[PDF] - Dynamics Newton's Three Laws of Motion Inertial Reference Frames PDF Document

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Dynamics in Two Dimensions

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· 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 / 103 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 / 103

Newton's laws are only valid in inertial reference frames: In an inertial frame of reference, all motion has a constant direction and magnitude. This is not the case in rotating and accelerating frames.

Inertial Reference Frames Slide 5 / 103

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 6 / 103 Normal Force and Weight

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

SLIDE 2

Slide 7 / 103 Kinetic Friction

fK 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 v

Slide 8 / 103 Static Friction

fS Static friction is always equal and

pposite the Net

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

Slide 9 / 103

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 10 / 103

Two Dimensions

Slide 11 / 103

Since forces are vectors, they may have both a horizontal and vertical influence on an object. In order to solve problems using forces acting at an angle, we must find the horizontal (x) and vertical (y) components of the forces using trigonometry (right triangles/SohCahToa).

Resolving Forces Slide 12 / 103

Consider a child pulling a wagon down the street. The wagon has a handle that is not vertical, not horizontal, but at an angle. This means the child is pulling UP and OVER at the same time.

Resolving Forces

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Slide 13 / 103

A free body diagram would include this, and all

ther forces, as seen below.

Resolving Forces

Ff mg FN Fapp Fapp FN Ff mg

Slide 14 / 103

# x (horizontal) component y (vertical) component

Fy Fx F = 50 N To find the net force on the object, we consider each component separately. Let's assume the force the child pulled with was 50 Newtons at 30o

# x (horizontal) component y (vertical) component

Slide 15 / 103

30o x (horizontal) component y (vertical) component

Fy Fx F = 50 N We can use COSINE = Adjacent / Hypotenuse to find Fx So Fx = 43.3 N

Slide 16 / 103

30o x (horizontal) component y (vertical) component

Fy Fx F = 50 N = 43.3 N The horizontal (x) component of the force is equal to 43.3 N. We can include this on a free body diagram:

Ff mg FN Fx Fx FN Ff mg

But if we do that, we lose the vertical (y) component

f the original force... so we must find that next:

Slide 17 / 103

30o x (horizontal) component y (vertical) component

Fy Fx F = 50 N We can use SINE = Opposite/ Hypotenuse to find Fy So Fy = 25 N

Slide 18 / 103

30o x (horizontal) component y (vertical) component

Fy Fx F = 50 N = 43.3 N The vertical (y) component of the force is equal to 25 N. We can now add this to complete the free body diagram:

Ff mg FN Fx Fx FN Ff mg

= 25 N

Fy Fy

Notice that our original force Fapp is no longer shown... it can be replaced by the x and y components!

SLIDE 4

Slide 19 / 103 Resolving forces practice:

Resolve each of the forces into x and y components, and then show the components on a free body diagram.

40o 20N Ex.

Slide 20 / 103

45o 100N 15o 500N

25o

330N 22o 250N 1. 2. 3. 4.

Resolving forces practice:

Resolve each of the forces into x and y components, and then show the components on a free body diagram. Show your work on a separate page!

Slide 21 / 103

40o 80N 35o 600N

60o

24N 12o 1500N 1. 2. 3. 4.

Resolving forces Homework:

Resolve each of the forces into x and y components, and then show the components on a free body diagram. Show your work on a separate page!

Slide 22 / 103 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 23 / 103

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

Force and friction acting on an object Slide 24 / 103

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 5

Slide 25 / 103 Forces at angles acting on an object

Now we'll solve problems where the forces act at an angle so that it is not parallel

r perpendicular with
ne another.

First we do a free body diagram, just as we did previously.

Slide 26 / 103 Forces at angles acting on an object

FN mg FAPP f

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 27 / 103 Forces at angles acting on an object

FN mg FAPP f

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'll get answers that don't make sense, and we'll have to reconsider

ur choice.

y

X

Slide 28 / 103 Forces at angles acting on an object

FN mg FAPP f

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

y

X

Slide 29 / 103 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, just using each component appropriately.

Slide 30 / 103

Let's use our work to find the acceleration if the applied force is 20N at 37o above horizontal, the box has a mass of 3.0kg, and the coefficient of kinetic friction is 0.20.

Force and friction acting on an object

FN mg f

y

X

Fy F

x

SLIDE 6

Slide 31 / 103

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 - Fsin# 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 Fcos# - μkmg = ma a = (Fcos# - 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 32 / 103 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 - Fsin#

Slide 33 / 103 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.

Slide 34 / 103 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 35 / 103 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 + Fsin#

Fy

Slide 36 / 103

1 The normal force on the box is:

Fapp θ

A

mg

B

mg sin#

C mg cos# D

mg + F sin#

E mg - F sin#

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SLIDE 7

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2 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

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Slide 38 / 103

3 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:

Fapp θ

v m A mg - Fapp cos# B mg - Fapp sin# C mg D mg + Fapp sin# E mg + Fapp cos#

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Slide 39 / 103

Fapp θ

v m

4 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 cos#) B µ(mg - Fapp sin#) C μmg D µ(mg + Fapp sin#) E µ(mg + Fapp cos#)

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Slide 40 / 103 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 41 / 103

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

Inclined Plane Slide 42 / 103 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 8

Slide 43 / 103 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 will change our axes so that the acceleration is all in

ne dimension. To do

this, we will call the surface our x-axis.

Slide 44 / 103 Choosing Axes

FN mg In this case, the block can only accelerate along the surface of the plane. Even if there is no acceleration in a problem, we will use the surface as the 'x- axis'. So we rotate our x-y axes to line up with the surface of the plane.

a

Slide 45 / 103 Choosing Axes

FN mg In this case, the block can only accelerate along the surface of the plane. So we rotate our x-y axes to line up with the surface of the plane.

y

X

a

Slide 46 / 103

Unlike the last section, in this case # is both the angle of incline and the angle between mg and the new y-axis. This means we will need to use different functions to find Force in the x and Force in the y dimension.

Inclined Plane Problems

y-axis

# #'

Slide 47 / 103 Inclined Plane Problems

y-axis #

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

Slide 48 / 103

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

Inclined Plane Problems

y-axis #

Since we have a right angle between mg and the surface, the angle # in the triangle complements the angle #' from the y-axis. #' + # = 90o But we already showed that # + # = 90o So we can conclude... #' = # # #'

SLIDE 9

Slide 49 / 103

mg #

a

We need to account for gravity going INTO the surface (y) and ALONG the surface (x). In order to do this, we resolve mg into its x and y components.

Inclined Plane Problems

Fx Fy #

a

Fy

F

x

#

Slide 50 / 103

#

Fx Fy #

a

F

x

We need to find the x and y components of mg using SohCahToa.

Inclined Plane Problems

For Fx we have our OPPOSITE side and our HYPOTENUSE, so we will use SINE.

Slide 51 / 103

#

Fx Fy #

a

F

x

We need to find the x and y components of mg using SohCahToa.

Inclined Plane Problems

For Fy we have our ADJACENT side and our HYPOTENUSE, so we will use COSINE.

Slide 52 / 103 Example 1

A 20 kg mass sits on an inclined plane at an angle of

40o. Determine the forces ALONG (x) and INTO (y) the

surface of the inclined plane. 20kg 40o

Slide 53 / 103

Incline Plane Practice Problems Find Fx and F-y for each example below:

m = 80kg # = 25o m = 2.0 kg # = 37o m = 150kg # = 45o

1. 2. 3. Slide 54 / 103

Incline Plane Homework Problems Find Fx and F-y for each example below:

m = 40 kg # = 17o m = 8.0 kg # = 42o m = 10 kg # = 73o

1. 2. 3.

SLIDE 10

Slide 55 / 103 Putting it all together:

FN mg In order to study the motion of the block along the plane, we can now evaluate our free body diagram using Fx and Fy.

y

X

a

Slide 56 / 103

Now we just use Newton's Second Law, which is true for each axis. #Fx = max mgsin# = max a = gsin# down the plane #Fy = may FN - mgcos# = 0 FN = mgcos# FN

F

x

= mg sin # F

y

= mg cos # #

a

Inclined Plane Problems

x - axis y - axis

Slide 57 / 103

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)

# 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 # mg cos # #

Answer

Slide 58 / 103

FN mg # 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 sin# mg cos#

#

fk

a

Slide 59 / 103

#Fx = max mgsin# - fk = ma mgsin# - μkFN = ma mgsin# - μkmgcos# = ma gsin# - μkgcos# = a a = gsin# - μkgcos# a = g(sin# - μk cos#)

Inclined Plane Problems with Friction y x

FN

m g s i n # mg cos#

#

fk

a

#Fy = may FN = mg cos# x - axis y - axis

Slide 60 / 103

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

Inclined Plane Problems with Friction

SLIDE 11

Slide 61 / 103

a = g(sin# - μkcos#) If the object is sliding with constant velocity: a = 0

Inclined Plane Problems with Friction

a = g(sin# - μkcos#) 0 = g(sin# - μkcos#) 0 = sin# - μkcos# μk cos# = sin# μk = sin# / cos# μk = tan# a = 0

y x

FN

mg sin# mg cos#

#

fk

Slide 62 / 103 Inclined Plane with Static Friction

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

bject begins to slide) than:

μs = tan#max But this requires that the MAXIMUM ANGLE of incline, #max, be used to determine μs. a = 0

y x

FN

mg sin# mg cos#

#

fs

Slide 63 / 103

A 5 kg block slides down an 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) #

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

FN m g

m g s i n # m g c

s

# #

fk

Answer

Slide 64 / 103

A 5 kg block is pulled up an incline at an angle of 30 degrees 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 # m g c

s

# #

Fapp fK

a = 0

#

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 65 / 103

#

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 # mg cos # #

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 66 / 103

FN mg #

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 12

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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 # m g c

s

# #

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

#

Slide 68 / 103

5 A block with a mass of m slides down an incline as shown above with an acceleration a. Which choice represents the correct free-body diagram? f W N f W N f W N f W N f W N

A B C D E

Slide 69 / 103

6 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

55.25 N

B

62.5 N

C 100.25 N D

107.5 N

E 147 N

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Slide 70 / 103

7 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

55.25 N

B

62.5 N

C 100.25 N D

107.5 N

E 147 N

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Slide 71 / 103 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 72 / 103

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 13

Slide 73 / 103

8 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?

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Slide 74 / 103

9 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: 6 kg 4 kg F

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Slide 75 / 103

It is also possible for two (or more) ropes to support a stationary object (a = 0) by exerting forces at angles. In this case, since it is at rest, the #F on the

bject is zero.

Tension Force

mg T1 T2

Slide 76 / 103

Since the only other force on the object is gravity: The vertical components of the force exerted by each rope must add up to mg.

Tension Force

mg T1 T2

#Fy = T1y + T2y - mg 0 = T1y + T2y - mg mg = T1y + T2y

Slide 77 / 103

And the horizontal components must add to zero.

Tension Force

mg T1 T2

#Fx = -T1x + T2x 0 = -T1x + T2x T1x = T2x

Slide 78 / 103

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

Tension Force

mg T1 T2

mg T1x T2x T2y T1y

#2 #1 T2 T1

SLIDE 14

Slide 79 / 103

Let's calculate the tension in two ropes if the first, T1, is at an angle of 50o from the vertical and the second,T2, is at an angle of 20o from the vertical and they are supporting an 8.0 kg mass.

Tension Force

50o

20o

mg T1x T2x T2y T1y

Slide 80 / 103

Let's start with T1. To find the horizontal component, we will use Sin(#) = Opp/Hyp So... Sin(#) = T1x/T1 T1x = T1 sin(#)

Tension Force

50o

mg T1x T1y T1

=T1sin(#)

Slide 81 / 103

To find the vertical component, we will use Cos(#) = Adj/Hyp So... Cos(#) = T1y/T1 T1y = T1 cos(#)

Tension Force

50o

mg T1x T1y T1

=T1cos(#) =T1sin(#)

Slide 82 / 103

Moving on to T2: To find the horizontal component, we will AGAIN use Sin(#) = Opp/Hyp So... Sin(#) = T2x/T2 T2x = T2 sin(#)

Tension Force

20o

mg T2x T2y T2

=T2sin(#)

Slide 83 / 103

To find the vertical component, we will use Cos(#) = Adj/Hyp So... Cos(#) = T2y/T2 T2y = T2 cos(#)

Tension Force

20o

mg T2x T

2 y

T2

=T2sin(#) =T2cos(#)

Slide 84 / 103 Tension Force

x - axis y - axis

50o

20o

mg T1x T2x T2y T1y Now we can put it all together using our force equations based on the free body diagram!

SLIDE 15

Slide 85 / 103

Tension Force

#Fx = max = 0 T1x - T2x = 0 T1sin#1 = T2sin#2 T1 = T2sin#2/sin#1 T1 = T2 (sin20o/sin50o) T1 = T2 (0.34/0.77) T1 = 0.44 T2 (Solve for T2 in the y-direction) T1 = 0.44 (64N) T1 = 28N #Fy = may = 0 T1y + T2y - mg = 0 T1cos#1 + T2cos#2 = mg (Plug-in back into the x-direction)

x - axis y - axis

50o 20o

mg T1x T2x T2y T1y

Slide 86 / 103 Tension Force

T1 = 28N T2 = 64N

50o

20o

mg T1x T2x T2y T1y Note that the tension 2 at an angle of 20o is significantly larger than the tension 1 at an angle of 50o. This is because the y-component of the tension is 'more vertical' in T2 than in T1. This will always be the case... Tensions at a smaller angle from the vertical will be GREATER and tensions at a larger angle from the vertical will be SMALLER.

Slide 87 / 103 Tension Force

θ

mg Tx Tx Ty Ty

θ θ θ

*A SPECIAL CASE! If the ropes form equal angles to the vertical, the tension in each must also be equal, otherwise the x-components

f Tension would not add to zero.

Slide 88 / 103 Tension Force

θ

mg Tx Tx Ty Ty

θ θ θ

#Fx = max = 0 T1x - T2x = 0 Tsin# = Tsin# #Fy = may = 0 T1y + T2y - mg = 0 Tcos# + Tcos# = mg 2Tcos# = mg T = mg / (2cos#)

x - axis y - axis

Note that the tension rises as cos# becomes smaller...which occurs as # approaches 90

.

It goes to infinity at 90

, which shows that the ropes can never be perfectly horizontal.

This confirms that if the angles are equal, the tensions are equal

Slide 89 / 103

10 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? T1 T2

A

T1 < T2

B T1 = T2 C T1 > T2 D T1 + T2 = mg E T1 - T2 = mg

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Slide 90 / 103

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

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SLIDE 16

Slide 91 / 103 Torque and Rotational Equilibrium

Forces act on object and create motion in a LINEAR direction. When an action on an object causes it to move in a ROTATIONAL direction, it is called TORQUE. Rotational dynamics is a major topic of AP Physics C:

Mechanics. It isn't particularly difficult, but for AP B, we only

need to understand the static case, where all the torques cancel, add to zero. First we need to know what torque is.

Slide 92 / 103 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've 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 93 / 103 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 94 / 103 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: # = Frsin# # (tau) is the symbol for torque; F is the applied force r is the distance from the pivot # is the angle of the force to a line to the pivot

Slide 95 / 103 Torque and Rotational Equilibrium

# = Frsin# When r decreases, so does the torque for a given force. When r = 0, # = 0. We will only study cases in which the force is applied at

90o. In this case sin(90o) =

1, so our equation becomes...

r 90o F

# = Fr

Slide 96 / 103 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 offset a 20N force acting at a distance of 4.0m from the pivot of a door ?

SLIDE 17

Slide 97 / 103 Rotational Equilibrium

20N 4.0m F1 ## = 0 F1r1+ F2r2 =0 F1r1 = - F2r2 F1 = - F2r2 / r1 F1 = - (-20N)(4.0m) / (7m) F1 = (80Nm) / (7m) F1 = 11.4N 3.0m

Slide 98 / 103 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 99 / 103 Rotational Equilibrium

1.0m 3.0m 4.0m 4kg 2kg ## = 0 F1r1 + F2r2 - F3r3 = 0 +m1gr1 + m2gr2 - m3gr3 = 0 +m1r1 + m2r2 - m3r3 = 0 m3 = (m1r1 + m2r2) / r3 m3 = ((4kg)(3m) +(2kg)(1m)) / 4m F1 = (14kg-m)) / (4m) F1 = 3.2kg

Slide 100 / 103

60o 120N 1.

Dynamics Quiz 1:

A man pulls a heavy suitcase along at an angle of 60o from the horizontal with a force of 120 N, as shown below. Determine the horizontal and vertical components of the force applied to the suitcase.

Slide 101 / 103

Fapp θ

A force of 500 N is applied at an angle of 30

from the horizontal, as

shown below. Determine the Normal Force on the box that results from this situation.

2. Slide 102 / 103

A box with mass 14.0 kg sits on an incline plane at an angle of 37o. Determine the components of the force of gravity on the box ALONG and INTO the plane (Fx and Fy)

# = 37o

3.

14.0 kg

SLIDE 18