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Dynamics

Slide 2 / 139 Table of Contents

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· Sliding Blocks · Fixed Axis Pulley · Suspended Pulleys · The Banked Curve · Plumb bob in car · Inclined Plane and a Pulley · Falling objects with Air Resistance · Non Uniform Circular Motion · Introduction

Slide 3 / 139

Introduction

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Slide 4 / 139 Introduction

This is not a typical chapter presentation. There are no multiple choice or free response questions to test student knowledge. Rather, eight comprehensive Dynamics questions are solved, step by step. Questions are built into the material - you should try to answer them before you move onto the next slide. These can be done in class, led by the teacher, or they can be done by the students outside of class.

Slide 5 / 139

Sliding Blocks

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Slide 6 / 139

Sliding Blocks

B Ground A Block B, of mass 36 kg, is about to be pulled across the ground by a rope with a force of Fapp. The coefficient of kinetic friction between block B and the ground is 0.27 . Block A, of mass 42 kg is resting on Block B with a coefficient of static friction of 0.71 between the two blocks.

Slide 7 / 139 Sliding Blocks

B Ground A We're going to use dynamics concepts to find the maximum force that may be applied by the rope before block A starts sliding on block B.

Slide 8 / 139 Sliding Blocks

B Ground A Once someone starts pulling on the rope, what do we expect will happen?

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Sliding Blocks

B Ground A Block B will start accelerating to the right, as it now as a net applied force to the right. The friction between the ground and block B will oppose the applied force, so that will reduce the acceleration. What about block A?

Slide 10 / 139 Sliding Blocks

B Ground A The most important thing to first notice about block A is that the applied force is NOT acting on block A, and when the free body diagrams are drawn, this force will not be shown on block A! Intuitively, we know that something will be acting in the x direction on block A. Block A will either stay right where it is, relative to block B, and move to the right, or it will start sliding to the left. Relative to the ground, it will be moving to the right in both cases. If not the force due to the rope, what is this force?

Slide 11 / 139 Sliding Blocks

B Ground A The static friction force. Let's take two limiting cases to explain this. If μs = 0, then that implies a frictionless surface between blocks A and B. When B is pulled to the right, block A would just stay where it is relative to the ground, and would slide to the left on block B. If μs were infinitely large, then it would just stick to block B and slide with it to the right.

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Sliding Blocks

B Ground A

Let's formally start the problem: Given: where μs is between blocks A and B and μk is between block B and the ground. Find: the maximum Fapp before Block A starts sliding on Block B.

Slide 13 / 139 Sliding Blocks

B Ground A

As in all dynamics problems, the first step is to draw the free body

• diagram. In this case, there are two objects of interest, blocks A and

B, so there should be two free body diagrams (FBD). But, they will not be the ones you'd expect! Here's a clue - there will not be a FBD for block A and a FBD for block B. We will define different systems - two of them. And draw FBDs for each one. Take a few minutes with your group and propose two systems.

Slide 14 / 139 Sliding Blocks

B Ground A

System 1 will be Block A. System 2 will be Blocks A and B. Given this starting point, try drawing free body diagrams for both systems. Remember - when working with a system, only the external forces on the system are shown.

A B A

System 1 System 2

Slide 15 / 139

Sliding Blocks

Here they are. Don't worry if you didn't get these exactly right. Each system will be now be analyzed in detail. FNA mAg fs a FN(A+B) (mA+mB)g Fapp fk a System 1 System 2

A B A

System 1 System 2

Slide 16 / 139 Sliding Blocks

Let's look at System 1 first. The Normal force and the force due to gravity are pretty straight forward. Gravity pulls down on block A (m

Ag), and block B

exerts an upwards Normal force (FNA) on block A. Note the additional subscript on the Normal force. That is to distinguish it from the Normal force acting on System 2. FNA mAg fs a But what of the static friction force? Why is it pointed to the right? Doesn't the static friction force oppose motion? Isn't block A moving to the right?

A

Slide 17 / 139 Sliding Blocks

B Ground A

Static friction seeks to maintain the relative position of the objects that are initially at rest. In most cases, this means if a force is applied to an object in one direction, then the static friction force is in the opposite direction. But, here, the applied force is not on block A, it is on block B. And for block A to maintain its position relative to block B, as block B starts moving to the right, the static friction force acts to the right. This results in block A staying on the right edge of block B - their relative positions don't change. FNA mAg fs a

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Sliding Blocks

B Ground A

Another way to look at this problem is using your intuition again. Again, that doesn't always work in physics, but it helps here. If you pull block B, your experience tells you that block A will move to the right. Maybe not as fast as block B is accelerating, but it will move to the right. That requires a Force (Newton's Second Law). The only force available in the x direction is the static friction force between block A and B. So, f

s goes on the free body diagram

pointing to the right! FNA mAg fs a

Slide 19 / 139 Sliding Blocks

B A

Time for System 2 now. Gravity is acting on the combination of blocks A and B, so it is equal to (mA + mB)g. The ground exerts an upward Normal force on the blocks, and is represented as FN(A+B). Note the additional subscript on the Normal force. That is to distinguish it from the Normal force in System 1. FN(A+B) (mA+mB)g Fapp fk a Next, we'll look at the forces in the x direction.

Slide 20 / 139 Sliding Blocks

B A

Fapp is acting to the right, and fk acts between the ground and block B and is directed to the left. The static friction force between blocks A and B is not drawn here. Why? FN(A+B) (mA+mB)g Fapp fk a

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Sliding Blocks

B A

System 2 was defined as the combination of blocks A and B. Therefore, the static friction force between blocks A and B is an internal force. Internal forces are not shown on FBDs (Discuss in your groups how Newton's Third Law comes into play here). FN(A+B) (mA+mB)g Fapp fk a

Slide 22 / 139 Sliding Blocks

Time to apply Newton's Second Law to the FBDs. We'll start with System 1. FNA mAg fs a

A

System 1 x direction y direction Took a little shortcut on the notation - since the objects are moving in the x direction, the subscript, x, will be

• mitted on the acceleration vector.

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Slide 25 / 139 Sliding Blocks

System 2 FN(A+B) (mA+mB)g Fapp fk a FNA mAg fs a System 1

System 2 equations: This gives us an expression for the acceleration of the total system (blocks A and B). If we want to find the point at which block A starts to slide, what assumption can we make about the acceleration of block A?

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System 2 FN(A+B) (mA+mB)g Fapp fk a FNA mAg fs a System 1

Switching to the System 1 equations, for block A not to slide, its acceleration must be equal to the acceleration of the total system, so aA = aA+B. Why? If aA is greater than aA+B, then block A would slide forward

• n block B. If aA is less than aA+B then it would slide
• backwards. So, since we're looking for the exact point

where the applied force causes A to slide, we have:

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Sliding Blocks

System 2 FN(A+B) (mA+mB)g Fapp fk a FNA mAg fs a System 1

Calculating fs from the Normal force: Putting these two equation together and canceling mA:

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Slide 31 / 139

Fixed Axis Pulley

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Slide 32 / 139 Fixed Axis Pulley

So far, pulleys have been used in constructions like Atwood's Machine and to allow for the horizontal and vertical motion of two objects in a system (the force is redirected) as shown below. Both of these systems used "fixed axis pulleys" as the pulley's axis of rotation does not change relative to a reference frame attached to the ground or table or ceiling.

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Fixed Axis Pulley

Assume a massless pully and string, and that there is no friction between the pulley and its axle and the string. This allows us to ignore the effects of rotational motion in the pulley. In both cases below, the tension force on either side of the pulley is the same. And, for the Atwood's machine - if the block on the left weighed 100 N, you would need a force on the right side of 100 N to keep it stationary.

Slide 34 / 139 Fixed Axis Pulley

Let's use a fixed axis pulley, but try something different. In the below picture, two window cleaners are sitting in harnesses. The picture doesn't show it, but the two lines from each harness loop over a fixed pulley on top of the building.

Image:Seattle_Public_Library_window_washers_06.jpg

Assuming the window cleaner on the left has a mass of 82.0 kg and the harness is 21.0 kg what force does he need to exert to stay stationary? (Of course, there are safety features associated with the harness, but to simplify the problem, we'll assume that the cleaner needs to exert a force to stay where he is).

Slide 35 / 139 Fixed Axis Pulley

As in all dynamics problems, we'll start with free body diagrams. What forces are acting on the window cleaner and the harness? The rope that he is holding with his hands is pulling him up. The harness seat is pushing him up with a Normal force (and the window cleaner is pushing down on the harness with a normal force). The rope that is attached to the harness is pulling up on the harness and the window cleaner. Gravity is pulling down on the window cleaner and the harness. Now, draw FBDs for the window cleaner, the harness and the combined system of the two.

Slide 36 / 139

Fixed Axis Pulley

In order from left to right, the FBDs for: harness, window cleaner, combined system of harness plus window cleaner. The Normal forces between the window cleaner and the harness cancel out in the combined system - they are internal forces and do not affect the motion of the system. The tension force is the force the window cleaner needs to exert to stay motionless - what do you notice about this force?

FT mhg FN FT mwcg FN FT mwc+hg FT

Slide 37 / 139 Fixed Axis Pulley

There are two Tension forces acting on the system! The window cleaner is holding on to the right rope with a force that, by Newton's Third Law, is balanced by the Tension force in the rope. The two Tension forces are equal to each other, since the rope is assumed to be massless and is not stretching or compressing - so the force that the window cleaner is pulling down on the right transmits itself throughout the rope. What's the next step? Think Newton.

FT mwc+hg FT

Slide 38 / 139 Fixed Axis Pulley

Apply Newton's Second Law to the system. Pretty straightforward. But what's very interesting about this result? What's the weight

• f the window cleaner and the harness?

FT mwc+hg FT

Slide 39 / 139

Fixed Axis Pulley

The weight of the system is 1009 N (W = mg). But, because of the pulley configuration, the window cleaner only needs to exert half of the force to overcome the gravitational force. Why? Doesn't this look like a violation of the conservation of energy? The work required to lift an object a distance Δy is: But with this configuration, the window cleaner is exerting half of the required force - F/2. How is the window cleaner still rising a distance of Δy? Think about how far the rope is travelling, d (as that is the distance the force is acting over).

Slide 40 / 139 Slide 41 / 139 Fixed Axis Pulley

Now that we've shown that the reduced force obtained by this arrangement requires that it be exerted over a greater distance (and energy is conserved), let's move on to another question: The window cleaner pulls down on the rope, and along with the harness, he accelerates upward at a rate of 0.330 m/s2. What force must he pull with, and what is the force that he now exerts on the harness?

FT mwc+hg FT

Slide 42 / 139

Fixed Axis Pulley

Let's start with the system FBD to find the Tension in the rope (which is equal to the window cleaner's pulling force) and then pick either of the individual FBDs to calculate the Normal force (the force that the window cleaner exerts on the harness).

FT mhg FN FT mwcg FN FT mwc+hg FT

Slide 43 / 139 Slide 44 / 139 Fixed Axis Pulley

We'll now solve both the harness and the window cleaner FBD to show we get the same result for the normal force. First, the harness system.

FT mhg FN FT mwcg FN FT mwc+hg FT

Slide 45 / 139

Fixed Axis Pulley

Now, the window cleaner system.

FT mhg FN FT mwcg FN FT mwc+hg FT

That's as we expected, FN = 309 N as calculated for both systems.

Slide 46 / 139 Fixed Axis Pulley

Where the pulley merely redirected the force, there was no benefit, in terms of reduced force required, obtained. The benefit was that the person lifting the object could stand in the best place to move the object - he didn't have to stand under it. But, an advantage was gained in the second case - the force required to lift the object was reduced - but had to be exerted

• ver a greater distance.

We're now going to move on to the case of the suspended pulley where the pulley's axis is not fixed and see if it can reduce the lifter's required force even more.

Slide 47 / 139

Suspended Pulleys

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Slide 48 / 139

Suspended Pulleys

First, let's define what a Suspended Pulley is! The fixed axis pulley's axis of rotation (axle) did not change relative to the ground or other stationary reference frame. In

• ther words, the pulley didn't move.

The rope moved, the load moved, but not the pulley. So, what could a suspended pully do?

Slide 49 / 139 Suspended Pulleys

It can move. It will move up or down and take the load with it. There are two systems of multiple pulleys here. Picture a crate attached to the ring on the bottom of the bottom pulley. Which is the fixed pulley? Which is the supended pulley?

Slide 50 / 139 Suspended Pulleys

It can move. It will move up or down and take the load with it. The top system is attached to the boom overhead and is a fixed pulley. The lower one is the one with suspended pulleys.

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Suspended Pulleys

Below is a schematic of fixed and suspended pulleys that will be used to lift the mass at the bottom by applying the force to the right. M A B C Fapp Identify the fixed and the suspended pulleys: Pulleys A and C are fixed. Pully B is suspended, as it will travel with the mass.

Slide 52 / 139 Suspended Pulleys

The problem at hand is to find the required applied force to lift the mass M. M A B C Fapp This may look tricky, but we'll be using free body diagrams to figure out the required force. First, label this picture with all the forces you can think of. Remember, we're dealing with massless pulleys and strings.

Slide 53 / 139 Suspended Pulleys

Here they are - the lines do not represent the magnitudes

• f the various forces - they're

just a schematic diagram - we don't have a free body diagram yet. The next slide will show the

• FBDs. Give them a try before

the slide changes. M A B C FT1 FT2 FT3 FT5 FT4 FT6 FT7 Fapp Mg

Slide 54 / 139

Suspended Pulleys

M A B C FT1 FT2 FT3 FT5 FT4 FT6 FT7 Fapp Mg A FT1 FT2 FT3 M FT4 Mg B FT5 FT7 FT4 FT3 C FT6 FT5 FT7 Here they are! Next, Newton's Second and Third Laws and the massless string and frictionless pulley assumptions will be used to relate the forces to each

• ther.

Slide 55 / 139 Suspended Pulleys

M A B C FT1 FT2 FT3 FT5 FT4 FT6 FT7 Fapp Mg The massless string and frictionless pulley assumptions allow us to say: Fapp = FT1 = FT3 = FT5 = FT7 Why? Because the string is not stretching or compressing, it has no mass to consider in the FBD, and the pulleys have no mass, nor do they use any energy in their rotation. So, the tension is the same everywhere in the string.

Slide 56 / 139 Suspended Pulleys

A FT1 FT2 FT3 M FT4 Mg B FT5 FT7 FT4 FT3 C FT6 FT5 FT7 Assume the mass is stationary - it is not

• accelerating. Write Newton's Second Law

equations for each FBD. Note how Newton's Second Law tells us which action-reaction forces are equal. The equations are on the next slide. Try them yourself first.

Slide 57 / 139

Suspended Pulleys

A FT1 FT2 FT3 M FT4 Mg B FT5 FT7 FT4 FT3 C FT6 FT5 FT7 FT6 - FT5 - FT7 = may = 0 FT2 - FT3 - FT1 = may = 0 FT3 + FT5 + FT7 - FT4 = may = 0 FT4 - Mg= may = 0

Slide 58 / 139 Suspended Pulleys

M A B C FT1 FT2 FT3 FT5 FT4 FT6 FT7 Fapp Mg Fapp = FT1 = FT3 = FT5 = FT7 FT4 - Mg = may = 0 FT3 + FT5 + FT7 - FT4 = may = 0 We actually only need three of these equations: Combine the first and third to get: Substitute this into the second equation: FT4 = 3Fapp 3Fapp = Mg

Slide 59 / 139 Suspended Pulleys

M A B C FT1 FT2 FT3 FT5 FT4 FT6 FT7 Fapp Mg One more step: Fapp = Mg/3 Thus it only takes one third of the force to hold mass M stationary than if it was being held up just by a rope.

Slide 60 / 139

Plumb bob in car

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Slide 61 / 139 Plumb bob in car

First, we need to describe a Plumb bob. It is a weight (normally pointed at the bottom) that is attached to a string that is then affixed to a horizontal support. What would be the purpose of such an object? A free body diagram (FBD) is a great way to figure that out. What are the forces acting on the Plumb bob?

Slide 62 / 139 Plumb bob in car

The only forces acting on the weight (bob) are the gravitational force, mg, and the Tension in the string, F

• T. Both forces act in

the y direction, and there are no forces in the x direction. So, the point of the bob points down, in a perfectly vertical direction, which gives a perfect reference line. This is helpful to people who want to build a house that is straight or any other constructions that need to be aligned with the y axis (wallpaper in a room, a skyscraper, etc.). FT mg

Slide 63 / 139

Plumb bob in car

FT mg Assume the bob is not moving, and is ready to be used as a reference line. Let's apply Newton's Second Law.

Slide 64 / 139 Plumb bob in car

For physics purposes, let's now attach this plumb bob to the

• verhead light in a car and accelerate the car forward in a safe
• manner. Will the plumb bob stay pointing in a vertical direction?

Think of what happens to an object on the center console of a car when the car accelerates forward.

Slide 65 / 139 Plumb bob in car

From the driver's perspective, an object on a center console will slide backwards. Since the plumb bob is attached to the

• verhead light, it will swing backwards, as seen by the driver.

An inertial observer, standing outside on the road, will see the plumb bob swing backwards. The problem will now be solved from the inertial observer's point reference frame, where the car is accelerating. What does the FBD look like now? a

Slide 66 / 139

Plumb bob in car

The gravitational force is still the same. FT has x and y

• components. The bob is no longer pointed vertically down.

What is the source of this force in the x direction? When the car was not moving, the bob pointed straight down. Now that the car is accelerating to the left, the bob has moved. So, the force is related to the car's acceleration and the original Tension force. Since there are now non perpendicular forces in two dimensions, it's time for vector resolution. FT mg a a

Slide 67 / 139 Plumb bob in car

We're going to find another use for the plumb bob - once the forces are resolved, the acceleration of the car will be found without using any other car gauges! The angle θ, which is the angle between the vertical and the string of the Plumb bob is measured. What are the x and y components of the Tension? Assume the acceleration is constant, so the bob is stationary in this new position.

FT mg

θ

FTy FTx a

Slide 68 / 139 Plumb bob in car

FT mg

θ

FTy FTx a

x direction y direction Let's stack these two equations on top of each other: What can you say about these two equations, and does anything about the trigonometry jump out at you?

Slide 69 / 139

Plumb bob in car

FT mg

θ

FTy FTx a

These equations define the motion of the plumb bob in two dimensions, at the same point of time, so they must both be true. Hence, they are simultaneous equations. Many times, you add/subtract simultaneous equations to get the solution. But, in this case - what about dividing? This will consolidate the trigonometric functions, and eliminate the mass of the bob from the solution.

Slide 70 / 139 Plumb bob in car

FT mg

θ

FTy FTx a

Whenever you see a sine and cosine function - think of dividing the equations. This is a very handy trick.

Slide 71 / 139 Plumb bob in car

FT mg

θ

FTy FTx a

Just by measuring the angle that the string makes with the vertical, the acceleration of the car can be calculated. Note how the mass of the plumb bob does not matter. If θ = 110, what is the acceleration of the car?

Slide 72 / 139

Plumb bob in car

FT mg

θ

FTy FTx a

This is about as fast as a four cylinder compact car can accelerate when it is trying as hard as it can. Hopefully, you'll never push the car to this acceleration, so if you have a plumb bob attached to the interior overhead light, it should never exceed an angle of 110.

Slide 73 / 139

The Banked Curve

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Slide 74 / 139 The Banked Curve

An interesting problem that involves centripetal acceleration, a non rotated coordinate system that looks like it should be rotated and static and kinetic friction, is the case of a car on a banked, curved roadway. θ

http://commons.wikimedia.org/wiki/File: 1968_AMC_Ambassador_DPL_station_w agon_FL-r1.jpg

Slide 75 / 139

The Banked Curve

First thing that we'll do is to prepare for the FBD. The photograph

• f the 1968 station wagon (ask your parents what those were) will

be replaced by a shape. The car has just entered a turn - which can be modeled as a circular orbit when seen from above. It is also on a banked curve - and the bank makes an angle θ with the flat ground. θ v Car making a turn as seen from above Car seen from behind on banked curve Draw the accleration vector on each sketch.

Slide 76 / 139 The Banked Curve

θ Car seen from behind on banked curve a v Car making a turn as seen from above a First, assume there is no static friction between the car and the road and draw the FBD for the sketch on the right. This is done not just to make the math easier, but it actually is the most conservative case to ensure the car does not slide up or down the incline (both cases involve leaving the roadway, which is clearly not safe).

Slide 77 / 139 The Banked Curve

θ a FN mg

x y

The first surprise, is even though we're dealing with an inclined plane, we're not going to rotate the coordinate system! Please discuss this within your group and suggest an explanation. Also, what kind of acceleration is represented by the vector, a, above?

Slide 78 / 139

The Banked Curve

θ a FN mg

x y

Certainly, one of the advantages of a rotated coordinate system is to reduce the amount of vectors that need to be resolved into component parts. But, even more important, is to line up the coordinate system so the acceleration is along one of its axes. And that's what's done above. Since the car is going in a circle - it's Centripetal Acceleration. What force is causing the centripetal acceleration?

Slide 79 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

The Banked Curve

By resolving the Normal force into its x and y components, it can be seen that Nsinθ points towards the center of the circle that the car is driving in, and is causing the centripetal

• acceleration. Take a minute to verify that the angle in the

triangle made by the Normal force is the same as the angle of inclination of the banked road. Next step? Newton's Second Law. In the x and the y directions.

Slide 80 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

The Banked Curve

First we'll write Newton's Second Law for the y direction, and assume the car stays on the road surface and isn't bouncing up and down so ay = 0. Why was the y direction solved first?

Slide 81 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

The Banked Curve

Because we'll need the value of FN for Newton's Second Law in the x direction. What is the significance of this result? And what property of the car is missing?

Slide 82 / 139 Slide 83 / 139 The Banked Curve

Now, let's bring in friction. When a car negotiates a turn, it's best that it stays in the middle of the road. If it moves up the ramp - well, then it can fail to make the turn and it will go flying off where it could hit trees or other

• bstructions or be stopped by a barrier of some sort.

If it moves down the ramp, then it could hit whatever problems await it on the inner curve - such as barriers, parked cars or the like. What type of friction is relevant here?

Slide 84 / 139

The Banked Curve

Static Friction. This is covered in the Rotational Motion unit of this course. But here's a quick explanation. A tire rotates as it moves the car forward. When the tire comes in contact with the road, the linear velocity of the tire is pointed towards the rear of the car. It is exactly equal to the linear velocity of the entire car. These velocities add to zero - so the tire is at rest with the roadway at the contact point. Hence - no relative motion - static friction!

Slide 85 / 139 The Banked Curve

First, find the maximum speed that a car can travel at without sliding up the bank as it goes around a banked curve. After this is worked out, the problem of what minimum speed a car will need to prevent it from slipping down the bank will be presented. Let's build on the previous work that left out friction. Here it is: θ mg

x y

FN a θ

FNcosθ FNsinθ

Is this still relevant? Does anything need to be added or subtracted?

Slide 86 / 139 The Banked Curve

We don't need to subtract anything. The acceleration vector still points towards the center of the curve. The non rotated coordinate system will still work. The Normal vector is still

• valid. But, the frictional force needs to be added.

The car needs to be prevented from sliding up the bank. What direction is the force of friction? θ mg

x y

FN a θ

FNcosθ FNsinθ

Slide 87 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf

The Banked Curve

Static friction works to maintain the relative position of the two

• bjects. So, to prevent the car from moving up the bank, the

friction force must point down the bank as shown below. What angle does the static friction force make with the negative x axis? Once you figure this out, resolve the force into its x and y components.

Slide 88 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf θ fsfcosθ fsfsinθ The Banked Curve

The angle is θ. Here's the completed FBD. Next, work out Newton's Second Law equations, starting, as always, with the y direction.

Slide 89 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf θ fsfcosθ fsfsinθ

The Banked Curve

y direction: assume the car does not bounce up and down, and fsf = μsfFN. The inequality is removed here, since the maximum value of the static friction force is the point at which the car will start slipping.

Slide 90 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf θ fsfcosθ fsfsinθ

The Banked Curve

x direction: work out Newton's Second Law, and then substitute in the value of FN obtained from the y direction equation.

Slide 91 / 139 The Banked Curve

Now, let's consider the minimum speed the car needs to travel at to prevent it from sliding down the bank and running off the road there. As in the previous case, its the force of static friction. But, which way is the static friction force this time? θ mg

x y

FN a θ

FNcosθ FNsinθ

Slide 92 / 139

θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf

The Banked Curve

Static friction works to maintain the relative position of the two

• bjects. So, to prevent the car from moving down the bank,

the friction force must point up the bank as shown below. It' now time to resolve the friction force into its x and y components.

Slide 93 / 139

The Banked Curve θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf θ fsfcosθ fsfsinθ

Here's the completed FBD. Next, work out Newton's Second Law equations, starting, as always, with the y direction. Can you see why the angle between fsf and the x axis is θ?

Slide 94 / 139 The Banked Curve

y direction: assume the car does not bounce up and down, and fsf = μsfFN. The inequality is removed here, since the maximum value of the static friction force is the point at which the car will start slipping. θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf θ fsfcosθ fsfsinθ

Slide 95 / 139 The Banked Curve

x direction: work out Newton's Second Law, and then substitute in the value of FN obtained from the y direction equation. θ mg

x y

FN a θ

FNcosθ FNsinθ

fsf θ fsfcosθ fsfsinθ

Slide 96 / 139

Inclined Plane and a Pulley

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Slide 97 / 139 Inclined Plane and a Pulley

#

Let's put together two concepts that have been covered in the Dynamics Chapter notebook (inclined plane) and in this Applications of Newton's Laws notebook (the pulley).

Slide 98 / 139 Inclined Plane and a Pulley

#

Assume the two masses are connected by a string of negligible mass and does not stretch or compress. Also assume a frictionless and massless pulley, so that rotational effects need not be considered. Find the acceleration of the two masses on the incline, where there is kinetic friction between the box and the incline.

Slide 99 / 139

Inclined Plane and a Pulley

#

Draw all the forces you can think of on the two boxes and show the acceleration vectors for each box. What if you choose the wrong direction for the acceleration?

Slide 100 / 139 Inclined Plane and a Pulley

FN m2g #

a

fk m1g

a

FT FT

Here it is. And if you chose a different direction for the acceleration than the actual acceleration - no problem - you will get a negative value for the acceleration after the equations are solved, which tells you it's opposite to your choice. Next step - draw the free body diagrams for each box.

Slide 101 / 139 Inclined Plane and a Pulley

FT a Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk

FT

There is a lot going on here. The next few slides will explain the free body diagrams in more detail. Each mass will be treated as its own system, and then it will be shown how they can be considered as one system of two masses. System 1 System 2

Slide 102 / 139

Inclined Plane and a Pulley

FT a

System 1 is the m1 block. We're assuming that the block is falling, hence the direction of a is down. Once we solve the Newton's Second Law equations, if we get a negative value for a, that just means it's actually accelerating in the opposite direction (up). This one is fairly straightforward - gravity is pulling down on the mass and the tension in the string is pulling it up. System 1

Slide 103 / 139 Inclined Plane and a Pulley

Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk

FT

This system is a bit more complex. It involves a rotated coordinate system and the resolution of the gravitational force along the new rotated axes. For a review of this algebra, please refer to the Dynamics Chapter Notebook presentation on the Inclined Plane. Acceleration is chosen in the up direction on the plane. While it isn't important which direction is chosen, it is important to choose the directions for both boxes that are consistent! If box m1 is accelerating downwards, then box m2 must be accelerating up the plane as the boxes are connected. System 2

Slide 104 / 139 Inclined Plane and a Pulley

Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk

FT

The friction force is shown opposite the direction of acceleration - of the relative motion of the box and the plane. The tension force is shown acting upwards. What is the relationship of this tension force with the tension force shown for System 1? System 2 Since the string is massless and it is assumed not to stretch, the two tension forces are equal.

Slide 105 / 139

Inclined Plane and a Pulley

FT a Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk

FT

Time for the application of Newton's Second Law and writing the equations for these free body

• diagrams. Try System 1 first.

System 1 System 2

Slide 106 / 139 Inclined Plane and a Pulley

FT a

System 1 That's all for this system. The only thing to watch out for is that this is a non rotated coordinate system - and ay for this system will actually equal ax for the box in System 2! Flip back to the previous slide to see this.

Slide 107 / 139 Inclined Plane and a Pulley

Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk

FT

y-axis: This will be used to calculate fk: x-axis: System 2 We now have two simultaneous equations for FT and a. What are they?

Slide 108 / 139

Inclined Plane and a Pulley

Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk FT System 2

FT a

System 1

Here they are. Two equations. Earlier, it was stated that the acceleration of both boxes was the same - even though they are named differently due to the coordinate system being rotated in System 2. So, ax = ay = a. What's next?

Slide 109 / 139 Inclined Plane and a Pulley

Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk FT System 2

FT a

System 1

Substitute a in for ax and ay and add the equations together. FT cancels out!

Slide 110 / 139 Inclined Plane and a Pulley

#

We've now found the acceleration of the two box system - and without using a single number! That's the way to solve physics problems - work out the algebra first, and then substitute in the

• numbers. This has a few benefits - one of which is your

teacher is more apt to give you partial credit if the work is

• shown. In class, the work is more important than the

numerical answer. It's only when rockets and bridges are built that numbers are very important!

Slide 111 / 139

Inclined Plane and a Pulley

#

Another benefit is that limiting cases can be taken to both validate the algebra and to observe interesting phenomena. What if the angle θ was set equal to 00 or 900 (the two limiting cases)? What would the system look like? Imagine rotating the incline clockwise until the incline is vertical for the 900 case

• r rotating it counter-clockwise until the incline is horizontal for

the 00 case. Start with the 00 system.

Slide 112 / 139 Inclined Plane and a Pulley

#

m1 m2

Did you come up with the physical picture for the 00 case? If not, start with the algebra and set θ = 00: Is this a familiar equation?

Slide 113 / 139 Inclined Plane and a Pulley

#

m1 m2 m1 m2

We've got the case of a block pulling another block off of a table where there is friction between the block and the surface. Set θ = 00

Slide 114 / 139

Inclined Plane and a Pulley

#

m1 m2

Time for the 900 system. Any luck? If not, start with the algebra and set θ = 900: The coefficient of kinetic friction is gone - because m2 is barely in contact with the rotated incline. Can you visualize it now?

Slide 115 / 139 Inclined Plane and a Pulley

Atwood Machine

m1 m2

FT1 = FT2

Set θ = 900 The inclined plane becomes an Atwood Machine!

#

m1 m2

Slide 116 / 139 Inclined Plane and a Pulley

#

m1 m2

One more thing to talk about for this problem. Look at all the work above. You can see the forces acting on the two masses in the acceleration equation. But which force is conspicuously missing? Does that give you an idea about a different way to solve the problem?

Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk FT System 2

FT a

System 1

Slide 117 / 139

Inclined Plane and a Pulley

#

m1 m2

What if we considered just ONE system - the system of both masses? FT is now an internal force - and it is left

• ut of the FBD because it consists of an infinite number
• f action-reaction forces within the string that cancel out.

Only external forces contribute to the motion of a system.

Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk FT System 2

FT a

System 1

The Tension force, FT doesn't appear!

Slide 118 / 139 Inclined Plane and a Pulley

#

m1 m2 Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk FT System 2

FT a

System 1

Look at the above two FBDs and the sketch. The pully's only purpose is to redirect m1g so it can pull m2 up the incline. Or, m1 pulls on m2 with FT, and m2 pulls on m1 with FT. So what can we call that?

Slide 119 / 139 Inclined Plane and a Pulley

#

m1 m2 Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk m1g System of both masses

That's an action-reaction pair, and FT is an internal force to the system and will not take part in the

• FBD. Since the pulley is merely redirecting the

force, m1g, we draw it on the new "System of both masses" FBD, as shown above.

Slide 120 / 139

Inclined Plane and a Pulley

#

m1 m2 Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk m1g System of both masses

y-axis: This will be used to calculate fk: x-axis: We did this before - the only change is replacing FT with m1g, and not having to solve simultaneous equations.

Slide 121 / 139 Inclined Plane and a Pulley

#

m1 m2 Fx=m2gsinθ Fy=m2gcosθ

#

a

FN

fk m1g System of both masses

One last step is expressing ax as a, since that is the only motion we're observing, and we have:

Slide 122 / 139

Falling Objects with Air Resistance

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• f Contents

Slide 123 / 139

Falling Objects with Air Resistance

The Kinematics unit covered objects in free fall - where the only force acting was the gravitational force. The objects fall with a constant acceleration, and the velocity increases as a linear function of time, or as the square root of the distance fallen. Near the surface of the earth, there is an opposing force to gravity for falling objects - the air. Air is a fluid, and it provides a resistance force (also called drag) that opposes the object's downward motion. This force increases as a factor of the object's velocity squared. At the point where this force equals the gravitational force, the net force will be zero, and the object will fall with a constant velocity - terminal velocity.

Slide 124 / 139 Falling Objects with Air Resistance

The resistance force can also be increased by increasing the surface area of the object falling (try this yourself with a sheet of paper, and then crumple it up and let it fall). This is part of the theory behind parachutes - which reduce the terminal velocity of the parachutist - a good thing. Newton's Second Law for an object with terminal velocity (a=0):

mg f = kv2

terminal velocity

Slide 125 / 139 Falling Objects with Air Resistance

A more interesting problem is finding the velocity at any point as the

• bject falls. This requires the use of integral calculus, specifically,

separation of variables and change of variables. We're going to change the frictional force to a linear dependence on velocity(f = kv), so that the integration is simpler.

mg f = kv

Motion in y direction is understood - drop subscript Differential Equation Separation of variables - v on the left, and t on the right Integrating from time, 0, to the velocity at a specific time, t

Slide 126 / 139

Falling Objects with Air Resistance

mg f = kv

Integrating from time, 0, to the velocity at a specific time, t. Use change of variables, to turn the integral into something we can work with Not mathematically rigorous here - we should've changed the limits of integration for u, but we'll fix that later on (a shortcut)

Slide 127 / 139 Falling Objects with Air Resistance

mg f = kv

Go back a page, and see what would have happened if the friction force was kv2. This integral would have required a table of integrals

• r solving by trigonometric substitution.

Now, we substitute u=mg-kv back, and we can use the original limits of integration (this was the shortcut) Take the exponential of each side

Slide 128 / 139 Falling Objects with Air Resistance

mg f = kv

Calculate the terminal velocity for this linear frictional force (drag): Do you see this term to the left? Make the substitution, vT = mg/k This is good - as t approaches ∞, v approaches vT.

Slide 129 / 139

Graphical Representation of Falling Objects with air resistance (drag)

• y

a t v t vT

Object starts at height, h, and falls - as f(t2) until it reaches vT; then f(t).

h

linear quadratic

g

a = 0 v = vT v = vT

t

Slide 130 / 139

Non Uniform Circular Motion

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• f Contents

Slide 131 / 139 Non-Uniform Circular Motion Vertical Loop

UCM assumes a constant speed motion - which is achieved by having only

• ne force act on the object -

for example, a horizontal tension force, which provides the centripetal motion. This works well for a horizontal rotation. But what if you are swinging a ball in a vertical circle?

atan arad atan atan atan arad arad arad arad arad

Slide 132 / 139

atan arad atan atan atan arad arad arad arad arad

Non-Uniform Circular Motion Vertical Loop

Gravity provides a downward force at all times. This force resolves into a tangential and radial component. The tangential force changes the speed, and the radial force combines with the tension force to change the centripetal acceleration. The non zero tangential force is added to the radial force to create an acceleration vector that points inside the circle - but not at the center. This is not UCM.

Slide 133 / 139 Non-Uniform Circular Motion Vertical Loop

θ θ mg mg mg Ttop Tbottom T r mg cosθ mg sinθ v v

Here is the most general free body diagram. We are not assuming that the tension force stays constant - it certainly can if a machine is providing the force, but if it's a person just swinging a ball, it probably won't be. Since this is radial motion, the most natural coordinate system is polar - the axes are lined up along the radial vector and tangent to the circle.

Slide 134 / 139 Non-Uniform Circular Motion Vertical Loop

θ θ mg mg mg Ttop Tbottom T r mg cosθ mg sinθ v v

Newton's Second Law: Radial: Tangential:

Slide 135 / 139

Non-Uniform Circular Motion Vertical Loop

atan arad atan atan atan arad arad arad arad arad

A little more complex than UCM, as the acceleration and velocity is different at every point on the circle, depending on θ and T.

Slide 136 / 139 Static Equilibrium

There is a whole field of problems in engineering and physics called "Statics" that has to do with cases where no acceleration occurs and objects remain at rest. Anytime we construct bridges, buildings or houses, we want them to remain stationary, which is only possible if there is no acceleration or no net force. There are two types of motion that we need to consider (and in both cases, motion is to be prevented!). What are they?

Slide 137 / 139 Torque and Rotational Equilibrium

A good example is opening a door, making a door rotate. The door does not accelerate in a straight line, it rotates around its hinges. Think of the best direction and location to push on a heavy door to get it to rotate and you'll have a good sense of how torque works.

Slide 138 / 139

Tension Force

θ

mg

Tx Tx Ty Ty

θ θ θ

#Fx = max = 0 T1x - T2x = 0 Tsin# = Tsin# which just confirms that if the angles are equal, the tensions are equal #Fy = may = 0 T1y + T2y - mg = 0 Tcos# + Tcos# = mg 2Tcos# = mg T = mg / (2cos#) Note that the tension rises as cos# becomes smaller...which occurs as # approaches 90o. It goes to infinity at 90o, which shows that the ropes can never be perfectly horizontal.

x - axis y - axis

Slide 139 / 139