dynamics in two dimensions -...
TRANSCRIPT
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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
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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
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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
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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
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Normal Force and WeightFN
mg
The Normal Force, FN, is ALWAYS perpendicular to the surface.
Weight, mg, is ALWAYS directed downward.
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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
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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
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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
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Two Dimensions
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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
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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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A free body diagram would include this, and all other forces, as seen below.
Resolving Forces
Ff
mg
FN
Fapp
Fapp
FN
Ff
mg
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#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
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30o
x (horizontal) component
y (vertical) componentFy
Fx
F = 50 N
We can use COSINE = Adjacent / Hypotenuse to find Fx
So Fx = 43.3 N
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30o
x (horizontal) component
y (vertical) componentFy
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:
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30o
x (horizontal) component
y (vertical) componentFy
Fx
F = 50 N
We can use SINE = Opposite/ Hypotenuse to find Fy
So Fy = 25 N
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30o
x (horizontal) component
y (vertical) componentFy
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!
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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.
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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!
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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!
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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.
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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
Force and friction acting on an object
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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
Force and friction acting on an object
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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.
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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 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?
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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 our choice.
yX
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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
yX
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Forces at angles acting on an object
FN
mg
f
yX
Fy
Fx
Once we do that, we can now proceed as we did previously, just using each component appropriately.
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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
yX
Fy
Fx
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FAPP = 20N at 37o m = 3.0kgμk = 0.20
Force and friction acting on an object
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
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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#
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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.
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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
yX
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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
yX
Fy
Fx
FN
mg
Just looking at the y-axis
ΣF = ma
FN - Fy - mg = 0
FN = mg + Fy
FN = mg + Fsin# Fy
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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
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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
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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
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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
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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.
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On the picture, draw the free body diagram for the block.
Show the weight and the normal force.
Inclined Plane
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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.
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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.
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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
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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
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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
#
#'
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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
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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...
#' = #
#
#'
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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
F y
#aFy
Fx
#
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Slide 50 / 103
#Fx
F y#
aFx
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.
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Slide 51 / 103
#Fx
F y#
aFx
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.
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Slide 52 / 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
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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.
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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.
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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
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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
Fx = mg sin #
F y = m
g co
s #
#
a
Inclined Plane Problems
x - axis y - axis
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Slide 57 / 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
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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
yx
FN
mg sin#
mg
cos#
#
fk a
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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
yx
FN
mg sin#
mg
cos#
#
fk a#Fy = mayFN = mg cos#
x - axis y - axis
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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
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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
yx
FN
mg sin#
mg
cos#
#
fk
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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 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
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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
FNm
g
mg sin #
mg cos #
#fk
Answer
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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
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
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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
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
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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 our reference frame so that the +x axis is parallel to the inclined plane and the +y axis points in the direction of FN.
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Slide 67 / 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
#
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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
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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 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
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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
http://njc.tl/70
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Slide 71 / 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.
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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
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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?
http://njc.tl/71
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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
http://njc.tl/72
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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 object is zero.
Tension Force
mg
T1T2
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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
T1T2
#Fy = T1y + T2y - mg
0 = T1y + T2y - mg
mg = T1y + T2y
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Slide 77 / 103
And the horizontal components must add to zero.
Tension Force
mg
T1T2 #Fx = -T1x + T2x
0 = -T1x + T2x
T1x = T2x
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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
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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
T2yT1y
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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
T1yT1
=T1sin(#)
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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
T1yT1=T1cos(#)
=T1sin(#)
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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(#)
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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(#
)
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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!
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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
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Slide 86 / 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.
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Slide 87 / 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.
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Slide 88 / 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
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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
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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)
http://njc.tl/74
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Slide 91 / 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.
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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.
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Slide 93 / 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.
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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
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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...
r90o
F
# = Fr
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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 ?
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Slide 97 / 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
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Rotational Equilibrium1.0m3.0m 4.0m
4kg 2kg
What mass must be added at distance 4.0m to put the above apparatus into equilibrium?
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Slide 99 / 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
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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.
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Slide 101 / 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.
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Slide 102 / 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
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Slide 103 / 103