Dynamics in Two Dimensions -...

/ 103

Dynamics in

Two Dimensions

/ 103

· 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

/ 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 object, the second object exerts an equal force in the opposite direction on the first object.

ΣF = ma

/ 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

/ 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

/ 103

Normal Force and WeightFN

mg

The Normal Force, FN, is ALWAYS perpendicular to the surface.

Weight, mg, is ALWAYS directed downward.

/ 103

Kinetic Friction

fK

Friction forces are ALWAYS parallel to the surface exerting them.

Kinetic friction is always directed opposite to direction the object is sliding and has magnitude:

fK = μkFN

v

/ 103

Static Friction

fS

Static friction is always equal and opposite the Net Applied Force acting on the object (not including friction).

Its magnitude is:

fS ≤ μSFN

FAPP

/ 103

When a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force, FT.

Tension Force

FT

mg

a

/ 103

Two Dimensions

/ 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

/ 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

/ 103

A free body diagram would include this, and all other forces, as seen below.

Resolving Forces

Ff

mg

FN

Fapp

Fapp

FN

Ff

mg

/ 103

#x (horizontal) component

y (vertical) componentFy

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

/ 103

30o

x (horizontal) component


Fx

F = 50 N

We can use COSINE = Adjacent / Hypotenuse to find Fx

So Fx = 43.3 N

/ 103

30o



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

FxFx

FN

Ff

mg

But if we do that, we lose the vertical (y) component of the original force... so we must find that next:

/ 103

30o



Fx

F = 50 N

We can use SINE = Opposite/ Hypotenuse to find Fy

So Fy = 25 N

/ 103

30o



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!

/ 103

Resolving forces practice:

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

40o20N

Ex.

/ 103

45o100N

15o

500N

-25o

330N

22o250N

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!

/ 103

40o80N

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!

/ 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.

/ 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.

FAPPf


/ 103

FN

mg

FAPPf

FAPP = 20Nm = 3.0kgμk = 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)/ma = (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


/ 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 or perpendicular with one another.

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

/ 103


FN

mg

FAPP

f

The next, critical, step is to choose axes.

Previously, we always used vertical and horizontal axes, since one 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 other perpendicular to that.

What's the answer in this case?

/ 103


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 our choice.

yX

/ 103


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

yX

/ 103


FN

mg

f

yX

Fy

Fx

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

/ 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.


FN

mg

f

yX

Fy

Fx

/ 103

FAPP = 20N at 37o m = 3.0kgμk = 0.20


FN

mg

f

yX

FAPPy

FAPPx

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)/ma = (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

/ 103

Normal Force and FrictionFriction 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

FAPPyFN

FN

Fy

mg

Just looking at the y-axis

ΣF = ma

FN + Fy - mg = 0

FN = mg - Fy

FN = mg - Fsin#

/ 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.

/ 103


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

yX

/ 103


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

yX

Fy

Fx

FN

mg

Just looking at the y-axis

ΣF = ma

FN - Fy - mg = 0

FN = mg + Fy

FN = mg + Fsin# Fy

/ 103

1 The normal force on the box is:

Fappθ

A mg

B mg sin#

C mg cos#

D mg + F sin#

E mg - F sin#

http://njc.tl/6v

http://njc.tl/6v

/ 103

2 The frictional force on the box is:

Fappθ

A μ(mg + Fsin(θ))

B μ(mg - Fsin(θ))

C μ(mg + Fcos(θ))

D μ(mg - Fcos(θ))

E μmg

http://njc.tl/6w

http://njc.tl/6w

/ 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

θv

m

A mg - Fapp cos#

B mg - Fapp sin#

C mg

D mg + Fapp sin#

E mg + Fapp cos# http://njc.tl/6x

http://njc.tl/6x

/ 103

Fapp

θv

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#) http://njc.tl/6y

http://njc.tl/6y

/ 103

Normal Force and WeightFN

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.

/ 103

On the picture, draw the free body diagram for the block.

Show the weight and the normal force.

Inclined Plane

/ 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.

/ 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 one dimension. To do this, we will call the surface our x-axis.

/ 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

/ 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

/ 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

#

#'

/ 103


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

/ 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...

#' = #

#

#'

/ 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.


Fx

F y

#aFy

Fx

#

/ 103

#Fx

F y#

aFx

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


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

/ 103

#Fx

F y#

aFx

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


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

/ 103

Example 1A 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

/ 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.

/ 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.

/ 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

/ 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

Fx = mg sin #

F y = m

g co

s #

#

a


x - axis y - axis

/ 103

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

/ 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

yx

FN

mg sin#

mg

cos#

#

fk a

/ 103

#Fx = max

mgsin# - fk = ma

mgsin# - μkFN = ma

mgsin# - μkmgcos# = ma

gsin# - μkgcos# = a

a = gsin# - μkgcos#

a = g(sin# - μk cos#)


yx

FN

mg sin#

mg

cos#

#

fk a#Fy = mayFN = mg cos#

x - axis y - axis

/ 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#


/ 103


If the object is sliding with constant velocity: a = 0



0 = g(sin# - μkcos#)

0 = sin# - μkcos#

μk cos# = sin#

μk = sin# / cos#

μk = tan#

a = 0

yx

FN

mg sin#

mg

cos#

#

fk

/ 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 object begins to slide) than:

μs = tan#max

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

a = 0

yx

FN

mg sin#

mg

cos#

#

fs

/ 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

FNm

g

mg sin #

mg cos #

#fk

Answer

/ 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

mg

mg sin #

mg cos #

#

Fapp

fKa = 0

#

y-directionΣF = ma

FN - mg cos θ = 0

FN = mg cos θ

x-directionΣF = 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

/ 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

mgm

g sin #

mg cos #

#

Fapp

fKa

y-axisΣF = ma

FN - mg cos θ = 0

FN = mg cos θ

x-axisΣF = 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

/ 103

FN

mg

#

a

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


y

x

It is helpful to rotate our reference frame so that the +x axis is parallel to the inclined plane and the +y axis points in the direction of FN.

/ 103

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)

FN

mg

mg sin #

mg cos #

#

fsa = 0

y-directionΣF = ma

FN - mg cos θ = 0

FN = mg cos θ

x-directionΣF = ma

fs = mg sin θ

μs mg cos θ = mg sin θ

μs cos θ = sin θ

μs = sin θ / cos θμs = tan θ

θ = tan-1 μs

θ = 21.8o

#

/ 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

/ 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 appliedby 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

http://njc.tl/6z

http://njc.tl/6z

/ 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

http://njc.tl/70

http://njc.tl/70

/ 103

Static EquilibriumThere 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.

/ 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

/ 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?

http://njc.tl/71

http://njc.tl/71

/ 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

http://njc.tl/72

http://njc.tl/72

/ 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 object is zero.

Tension Force

mg

T1T2

/ 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

T1T2

#Fy = T1y + T2y - mg

0 = T1y + T2y - mg

mg = T1y + T2y

/ 103

And the horizontal components must add to zero.

Tension Force

mg

T1T2 #Fx = -T1x + T2x

0 = -T1x + T2x

T1x = T2x

/ 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

/ 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

T2yT1y

/ 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

T1yT1

=T1sin(#)

/ 103

To find the vertical component, we will use Cos(#) = Adj/Hyp

So...

Cos(#) = T1y/T1

T1y = T1 cos(#)

Tension Force

50o

mg

T1x

T1yT1=T1cos(#)

=T1sin(#)

/ 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(#)

/ 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 2y T2

=T2sin(#)

=T2c

os(#

)

/ 103

Tension Force

x - axis y - axis50o 20o

mg

T1x

T2x

T2yT1y

Now we can put it all together using our force equations based on the free body diagram!

/ 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 = 0T1y + T2y - mg = 0T1cos#1 + T2cos#2 = mg(Plug-in back into the x-direction)

x - axis y - axis

50o 20o

mg

T1x

T2x

T2yT1y

/ 103

Tension Force

T1 = 28NT2 = 64N

50o 20o

mg

T1x

T2x

T2yT1y

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.

/ 103

Tension Forceθ

mg

Tx Tx

TyTyθ

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 of Tension would not add to zero.

/ 103

Tension Forceθ

mg

Tx Tx

TyTyθ

θ

θ

#Fx = max = 0

T1x - T2x = 0

Tsin# = Tsin#

#Fy = may = 0T1y + T2y - mg = 0Tcos# + Tcos# = mg 2Tcos# = mg T = mg / (2cos#)

x - axis y - axis

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.

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

/ 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?

T1T2

A T1 < T2

B T1 = T2

C T1 > T2

D T1 + T2 = mg

E T1 - T2 = mg

http://njc.tl/73

http://njc.tl/73

/ 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)

http://njc.tl/74

http://njc.tl/74

/ 103

Torque and Rotational EquilibriumForces 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.

/ 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 object to rotate around its center of mass...or around a pivot point, that force has exerted a torque on the object.

/ 103

Torque and Rotational EquilibriumA 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.

/ 103

Torque and Rotational EquilibriumThe 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 forcer is the distance from the pivot# is the angle of the force to a line to the pivot

/ 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...

r90o

F

# = Fr

/ 103

Rotational Equilibrium

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

/ 103

Rotational Equilibrium

20N

4.0mF1

## = 0

F1r1+ F2r2 =0

F1r1 = - F2r2

F1 = - F2r2 / r1

F1 = - (-20N)(4.0m) / (7m)

F1 = (80Nm) / (7m)

F1 = 11.4N

3.0m

/ 103

Rotational Equilibrium1.0m3.0m 4.0m

4kg 2kg

What mass must be added at distance 4.0m to put the above apparatus into equilibrium?

/ 103

Rotational Equilibrium1.0m3.0m 4.0m

4kg 2kg

## = 0

F1r1 + F2r2 - F3r3 = 0

+m1gr1 + m2gr2 - m3gr3 = 0

+m1r1 + m2r2 - m3r3 = 0m3 = (m1r1 + m2r2) / r3 m3 = ((4kg)(3m) +(2kg)(1m)) / 4m

F1 = (14kg-m)) / (4m)

F1 = 3.2kg

/ 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.

/ 103

Fappθ

A force of 500 N is applied at an angle of 30o from the horizontal, as shown below. Determine the Normal Force on the box that results from this situation.

2.

/ 103

A box with mass 14.0 kg sits on an incline plane at anangle 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

Dynamics in Two Dimensions -...

Documents

Transcript of Dynamics in Two Dimensions -...