Dynamics Newton's Three Laws of Motion Inertial Reference Frames - - PDF document

SLIDE 1

Slide 1 / 156

Dynamics

Slide 2 / 156

· Newton's Three Laws of Motion · Inertial Reference Frames · Mass vs. Weight · Forces we studied: weight / gravity normal force tension friction (kinetic and static) · Drawing Free Body Diagrams · Problem Solving

Things to Remember from Last Year Slide 3 / 156 Newton's Laws of Motion

• 1. An object maintains its velocity (both speed and

direction) unless acted upon by a nonzero net force.

• 2. Newton’s second law is the relation between

acceleration and force.

• 3. Whenever one object exerts a force on a second
• bject, the second object exerts an equal force in

the opposite direction on the first object.

ΣF = ma Slide 4 / 156

1 Which of Newton's laws best explains why motorists should buckle-up?

A

First Law

B

Second Law

C

Third Law

D

Law of Gravitation

Slide 5 / 156

2 You are standing in a moving bus, facing forward, and you suddenly fall forward. You can infer from this that the bus's

A

velocity decreased.

B

velocity increased.

C

speed remained the same, but it is turning to the right.

D

speed remained the same, but it is turning to the left.

Slide 6 / 156

3 You are standing in a moving bus, facing forward, and you suddenly fall forward as the bus comes to an immediate stop. What force caused you to fall forward?

A

gravity

B

normal force due to your contact with the floor

• f the bus

C

force due to friction between you and the floor

• f the bus

D

There is not a force leading to your fall.

SLIDE 2

Slide 7 / 156

4 When the rocket engines on the spacecraft are suddenly turned off, while traveling in empty space, the starship will

A

stop immediately.

B

slowly slow down, then stop.

C

go faster.

D

move with constant speed

Slide 8 / 156

5 A net force F accelerates a mass m with an acceleration a. If the same net force is applied to mass 2m, then the acceleration will be

A

4a

B

2a

C

a/2

D

a/4

Slide 9 / 156

6 A net force F acts on a mass m and produces an acceleration a. What acceleration results if a net force 2F acts on mass 4m?

A

a/2

B

8a

C

4a

D

2a

Slide 10 / 156

7 An object of mass m sits on a flat table. The Earth pulls on this object with force mg, which we will call the action force. What is the reaction force?

A

The table pushing up on the object with force mg

B

The object pushing down on the table with force mg

C

The table pushing down on the floor with force mg

D

The object pulling upward on the Earth with force mg

Slide 11 / 156

8 A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. Since a lot of damage is done on the car

A

the force on the truck is greater then the force on the car

B

the force on the truck is equal to the force

• n the car

C

the force on the truck is smaller than the force on the car

D

the truck did not slow down during the collision

Slide 12 / 156

Newton's laws are only valid in inertial reference frames: An inertial reference frame is one in which Newton’s first law is valid. This excludes rotating and accelerating frames.

Inertial Reference Frames

SLIDE 3

Slide 13 / 156

MASS is the measure of the inertia of an object, the resistance of an object to accelerate. WEIGHT is the force exerted on that object by

• gravity. Close to the surface of the Earth, where the

gravitational force is nearly constant, the weight is: Mass is measured in kilograms, weight Newtons.

Mass and Weight FG = mg Slide 14 / 156 Normal Force and Weight

FN mg The Normal Force, FN, is ALWAYS perpendicular to the surface. Weight, mg, is ALWAYS directed downward.

Slide 15 / 156

9 The acceleration due to gravity is lower on the Moon than on Earth. Which of the following is true about the mass and weight of an astronaut on the Moon's surface, compared to Earth?

A

Mass is less, weight is same

B

Mass is same, weight is less

C

Both mass and weight are less

D

Both mass and weight are the same

Slide 16 / 156

10 A 14 N brick is sitting on a table. What is the normal force supplied by the table?

A

14 N upwards

B

28 N upwards

C

14 N downwards

D

28 N downwards

Slide 17 / 156 Kinetic Friction

fK v Friction forces are ALWAYS parallel to the surface exerting them. Kinetic friction is always directed

• pposite to

direction the object is sliding and has magnitude: fK = μkFN

Slide 18 / 156

11 A 4.0kg brick is sliding on a surface. The coefficient of kinetic friction between the surfaces is 0.25. What it the size of the force of friction?

A

8.0

B

8.8

C

9.0

D

9.8

SLIDE 4

Slide 19 / 156

12 A brick is sliding to the right on a horizontal surface. What are the directions of the two surface forces: the friction force and the normal force?

A

right, down

B

right, up

C

left, down

D

left, up

Slide 20 / 156 Static Friction

fS FAPP Static friction is always equal and

• pposite the Net

Applied Force acting on the object (not including friction). Its magnitude is: fS ≤ μSFN

Slide 21 / 156

13 A 4.0 kg brick is sitting on a table. The coefficient

• f static friction between the surfaces is 0.45.

What is the largest force that can be applied horizontally to the brick before it begins to slide?

A

16.33

B

17.64

C

17.98

D

18.12

Slide 22 / 156

14 A 4.0kg brick is sitting on a table. The coefficient

• f static friction between the surfaces is 0.45. If a

10 N horizontal force is applied to the brick, what will be the force of friction and will the brick move?

A

16.12, no

B

17.64, no

C

16.12, yes

D

17.64, yes

Slide 23 / 156

When a cord or rope pulls on an

• bject, it is said to be under

tension, and the force it exerts is called a tension force, FT.

Tension Force

FT mg a

Slide 24 / 156

15 A crane is lifting a 60 kg load at a constant

• velocity. Determine the tension force in the cable.

A 568 N B

578 N

C

504 N

D

600 N

SLIDE 5

Slide 25 / 156

16 A system of two blocks is accelerated by an applied force of magnitude F on the frictionless horizontal surface. The tension in the string between the blocks is: A 6F B 4F C 3/5 F D 1/6 F E 1/4 F 6 kg 4 kg F

Slide 26 / 156

Two Dimensions

Slide 27 / 156

Since forces are vectors, they add as vectors. The simplest case is if the forces are perpendicular (orthogonal) with another. Let's add a 50 N force at 0o with a 40 N force at 90o.

Adding Orthogonal Forces Slide 28 / 156

• 1. Draw the first force

vector, 50 N at 0o, beginning at the origin.

Adding Orthogonal Forces Slide 29 / 156

• 1. Draw the first force

vector, 50 N at 0o, beginning at the origin.

• 2. Draw the second force

vector, 40 N at 90o, with its tail at the tip of the first vector.

Adding Orthogonal Forces Slide 30 / 156

• 1. Draw the first force

vector, 50 N at 0o, beginning at the origin.

• 2. Draw the second force

vector, 40 N at 90o, with its tail at the tip of the first vector.

• 3. Draw Fnet from the tail of

the first force to the tip of the last force.

Adding Orthogonal Forces

SLIDE 6

Slide 31 / 156

We get the magnitude of the resultant from the Pythagorean Theorem. c2 = a2 + b2 Fnet

2 = (50N)2 + (40N)2

= 2500N2 + 1600N2 = 4100N2 Fnet = (4100N2)1/2 = 64 N

Adding Orthogonal Forces Slide 32 / 156 Adding Orthogonal Forces

We get the direction of the net Force from the inverse tangent. tan(θ) = opp / adj tan(θ) = (40N) / (50N) tan(θ) = 4/5 tan(θ) = 0.8 θ = tan-1(0.80) = 39o Fnet = 64 N at 39o

Slide 33 / 156 Decomposing Forces into Orthogonal Components

Let's decompose a 120 N force at 34o into its x and y components. F θ

Slide 34 / 156 Decomposing Forces into Orthogonal Components

We have Fnet, which is the hypotenuse, so let's first find the adjacent side by using cosθ = adj / hyp cosθ = Fx / Fnet Fx = Fnet cosθ = 120N cos34o = 100 N F Fx θ

Slide 35 / 156 Decomposing Forces into Orthogonal Components

Now let's find the opposite side by using sinθ = adj / hyp sinθ = Fy / Fnet Fy = Fnet sinθ = 120N sin34o = 67 N Fx F Fy θ

Slide 36 / 156 Adding non-Orthogonal Forces

Now that we know how to add orthogonal forces. And we know now to break forces into orthogonal components, so we can make any force into two

• rthogonal forces.

We combine those two steps to add any number of forces together at any different angles.

SLIDE 7

Slide 37 / 156 Adding non-Orthogonal Forces

Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o First, we will do it graphically, to show the principle for how we will do it analytically. Then, we will do it analytically, which is easy once you see why it works that way.

Slide 38 / 156 Adding non-Orthogonal Forces

F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o First, here are the vectors all drawn from the origin.

Slide 39 / 156 Adding non-Orthogonal Forces

F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Now we arrange them tail to tip. F1 F2 F3

Slide 40 / 156 Adding non-Orthogonal Forces

F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Then we draw in the Resultant, the sum

• f the vectors.

F1 F2 F3 F1 F2 F3 F F

Slide 41 / 156

F1 F2 F3 F1x F2x F3x F1y F2y F3y F

Adding non-Orthogonal Forces

F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Now, we graphically break each vector into its orthogonal components.

Slide 42 / 156 Adding non-Orthogonal Forces

F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Then, we remove the original vectors and note that the sum of the components is the same as the sum of the vectors. F1x F2x F3x F1y F2y F3y F

SLIDE 8

Slide 43 / 156 Adding non-Orthogonal Forces

F1 = 4.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o This is the key result: The sum of the x- components of the vectors equals the x- component of the Resultant. As it is for the y- components. F1x F2x F3x F1y F2y F3y F Fy Fx

Slide 44 / 156 Adding non-Orthogonal Forces

This explains how we will add vectors analytically.

• 1. Write the magnitude and direction (from 0 to 360 degrees)
• f each vector.
• 2. Compute the values of the x and y components.
• 3. Add the x components to find the x component of the

Resultant.

• 4. Repeat for the y-components.
• 5. Use Pythagorean Theorem to find the magnitude of the

Resultant.

• 6. Use Inverse Tangent to find the Resultant's direction.

Slide 45 / 156 Adding non-Orthogonal Forces

First make a table and enter what you know. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 F2 F3 Fnet

Slide 46 / 156 Adding non-Orthogonal Forces

Then calculate the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 8.0 N 50o F2 6.5 N 75o F3 8.4 N 30o Fnet

Slide 47 / 156 Adding non-Orthogonal Forces

Then add up the x and y components. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 8.0 N 50o 5.14 N 6.13 N F2 6.5 N 75o 1.68 N 6.28 N F3 8.4 N 30o 7.27 N 4.2 N Fnet

Slide 48 / 156 Adding non-Orthogonal Forces

Then use Pythagorean Theorem and tan-1 to determine Fnet. F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 8.0 N 50o 5.14 N 6.13 N F2 6.5 N 75o 1.68 N 6.28 N F3 8.4 N 30o 7.27 N 4.20 N Fnet 14.09 N 16.61 N

SLIDE 9

Slide 49 / 156 Adding non-Orthogonal Forces

Let's add together these three forces: F1 = 8.0 N at 50o F2 = 6.5 N at 75o F3 = 8.4 N at 30o Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 8.0 N 50o 5.14 N 6.13 N F2 6.5 N 75o 1.68 N 6.28 N F3 8.4 N 30o 7.27 N 4.20 N Fnet 22.9 N 50o 14.09 N 16.61 N

Slide 50 / 156 Adding non-Orthogonal Forces

Now use the same procedure to add these forces. DO NOT IGNORE NEGATIVE SIGNS F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o

Slide 51 / 156

Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 21 N 60o F2 65 N 150o F3 45 N 330o Fnet

Adding non-Orthogonal Forces

F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o

Slide 52 / 156

Force Magnitude Direction x-comp. (Fcosθ) y-comp. (Fsinθ) F1 21 N 60o 10.50 N 8.66 N F2 65 N 150o

• 56.29 N

32.50 N F3 45 N 330o 38.97 N

• 22.50 N

Fnet 131 N 290o

• 6.82 N

18.66 N

Answers

F1 = 21 N at 60o F2 = 65 N at 150o F3 = 8.4 N at 330o

Slide 53 / 156

17 Three forces act on an object. Which of the following is true in order to keep the object in translational equilibrium?

• I. The vector sum of the three forces must equal zero.
• II. The magnitude of the three forces must be equal.
• III. All three forces must be parallel.

A I only B II only C I and III only D II and III only E I, II, and III

Slide 54 / 156 Force and friction acting on an object

Previously, we solved problems with multiple forces, but they were either parallel or perpendicular. For instance, draw the free body diagram of the case where a box is being pulled along a surface, with friction, at constant speed.

SLIDE 10

Slide 55 / 156

FN mg FAPP f Now find the acceleration given that the applied force is 20 N, the box has a mass of 3.0kg, and the coefficient of kinetic friction is 0.20.

Force and friction acting on an object Slide 56 / 156

FN mg FAPP f

FAPP = 20N m = 3.0kg μk = 0.20 ΣF = ma FN - mg = 0 FN = mg FN = (3.0kg)(10m/s2) FN = 30N ΣF = ma FAPP - fk = ma FAPP - μkFN = ma FAPP - μkmg = ma a = (FAPP - μkmg)/m a = (20N - (0.20)(30N))/3.0kg a = (20N - 6.0N)/3.0kg a = (14N)/3.0kg a = 4.7 m/s2 x - axis y - axis

Force and friction acting on an object Slide 57 / 156 Forces at angles acting on an object

Now we will solve problems where the forces act at an angle so that it is not parallel or perpendicular with one another. First we do a free body diagram, just as we did previously.

Slide 58 / 156

FN mg FAPP f

Forces at angles acting on an object

The next, critical, step is to choose axes. Previously, we always used vertical and horizontal axes, since

• ne axis lined up with

the forces...and the acceleration. Now, we must choose axes so that all the acceleration is along one axis, and there is no acceleration along the other. You always have to ask, "In which direction could this object accelerate?" Then make one axis along that direction, and the

• ther perpendicular to that.

What's the answer in this case?

Slide 59 / 156

FN mg FAPP f

y

X

Forces at angles acting on an object

This time vertical and horizontal axes still work...since we assume the box will slide along the surface. However, if this assumption is wrong, we will get answers that do not make sense, and we will have to reconsider our choice.

Slide 60 / 156 Forces at angles acting on an object

FN mg FAPP f

Now we have to break any forces that do not line up with our axes into components that do. In this case, FAPP, must be broken into Fx and Fy

y

X

SLIDE 11

Slide 61 / 156 Forces at angles acting on an object

FN mg f

y

X

Fy F

x

Once we do that, we can now proceed as we did previously, using each component appropriately.

Slide 62 / 156

Let's use our work to find the acceleration if the applied force is 20 N at 37o above horizontal, the box has a mass of 3.0kg, and the coefficient

• f kinetic friction is

0.20.

Force and friction acting on an object

FN mg f

y

X

Fy F

x

Slide 63 / 156

FAPP = 20N at 37o m = 3.0kg μk = 0.20

Force and friction acting on an object

FN mg f

y

X FAPPy F

APPx

ΣF = ma FN + Fy - mg = 0 FN = mg - Fy FN = mg - Fsinθ FN = (3.0kg)(10m/s2)

• (20N)(sin37o)

FN = 30N - 12N FN = 18N Note that FN is lower due to the force helping to support the object ΣF = ma Fx - fk = ma Fx - μkFN = ma Fcosθ - μkmg = ma a = (Fcosθ - μkFN)/m a = (20N cos37o

• (0.20)(18N))/3.0kg

a = (16N - 3.6N)/3.0kg a = (12.4N)/3.0kg a = 4.1 m/s2 x - axis y - axis

Slide 64 / 156 Normal Force and Friction

Friction was reduced because the Normal Force was reduced; the box's weight, mg, was supported by the y- component of the force plus the Normal Force...so the Normal Force was lowered...lowering friction.

mg FAPPy FN FN Fy mg

Just looking at the y-axis ΣF = ma FN + Fy - mg = 0 FN = mg - Fy FN = mg - Fsinθ

Slide 65 / 156 Normal Force and Friction

What would happen with both the Normal Force and Friction in the case that the object is being pushed along the floor by a downward angled force?

FN mg FAPP f

y

X

Slide 66 / 156 Normal Force and Friction

In this case the pushing force is also pushing the box into the surface, increasing the Normal Force as well as friction.

FN mg f

y

X

Fy F

x

FN mg

Just looking at the y-axis ΣF = ma FN - Fy - mg = 0 FN = mg + Fy FN = mg + Fsinθ

Fy

SLIDE 12

Slide 67 / 156

18 The normal force on the box is: A mg B mg sin(θ) C mg cos(θ) D mg + F sin(θ) E mg – F sin(θ)

Fapp θ

Slide 68 / 156

19 The frictional force on the box is:

Fapp θ

A μ(mg + Fsin(θ)) B μ(mg - Fsin(θ)) C μ(mg + Fcos(θ)) D μ(mg - Fcos(θ)) E μmg

Slide 69 / 156

20 A block of mass m is pulled along a horizontal surface at constant speed v by a force Fapp , which acts at an angle of θ with the horizontal. The coefficient of kinetic friction between the block and the surface is μ. The normal force exerted on the block by the surface is: A mg - Fapp cosθ B mg - Fapp sinθ C mg D mg + Fapp sinθ E mg + Fapp cosθ

Fapp θ

v m

Slide 70 / 156

Fapp θ

v m 21 A block of mass m is pulled along a horizontal surface at constant speed v by a force Fapp , which acts at an angle of θ with the horizontal. The coefficient of kinetic friction between the block and the surface is μ. The friction force on the block is: A μ(mg - Fapp cosθ) B μ(mg - Fapp sinθ) C μmg D μ(mg + Fapp sinθ) E μ(mg + Fapp cosθ)

Slide 71 / 156 Normal Force and Weight

FN mg Previously we dealt mostly with horizontal (or, rarely, vertical surfaces). In that case FN, and mg were always along the same axis. Now we will look at the more general case.

Slide 72 / 156

On the picture, draw the free body diagram for the block. Show the weight and the normal force.

Inclined Plane

SLIDE 13

Slide 73 / 156 Inclined Plane

FN mg FN is ALWAYS perpendicular to the surface. mg is ALWAYS directed downward. But now, they are neither parallel or perpendicular to one another.

Slide 74 / 156 Choosing Axes

FN mg Previously, we used vertical and horizontal

• axes. That worked

because problems always resulted in an acceleration that was along one of those axes. We must choose axes along which all the acceleration is along

• ne axis...and none

along the other.

Slide 75 / 156 Choose of Axes

FN mg

y

X

a

In this case, the block can only accelerate along the surface of the plane. Even if there is no acceleration in a problem, this is the ONLY POSSIBLE direction of acceleration. So we rotate our x-y axes to line up with the surface of the plane.

Slide 76 / 156

By the way, here's one way to see that θ is the both the angle of incline and the angle between mg and the new y-axis.

Inclined Plane Problems

y-axis α

Let's name the angle of the inclined plane θ and the angle between mg and the x-axis θ' and show that θ = θ' Since the angles in a triangle add to 180o, and the bottom left angle is 90o, that means: α + θ = 90o θ θ'

Slide 77 / 156

Now look at the angles in the upper left corner of the triangle.

Inclined Plane Problems

y-axis α

The surface of the inclined plane forms an angle of 180o, so 90o + θ' + α = 180o θ' + α = 90o But we already showed that α + θ = 90o θ' + α = 90o = α + θ θ' = θ θ θ'

Slide 78 / 156

FN mg θ

a

We need to resolve all forces which don't align with an axis into components that will. In this case, we resolve mg into its x and y components.

Inclined Plane Problems FN

m g s i n θ mg cos θ θ

a

y

X

θ

SLIDE 14

Slide 79 / 156

Now we just use Newton's Second Law, which is true for each axis. ΣFx = max mgsinθ = max a = gsinθ down the plane ΣFy = may FN - mgcosθ = 0 FN = mgcosθ FN

m g s i n θ mg cos θ θ

a

Inclined Plane Problems

x - axis y - axis

Slide 80 / 156

A 5 kg block slides down a frictionless incline at an angle

• f 30 degrees.

a) Draw a free body diagram. b) Find its acceleration. (Use g = 10 m/s2)

θ ΣFx = max mg sin θ = ma g sin θ = a a = g sin θ a = 10 m/s2 sin (30o) a = 10 m/s2 (0.5) a = 5 m/s2

y x FN mg

mg sin θ mg cos θ θ

Answer

Slide 81 / 156

FN mg θ fk

a We can add kinetic friction to our inclined plane example. The kinetic friction points opposite the direction of motion. We now have a second vector along the x axis. fk points in the negative direction (recall fk = μk FN)

Inclined Plane Problems with Friction

y x

FN

mg sinθ m g c

• s

θ

θ

fk

a

Slide 82 / 156

ΣFx = max mgsinθ - fk = ma mgsinθ - μkFN = ma mgsinθ - μkmgcosθ = ma gsinθ - μkgcosθ = a a = gsinθ - μkgcosθ a = g(sinθ - μk cosθ)

Inclined Plane Problems with Friction y x

FN

mg sinθ mg cosθ

θ

fk

a

ΣFy = may FN = mg cosθ x - axis y - axis

Slide 83 / 156

The general solution for objects sliding down an incline is: a = g(sinθ - μkcosθ) Note, that if there is no friction: μk = 0 and we get our previous result for a frictionless plane: a = gsinθ

Inclined Plane Problems with Friction Slide 84 / 156

a = g(sinθ - μkcosθ) If the object is sliding with constant velocity: a = 0

Inclined Plane Problems with Friction

a = g(sinθ - μkcosθ) 0 = g(sinθ - μkcosθ) 0 = sinθ - μkcosθ μk cosθ = sinθ μk = sinθ / cosθ μk = tanθ a = 0

y x

FN

mg sinθ m g c

• s

θ

θ

fk

SLIDE 15

Slide 85 / 156 Inclined Plane with Static Friction

We just showed that for an object sliding with constant velocity down an inclined plane that: μk = tanθ Similarly, substituting μs for μk (at the maximum static force; the largest angle of incline before the

• bject begins to slide) than:

μs = tanθmax But this requires that the MAXIMUM ANGLE of incline, θmax, be used to determine μs. a = 0

y x

FN

mg sinθ m g c

• s

θ

θ

fs

Slide 86 / 156

A 5 kg block slides down a incline at an angle of 30o with a constant speed. a) Draw a free body diagram. b) Find the coefficient of friction between the block and the incline. (Use g = 10 m/s2) θ

ΣF = ma mg sin θ - fK = 0 mg sin θ = fK mg sin θ = μ FN mg sin θ = μ mg cos θ μ = mg sin θ / mg cos θ μ = tan θ μ = tan 30 = 0.58

y x

F

N

mg

mg sin θ mg cos θ θ

fk

Answer

Slide 87 / 156

A 5 kg block is pulled up an incline at an angle of 30o at a constant velocity The coefficient of friction between the block and the incline is 0.3. a) Draw a free body diagram. b) Find the applied force. (Use g = 10 m/s2)

FN m g

mg sin θ m g c

• s

θ θ

Fapp fK

a = 0

θ

y-direction ΣF = ma FN - mg cos θ = 0 FN = mg cos θ x-direction ΣF = ma Fapp - mg sin θ - fk = ma Fapp - mg sin θ -μk FN = 0 Fapp = mg sin θ + μk mg cos θ Fapp = 38 N

Answer

Slide 88 / 156

θ

A 5 kg block is pulled UP an incline at an angle of 30 degrees with a force of 40 N. The coefficient of friction between the block and the incline is 0.3. a) Draw a free body diagram. b) Find the block's acceleration. (Use g = 10 m/s2)

FN m g

mg sin θ mg cos θ θ

F

a p p

fK

a y-axis ΣF = ma FN - mg cos θ = 0 FN = mg cos θ x-axis ΣF = ma Fapp - mg sin θ - fk = ma Fapp - mg sin θ - μk FN = ma Fapp - mg sin θ - μk mg cos θ = ma a = (Fapp/m) - g sin θ - μk g cos θ a = 0.4 m/s2

Slide 89 / 156

FN mg θ

a

If a mass, m, slides down a frictonless inclined plane, we have this general setup:

Inclined Plane Problems

y x

It is helpful to rotate

• ur reference frame so

that the +x axis is parallel to the inclined plane and the +y axis points in the direction

• f FN.

Slide 90 / 156

A 5 kg block remains stationary on an incline. The coefficients of static and kinetic friction are 0.4 and 0.3, respectively. a) Draw a free body diagram. b) Determine the angle that the block will start to move. (Use g = 10 m/s2)

F

N

mg

mg sin θ m g c

• s

θ θ

f

s

a = 0

y-direction ΣF = ma FN - mg cos θ = 0 FN = mg cos θ x-direction ΣF = ma fs = mg sin θ μs mg cos θ = mg sin θ μs cos θ = sin θ μs = sin θ / cos θ μs = tan θ θ = tan-1 μs θ = 21.8o

θ

SLIDE 16

Slide 91 / 156

22 A block of mass m slides down a rough incline as shown. Which free body diagram correctly shows the forces acting

• n the block?

A B C D E

f W N f W N f W N f W N f W N

Slide 92 / 156

23 A block with a mass of 15 kg slides down a 43° incline as shown above with an acceleration of 3 m/s2. What is the normal force N applied by the inclined plane on the block?

A

95 N

B

100 N

C

105 N

D

110 N

E

115 N

Slide 93 / 156

24 A block with a mass of 15 kg slides down a 43° incline as shown above with an acceleration of 3 m/s2. The magnitude of the frictional force along the plane is nearly:

A

56 N

B

57 N

C

58 N

D

59 N

E

60 N

Slide 94 / 156 Static Equilibrium

There is a whole field of problems called "Statics" that has to do with cases where no acceleration occurs, objects remain at rest. Anytime we construct something (bridges, buildings, houses, etc.) we want them to remain stationary, not accelerate. So this is a very important field. The two types of static equilibrium are with respect to linear and rotational acceleration, a balancing of force and of torque (forces that cause objects to rotate). We'll look at them in that order.

Slide 95 / 156

Previously, we did problems where a rope supporting an object exerted a vertical force straight upward, along the same axis as the force mg was pulling it down. That led to the simple case that if a = 0, then FT = mg

Tension Force

mg FT

Slide 96 / 156

25 A uniform rope of weight 20 N hangs from a hook as shown above. A box of mass 60 kg is suspended from the rope. What is the tension in the rope? A 20 N throughout the rope B 120 N throughout the rope C 200 N throughout the rope D 600 N throughout the rope E It varies from 600 N at the bottom of the rope to 620 N at the top.

SLIDE 17

Slide 97 / 156

But it is possible for two (or more) ropes to support an object (a = 0) by exerting forces at angles. In that case: The vertical components of the force exerted by each rope must add up to mg. And the horizontal components must add to zero.

Tension Force

mg T1 T2

Slide 98 / 156

So we need to break the forces into components that align with our axes.

Tension Force

mg T1 T2

Slide 99 / 156

One result can be seen just from this diagram. In order for the forces along the x-axis to cancel out, the more vertical Tension must be much larger than the Tension which is at a greater angle to the vertical. In problems with two supporting ropes or wires, the more vertical has a greater tension.

Tension Force

mg T1x T2x T2y T1y

Slide 100 / 156

Let's calculate the tension in two ropes if T1 is at an angle of 50o from the vertical and T2 is at an angle of 20o from the vertical and they are supporting a 8.0 kg mass. Note, that in this case the horizontal component is given by Tsinθ, where before it was given by Tcosθ. To know which function to use you MUST draw the diagram and find the sides opposite and adjacent to the given angle.

Tension Force

50o 50o 20o 20o

mg T1x T2x T2y T1y

Slide 101 / 156 Tension Force

ΣFx = max = 0 T1x - T2x = 0 T1sinθ1 = T2sinθ2 T1 = T2sinθ2/sinθ1 T1 = T2 (sin20o/sin50o) T1 = T2 (0.34/0.77) T1 = 0.44 T2 T1 = 0.44 (64N) T1 = 28N ΣFy = may = 0 T1y + T2y - mg = 0 T1cosθ1 + T2cosθ2 = mg (0.44T2)cos(50o) + T2cos(20o) = mg (0.44T2)(0.64) + T2(0.94) = (8kg)(9.8m/s2) 1.22 T2 = 78N T2 = 78N / 1.22 T2 = 64N

x - axis y - axis

50o 50o 20o 20o

mg T1x T2x T2y T1y

Slide 102 / 156 Tension Force

θ

mg Tx Tx Ty Ty

θ θ θ

If the ropes form equal angles to the vertical, the tension in each must also be equal, otherwise the x- components of Tension would not add to zero.

SLIDE 18

Slide 103 / 156 Tension Force

θ

mg Tx Tx Ty Ty

θ θ θ

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

x - axis y - axis

Slide 104 / 156

26 A lamp of mass m is suspended from two ropes of unequal length as shown above. Which of the following is true about the tensions T1 and T2 in the cables? A T1 > T2 B T1 = T2 C T1 < T2 D T1 - T2 = mg E T1 + T2 = mg T1 T2

Slide 105 / 156

27 A large mass m is suspended from two massless strings of an equal length as shown below. The tension force in each string is: A ½ mg cos(θ) B 2 mg cos(θ) C mg cos(θ) D mg/cos(θ) E mg/2cos(θ) θ θ m

Slide 106 / 156 Torque and Rotational Equilibrium

Forces causes objects to linearly accelerate. Torque causes objects to rotationally accelerate. Rotational dynamics is a major topic of AP Physics C:

• Mechanics. It isn't particularly difficult, but for AP B, we
• nly need to understand the static case, where all the

torques cancel and add to zero. First we need to know what torque is.

Slide 107 / 156 Torque and Rotational Equilibrium

Until now we've treated objects as points, we haven't been concerned with their shape or extension in space. We have assumed that any applied force acts through the center of the object and it is free to accelerate. That does not result in rotation, just linear acceleration. But if the force acts on an object so that it causes the

• bject to rotate around its center of mass...or around a

pivot point, that force has exerted a torque on the object.

Slide 108 / 156 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. Which force (blue arrow) placed at which location would create the most rotational acceleration of the green door about the black hinge?

SLIDE 19

Slide 109 / 156 Torque and Rotational Equilibrium

The maximum torque is obtained from: · The largest force · At the greatest distance from the pivot · At an angle to the line to the pivot that is closest to perpendicular Mathematically, this becomes: τ = Frsinθ τ (tau) is the symbol for torque; F is the applied force r is the distance from the pivot θ is the angle of the force to a line to the pivot

Slide 110 / 156 Torque and Rotational Equilibrium

τ = Frsinθ When r decreases, so does the torque for a given

• force. When r = 0, τ = 0.

When θ differs from 90o, the torque decreases. When θ = 0o or 180o, τ = 0.

r θ F

Slide 111 / 156 Two Physical Interpretations of Torque

τ = r(Fsinθ) The torque due to F acting at angle θ can be replaced by a smaller perpendicular force Fsinθ acting at r.

r θ F τ = F(rsinθ) The torque due to F acting at angle θ can be replaced by F acting at a smaller distance rsinθ. r θ F r θ Fsinθ rsinθ F rsinθ

These are all equivalent.

Slide 112 / 156 Rotational Equilibrium

When the sum of the torques on an object is zero, the

• bject is in rotational equilibrium.

Define counter clockwise (CCW) as the positive direction for rotation and clockwise (CW) as the negative. For instance, what perpendicular force, F, must be applied at a distance of 7.0 m for the pivot to exactly

• ffset a 20N force acting at a distance of 4.0m from the

pivot of a door at an angle of 30o?

Slide 113 / 156 Rotational Equilibrium

20N 30o 4.0m F1 Στ = 0 F1r1sinθ1 + F2r2sinθ2 =0 F1r1sinθ1 = - F2r2sinθ2 F1 = - F2r2sinθ2 / (r1sinθ1) F1 = - (-20N)(4.0m)sin(30o) / ((7m)sin90o) F1 = (80N-m)(0.50) / (7m) F1 = 5.7 N

Slide 114 / 156 Rotational Equilibrium

1.0m 3.0m 4.0m 4kg 2kg What mass must be added at distance 4.0m to put the above apparatus into equilibrium?

SLIDE 20

Slide 115 / 156 Rotational Equilibrium

1.0m 3.0m 4.0m 4kg 2kg Στ = 0 F1r1sinθ1 + F2r2sinθ2 - F3r3sinθ2 = 0 m1gr1 + m2gr2 - m3gr3 = 0 m1r1 + m2r2 - m3r3 = 0 m3 = (m1r1 + m2r2) / r3 m3 = ((4kg)(3m) +(2kg)(1m)) / 4m F1 = (14kg-m)) / (4m) F1 = 3.5 kg

Slide 116 / 156

A 12 kg load hangs from one end of a rope that passes over a small frictionless

• pulley. A 15 kg counterweight is suspended from the other end of the rope. The

system is released from rest.

• a. Draw a free-body diagram for each object showing all applied forces in

relative scale. Next to each diagram show the direction of the acceleration

• f that object.

b. Find the acceleration each mass. c. What is the tension force in the rope? d. What distance does the 12 kg load move in the first 3 s? e. What is the velocity of 15 kg mass at the end of 5 s?

12 kg 15 kg

Slide 117 / 156

12 kg 15 kg

m1 = 12 kg m2 = 15 kg

a. m1 m1g FT m2 m2g FT a a

• b. ΣF = ma

ΣF = FT - m1g = m1a FT = m1g + m1a ΣF = FT - m2g = -m2a FT = m2g - m2a FT = FT

m1g + m1a = m2g - m2a

m1a + m2a = m2g - m1g a(m1 + m2) = g(m2 - m1) a = (g(m2 - m1)) / (m1 + m2) = (10 m/s2 (15kg - 12kg)) / (12kg + 15kg) = 30 m/s2 * kg / 27kg a = 1.11 m/s2

• c. FT = m1g + m1a or FT = m2g - m2a

FT = m1g + m1a = 12kg(10 m/s2) + 12kg(1.11 m/s2) = 120 N + 13.33 N = 133.33 N FT = m2g - m2a = 15kg(10 m/s2) - 15kg(1.11 m/s2) = 150 N - 16.67 N = 133.33 N

Slide 118 / 156

12 kg 15 kg

m1 = 12 kg m2 = 15 kg

• d. x = xo + vot + 1/2 at2

xo = 0 vo = 0 x = 1/2 at2 = 1/2 (1.11 m/s2)(3s)2 = 5 m

• e. v = vo + at

vo = 0 v = at = (1.11 m/s2)(5s) = 5.55 m/s

Slide 119 / 156

A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest. a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g

• masses. Include all forces and draw them to relative scale. Draw the

expected direction of acceleration next to each free-body diagram. b. Use Newton’s Second Law to write an equation for the 500 g mass. c. Use Newton’s Second Law to write an equation for the 300 g mass. d. Find the acceleration of the system by simultaneously solving the system

• f two equations.

e. What is the tension force in the string?

500 g 300 g

Slide 120 / 156

a. FT FT FN m2g m1g f

a a

m1 = 500 g = 0.5 kg m2 = 300 g = 0.3 kg μ = 0.25

• b. ΣF = ma

x-direction ΣF = FT - f = m1a f = μFn f = μm1g y-direction ΣF = ma = 0 ΣF = FN - m1g = 0 FN = m1g

• c. ΣF = ma

ΣF = FT - m2g = - m2a

• d. FT - f = m1a

FT - m2g = - m2a FT = f + m1a FT = m2g - m2a f + m1a = m2g - m2a m1a + m2a = m2g - f a(m1 + m2) = m2g - f a = (m2g - f) / (m1 + m2) = (m2g - μm1g) / (m1 + m2) = ((0.3kg)(10 m/s2) - (0.25)(0.5kg)(10m/s2)) / (0.5kg +0.3kg) = (3 - 1.25) kg*m/s2 / (0.8 kg) = 2.1875 m/s2

• e. FT = f + m1a or FT = m2g - m2a

FT = μm1g + m1a FT = (0.25)(0.5kg)(10m/s2) + 0.5kg(2.1875 m/s2) = 2.34 N FT = m2g - m2a FT = 0.3kg(10m/s2) - 0.3kg(2.1875 m/s2) = 2.34 N

SLIDE 21

Slide 121 / 156

A 10-kilogram block is pushed along a rough horizontal surface by a constant horizontal force F as shown above. At time t = 0, the velocity v of the block is 6.0 meters per second in the same direction as the force. The coefficient of sliding friction is 0.2. Assume g = 10 meters per second squared. a. Calculate the force F necessary to keep the velocity constant.

F

Slide 122 / 156

m = 10 kg v = 6 m/s

μ = 0.2

g = 10 m/s2

• a. Calculate the force F necessary to keep the velocity constant.

Fapp fk FN mg

x y ΣF = ma ΣF = FN - mg = 0 Fapp - fk = 0

FN = mg

Fapp = fk Fapp = μFN Fapp = μmg Fapp = (0.2)(10kg)(10m/s2) Fapp = 20 N

Slide 123 / 156

A helicopter holding a 70-kilogram package suspended from a rope 5.0 meters long accelerates upward at a rate of 5 m/s2. Neglect air resistance on the package.

• a. On the diagram, draw and label all of the forces acting on the package.
• b. Determine the tension in the rope.
• c. When the upward velocity of the helicopter is 30 meters per second, the

rope is cut and the helicopter continues to accelerate upward at 5 m/s2. Determine the distance between the helicopter and the package 2.0 seconds after the rope is cut.

Slide 124 / 156

mg FT a

• a. On the diagram, draw and label all of the forces acting on the package.
• b. Determine the tension in the rope.

ΣF = ma FT - mg = ma FT = mg + ma FT = m (g+a) FT = (70kg) (10 m/s2 + 5 m/s2) FT = (70kg) (15 m/s2) FT = 1050 N m = 70 kg a = 5 m/s2 x = 5 m

Slide 125 / 156

m = 70 kg a = 5 m/s2 x = 5 m v = 30 m/s

• c. When the upward velocity of the helicopter is 30

meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s2. Determine the distance between the helicopter and the package 2 seconds after the rope is cut. ahelicopter apackage Method 1 Use the reference frame of the package after it is released. The relative acceleration of the helicopter is given by: arelative = ahelicopter - apackage arelative = 5 m/s2 - (-10 m/s2) arelative = 15 m/s2 Their relative initial velocity is zero, since they are connected together The helicopter is 5.0 m above the package when released, the length of the rope. y = y0 + v0 + 1/2at2 y = 5m + 0 + 1/2(15 m/s2)(2s)2 y= 35m

Slide 126 / 156

Method 2 Use the reference frame of the earth, but use the initial height of the package as zero. Package Helicopter y = y0 + v0t+ 1/2at2 y = y0 + v0t+ 1/2at2 y = 0 + (30 m/s)(2s) + 1/2(-10 m/s2)(2s)2 y = 5m + (30 m/s)(2s) + 1/2(+5 m/s2)(2s)2 y= 60 m - 20 m y= 5m + 60 m + 10 m y= 40 m y= 75 m Their separation is just the difference in their positions Δy = yhelicopter - ypackage Δy = 75m - 40m Δy = 35m m = 70 kg a = 5 m/s2 x = 5 m v = 30 m/s

• c. When the upward velocity of the helicopter is 30

meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s2. Determine the distance between the helicopter and the package 2 seconds after the rope is cut. ahelicopter apackage

SLIDE 22

Slide 127 / 156

Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings,

• ne of which passes over a frictionless pulley of negligible mass, as shown above.

Calculate each of the following.

• a. The acceleration of the 4-kilogram block
• b. The tension in the string supporting the 4-kilogram block
• c. The tension in the string connected to the l-kilogram block

1 kg 2 kg 4 kg

Slide 128 / 156

m1 = 1 kg m2 = 2 kg m3 = 4 kg g = 9.8 m/s2

• a. The acceleration of the 4-kilogram block

ΣF = ma m1: m2: m3: FT1 = m1a FT2 - FT1 = m2a FT2 - m3g = -m3a FT2 - m1a = m2a FT2 - m3g = -m3a FT2 = m1a + m2a FT2 = m3g - m3a m1a + m2a = m3g - m3a m1a + m2a + m3a = m3g a(m1 + m2 + m3) = m3g a = m3g / (m1 + m2 + m3) a = (4kg)(9.8 m/s2) / (1kg + 2kg + 4kg) a = 5.6 m/s2

4 kg 1 kg 2 kg FN FN m1g m2g m3g FT1 FT1 FT2 FT2 a a a

Slide 129 / 156

m1 = 1 kg m2 = 2 kg m3 = 4 kg g = 9.8 m/s2

• b. The tension in the string supporting the 4-kilogram block

FT2 = m1a + m2a FT2 = 1kg(5.6 m/s2) + 2kg(5.6 m/s2) FT2 = 16.8 N

• r

FT2 = m3g - m3a FT2 = 4kg(9.8 m/s2) - 4kg(5.6 m/s2) FT2 = 16.8 N

• c. The tension in the string connected to the l-kilogram block

FT1 = m1a FT1 = 1kg(5.6 m/s2) FT1 = 5.6 N

Slide 130 / 156

A 10-kilogram block rests initially on a table as shown in cases I and II above. The coefficient of sliding friction between the block and the table is 0.2. The block is connected to a cord of negligible mass, which hangs over a massless frictionless pulley. In case I a force of 50 newtons is applied to the cord. In case II an object of mass 5 kilograms is hung on the bottom of the cord. Use g = 10 meters per second squared.

• a. Calculate the acceleration of the 10-kilogram block in case I.
• b. On the diagrams below, draw and label all the forces acting on each block in

case II

• c. Calculate the acceleration of the 10-kilogram block in case II.
• d. Calculate the tension force in case II.

F Case I 10 kg 10 kg 5 kg Case II

Slide 131 / 156

m1 = 10 kg m2 = 5 kg μ = 0.2 FApp = 50 N g = 10 m/s2

• a. Calculate the acceleration of the 10-kilogram block in case I.

ΣF = ma FApp - fK = ma FApp - μmg = ma a = (FApp - μmg)/m a = (50N - 0.2(10kg)(10m/s2))/10kg a = 3 m/s2

y ΣF = ma ΣF = FN - mg = 0 FN = mg fk = μFN fk = μmg

• b. On the diagrams below, draw and label all the forces acting on each block in

case II

FN m1g f FT FT m2g 5 kg 10 kg

Slide 132 / 156

m1 = 10 kg m2 = 5 kg μ = 0.2 FApp = 50 N g = 10 m/s2

• c. Calculate the acceleration of the 10-kilogram block in case II.

ΣF = ma FT - f = m1a FT - m2g = -m2a FT - μm1g = m1a FT = m2g - m2a FT = μm1g + m1a μm1g + m1a = m2g - m2a m1a + m2a = m2g - μm1g a(m1 + m2) = g(m2 - μm1) a = g(m2 - μm1)/(m1 + m2) = 10 m/s2(5kg - 0.2*10kg)/(5kg +10kg) a = 2 m/s2

• d. Calculate the tension force in case II.

FT - m2g = -m2a FT = m2g - m2a FT = m2(g - a) = 5kg (10 m/s2 - 2 m/s2) = 40 N

SLIDE 23

Slide 133 / 156

In the system shown above, the block of mass M1 is on a rough horizontal table. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction mk between M1 and the table is less than the coefficient of static friction ms

• a. On the diagram below, draw and identify all the forces acting on the block of mass

M1.

• b. In terms of M1 and M2 determine the minimum value of ms that will prevent the

blocks from moving. The blocks are set in motion by giving M2 a momentary downward push. In terms of M1, M2, mk, and g, determine each of the following:

• c. The magnitude of the acceleration of M1
• d. The tension in the string.

M1 M2

Slide 134 / 156

• a. On the diagram below, draw and identify all the forces acting on the block of

mass M1.

• b. In terms of M1 and M2 determine the minimum value of ms that will prevent the

blocks from moving. The pulley system is not moving, so acceleration is zero. ΣF = ma x-direction for M1: FT - f = 0 FT = f y-direction for M1: FN - M1g = 0 FN = M1g y-direction for M2: FT - M2g = 0 FT = M2g f = μsFN or msFN in this case f = msM1g = M2g ms = M2 / M1 M1g FN FT f M1, M2, mk, ms, g

Slide 135 / 156

The blocks are set in motion by giving M2 a momentary downward push. In terms of M1, M2, mk, and g, determine each of the following:

• c. The magnitude of the acceleration of M1

ΣF = ma x-direction for M1: FT - f = M1a FT = f + M1a y-direction for M1: FN - M1g = 0 FN = M1g f = μkFN or mkFN in this case f = mkM1g y-direction for M2: FT - M2g = -M2a FT = M2g - M2a FT = f + M1a, FT = M2g - M2a mkM1g + M1a = M2g - M2a M1a + M2a = M2g - mkM1g a(M1 + M2) = M2g - mkM1g a = M2g - mkM1g / (M1 + M2)

• d. The tension in the string.

FT = f + M1a or FT = M2g - M2a FT = mkM1g + M1(M2g - mkM1g / (M1 + M2) ) or FT = M2g - M2(M2g - mkM1g / (M1 + M2) )

Slide 136 / 156

Two 10-kilogram boxes are connected by a massless string that passes over a massless frictionless pulley as shown above. The boxes remain at rest, with the

• ne on the right hanging vertically and the one on the left 2.0 meters from the

bottom of an inclined plane that makes an angle of 60° with the horizontal. The coefficients of kinetic friction and static friction between the Ieft-hand box and the plane are 0.15 and 0.30, respectively. You may use g = 10 m/s2, sin 60° = 0.87, and cos 60° = 0.50. a. What is the tension FT in the string. b. On the diagram above, draw and label all the forces acting on the box that is

• n the plane.

c. Determine the magnitude of the frictional force acting on the box on the plane. d. The string is then cut and the left-hand box slides down the inclined plane. What is the magnitude of its acceleration?

m1 m2 2 m

60o

Slide 137 / 156

m1 = 10 kg m2 = 10 kg x = 2 m θ = 60o

, sin 60° = 0.87, cos 60° = 0.50

μk = 0.15, μs = 0.30 g = 10 m/s2

• a. What is the tension FT in the string.

The acceleration is 0 because the system is not moving. ΣF = ma x-direction for m1: FT - f - m1gsinθ = 0 FT = f + m1gsinθ y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μsFN f = μsm1gcosθ FT = f + m1gsinθ FT = μsm1gcosθ + m1gsinθ FT = (0.30)(10kg)(10 m/s2)(0.50) + (10kg)(10 m/s2)(0.87) FT = 102 N

Slide 138 / 156

m1 = 10 kg m2 = 10 kg x = 2 m θ = 60o

, sin 60° = 0.87, cos 60° = 0.50

μk = 0.15, μs = 0.30 g = 10 m/s2

• b. On the diagram above, draw and label all the forces acting on the box that is on the

plane.

• c. Determine the magnitude of the frictional force acting on the box on the plane.

y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μsFN f = μsm1gcosθ f = 0.30(10kg)(10 m/s2)(0.50) f = 15 N

m1 FN f m1g FT

60o

SLIDE 24

Slide 139 / 156

m1 = 10 kg m2 = 10 kg x = 2 m θ = 60o

, sin 60° = 0.87, cos 60° = 0.50

μk = 0.15, μs = 0.30 g = 10 m/s2

• d. The string is then cut and the left-hand box slides down the inclined plane. What is

the magnitude of its acceleration? ΣF = ma x-direction of m1: f - m1gsinθ = -m1a y-direction for m1: FN - m1gcosθ = 0 FN = m1gcosθ f = μkFN = μkm1gcosθ m1a = m1gsinθ - f a = (m1gsinθ - μkm1gcosθ) / m1 a = ((10kg)(10 m/s2)(0.87) - (0.15)(10kg)(10 m/s2)(0.50)) / 10kg a = 7.95 m/s2

m1 m2

60o

a f FN m1g

Slide 140 / 156

Masses ml and m2, are connected by a light string. They are further connected to a block of mass M by another light string that passes over a pulley. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle θ with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f. a. On the figure above, draw and label all the forces on block m1, m2, and M. Express your answers to each of the following in terms of ml, m2, g, θ, and f. b. Determine the coefficient of kinetic friction between the inclined plane and block 1. c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane. d. The string between blocks 1 and 2 is now cut. Determine the acceleration

• f block 1 while it is on the inclined plane.

m1 m2 M v

θ

Slide 141 / 156

m1 m2 M v

θ

• a. On the figure above, draw and label all the forces on block m1, m2, and M.

m1 v m1g FT1 f FN a = 0 m2 v m2g FT2 f FN a = 0 FT1 M Mg FT2

θ θ v

a = 0

Slide 142 / 156

Express your answers to each of the following in terms of ml, m2, g, θ, and f.

• b. Determine the coefficient of kinetic friction between the inclined plane and block 1.

f = μFN ΣF = ma FN - m1gcosθ = 0 FN = m1gcosθ f = μm1gcosθ μ = f /(m1gcosθ)

• c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with

constant velocity down the plane. m1: ΣF = ma M: FT2 - Mg = 0 FT1 - m1gsinθ - f = 0 FT2 = Mg FT1 = m1gsinθ m2: ΣF = ma FT2 - FT1 - m2gsinθ - 2f = 0 Mg - m1gsinθ - f - m2gsinθ - 2f = 0 Mg = m1gsinθ + m2gsinθ + 3f M = (m1gsinθ + m2gsinθ + 3f) / g

Slide 143 / 156

• d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1

while it is on the inclined plane. x-direction: ΣF = ma f - m1gsinθ = -m1a m1gsinθ - f = m1a a = (m1gsinθ - f) / m1

m1 m1g f FN a

θ

Slide 144 / 156

Calculus Based

SLIDE 25

Slide 145 / 156 Fluid Resistance

v Fresistance When first learning dynamics discussed cases in which an

• bject moves with any friction. Then we began to take the

frictional force into account between an object and its

• surface. Now we are going to begin taking into account air

resistance.

Slide 146 / 156

a t

g t v y t

Fluid Resistance

No resistance v = vo + gt vo = 0 a is constant v increasing mg = ma a = g

Slide 147 / 156 Fluid Resistance

If we do not take into account air resistance then the

• bject will fall with a constant acceleration.

If air resistance however is taken into account then the objects acceleration will eventually be cancelled

• ut by the drag force.

Since the acceleration will essentially become zero, at some point in time your velocity will be constant.

Slide 148 / 156

With resistance

FR FR a=g vo=0

Fluid Resistance

When the falling object stops accelerating it reaches its terminal velocity. In this example the equation for the resistive force is given as F = kv, but in other problems the resisitive force equation can be different.

Slide 149 / 156

28

If the equation for the force due to air resistance is kv2. What is the terminal velocity reached by the object? A B C D E

Slide 150 / 156

Velocity with respect to Time

SLIDE 26

Slide 151 / 156 Graphical Representation

• f the Fluid Resistance

Equation

x t a t v t

Slide 152 / 156

v t

29 Which of these is the graph that best represents the general velocity versus time for air resistance? A B C D

v t v t

t v

Slide 153 / 156

We have learned before that distance is the integral of velocity so by integrating the velocity equation we were able to come up with the distance equation.

Slide 154 / 156

30 Which of these is the graph that best represents the general position versus time for air resistance? A

B C D

x t x t x t t x

Slide 155 / 156

The final equation shows how acceleration approaches zero.

Slide 156 / 156

31 Which of these is the graph that best represents the general acceleration versus time for air resistance? A

B C D

a t

a

t

a t

t a