

© 2014 John Wiley & Sons, Inc. All rights reserved. Example: 6.01 Angled Force to Stationary Block Force of 12.0 N applied to 8 kg block at downward angle of 30 degrees; static friction coefficient is ms = 0.700, and kinetic friction coefficient is mk = 0.400. Does it slide? If so, what is the acceleration? If not, what is the friction force? © 2014 John Wiley & Sons, Inc. All rights reserved.

Example: 6.01 Angled Force to Stationary Block Force of 12.0 N applied to 8 kg block at UPWARD angle of 30 degrees; static friction coefficient is ms = 0.700, and kinetic friction coefficient is mk = 0.400. Now does it slide? If so, what is the acceleration? If not, what is the friction force?

© 2014 John Wiley & Sons, Inc. All rights reserved. Sliding on ice… How far does it take for a car moving at a constant speed of 10 m/s to slide to a stop on a horizontal road if the coefficient of kinetic friction is 0.60? If it is on ice (mk = 0.10?) If it is on a icy HILL (with slope = 20 degrees)? © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 The Drag Force and Terminal Speed A fluid is anything that can flow (gas or liquid) When there is relative velocity between fluid and an object there is a drag force: Opposes the relative motion Points along direction of flow, relative to the body © 2014 John Wiley & Sons, Inc. All rights reserved. 62

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 The Drag Force and Terminal Speed Here we examine the drag force for Air With a body that is not streamlined For motion fast enough that the air becomes turbulent (breaks into swirls) © 2014 John Wiley & Sons, Inc. All rights reserved. 63

Fluid resistance and terminal speed Fluid resistance on a body depends on the speed of the body. Resistance can depend upon v or v2 and upon the shape moving through the fluid. Fresistance = -kv or - Dv2 These will result in different terminal speeds

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 The Drag Force and Terminal Speed For this case, the drag force is: Where: v is relative velocity ρ is air density (mass/volume) C is experimentally determined drag coefficient A is effective cross-sectional area of body ( area taken perpendicular to the relative velocity) Drag force Weight © 2014 John Wiley & Sons, Inc. All rights reserved. 65

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 The Drag Force and Terminal Speed For this case, the drag force is: Where: v is relative velocity (m/s) ρ is air density (kg/m3) C is drag coefficient (#) A is cross-sectional area (m2) Units: kg/m3 * m2 * (m/s)2 = kg m/s2 = N = Force! Drag force Weight © 2014 John Wiley & Sons, Inc. All rights reserved. 66

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 The Drag Force and Terminal Speed The drag force from the air opposes a falling object Once the drag force equals the gravitational force, the object falls at a constant terminal speed: Terminal speed can be increased by reducing A Terminal speed can be decreased by increasing A Skydivers use this to control descent © 2014 John Wiley & Sons, Inc. All rights reserved. 67

Fluid resistance and terminal speed A falling body reaches its terminal speed when resisting force equals weight of the body. If F = -kv for a falling body, vterminal = mg/k If F = - Dv2 for a falling body, vterminal = (mg/D)½

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 The Drag Force and Terminal Speed Example Speed of a rain drop: Spherical drop feels gravitational force F = mg: Express in terms of density of water Use A = πR2 for the cross-sectional area © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 The Drag Force and Terminal Speed Example Speed of a rain drop: So plug in to the terminal velocity equation using the values provided in the text: © 2014 John Wiley & Sons, Inc. All rights reserved.

Dynamics of circular motion If something is in uniform circular motion, both its acceleration and net force on it are directed toward center of circle. The net force on the particle is Fnet = mv2/r, always towards the center.

Using Newton’s First Law when forces are in equilibrium IF you see uniform circular motion… Radius r Tension T v

Using Newton’s First Law when forces are in equilibrium IF you see uniform circular motion… THEN remember centripetal force is NOT another force – it is the SUM of one or more forces already present! SFx = mv2/r NOT mv2/r + T – mg = ma Radius r Tension T v

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-3 Uniform Circular Motion Centripetal force is not a new kind of force, it is simply an application of force Eq. (6-18) For the puck on a string, the string tension supplies the centripetal force necessary to maintain circular motion Figure 6-8 © 2014 John Wiley & Sons, Inc. All rights reserved. 74

Avoid using “centrifugal force” Figure (a) shows the correct free-body diagram for a body in uniform circular motion.

Avoid using “centrifugal force” Figure (b) shows a common error. In an inertial frame of reference, there is no such thing as “centrifugal force.”

Force in uniform circular motion A 25 kg sled on frictionless ice is kept in uniform circular motion by a 5.00 m rope at 5 rev/minute. What is the force?

Force in uniform circular motion A 25 kg sled on frictionless ice is kept in uniform circular motion by a 5.00 m rope at 5 rev/minute. What is the force?

What if the string breaks? If the string breaks, no net force acts on the ball, so it obeys Newton’s first law and moves in a straight line.

A conical pendulum A bob at the end of a wire moves in a horizontal circle with constant speed.

A car rounds a flat curve A car rounds a flat unbanked curve. What is its maximum speed?

6-3 Uniform Circular Motion Recall that circular motion requires a centripetal acceleration Examples You are a passenger: For a car, rounding a curve, the car accelerates toward the center of the curve due to a centripetal force provided by the inward friction on the tires. Your inertia makes you want to go straight ahead so you may feel friction from your seat and may also be pushed against the side of the car. These inward forces keep you in uniform circular motion in the car. 82

A car rounds a flat curve A car rounds a flat unbanked curve. What is its maximum speed?

A car rounds a banked curve At what angle should a curve be banked so a car can make the turn even with no friction?

A car rounds a banked curve At what angle should a curve be banked so a car can make the turn even with no friction?

Maximum Speeds for Cornering on Banked Curves Maximum Cornering Speed for Banked Curves Radius (m) 30 40 50 60 Coefficient of Static Friction 0.8 0.6 0.4 0.2 Angle (degrees) 20 10 5   max speed (m/s) 27 22 18 15 23 19 16 21 11 max speed (mph) 48 34 51 43 35 41 46 24 © 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-3 Uniform Circular Motion Example Car in a banked circular turn: Sum components along the radial direction: Sum components along the vertical direction: Divide and replace (sin θ)/(cos θ) with tangent. Figure 6-11 Eq. (6-23) Eq. (6-24) © 2014 John Wiley & Sons, Inc. All rights reserved. 87

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-3 Uniform Circular Motion © 2014 John Wiley & Sons, Inc. All rights reserved. 88

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-3 Uniform Circular Motion Answer: (a) accel downward, FN upward (b) accel upward, FN upward (c) the magnitudes must be equal for the motion to be uniform (d) FN is greater in (b) than in (a) © 2014 John Wiley & Sons, Inc. All rights reserved. 89

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-3 Uniform Circular Motion Example Bicycle going around a vertical loop: At the top of the loop we have: Solve for v and plug in our known values, including FN = 0 for the minimum answer: Figure 6-9 Eq. (6-19) © 2014 John Wiley & Sons, Inc. All rights reserved. 90