Classical mechanics describes the motion of object and the forces acting on them. Classical...
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Transcript of Classical mechanics describes the motion of object and the forces acting on them. Classical...
Classical mechanics describes the motion of object and the forces acting on them.Classical mechanics is very accurate as long as we do not try to study something as small as an atom, or something moving close to the speed of light.
A force is an action exerted on an object which may change the object’s state of rest or motion.
EQ: What is a force?
Forces can cause a massive object to accelerate.
The SI unit of force is the newton, N.
Forces can act through contact or at a distance.
What happened to the door?
Newton’s Contributions Calculus Light is composed of
rainbow colors Reflecting Telescope
Your objectives are to learn.
Three Laws of Motion Theory of Gravitation
What is a force?
Here are some ways of describing forces: A push A pull A stretch A squeeze A catch A twist
We can’t see forces but we can see the effects of a force.
Acceleration
TensionCompression-DeformationDe-accelerationTorque
Any change in motion
Forces can act through contact or at a distance.
Forces can be labeled into two categories: field force and contact force.
N= kg * m s2
lb = slug * ft s2
X =
Mass vs. weight: Do you have the same amount of mass on the Moon as
you do the Earth?
Force = Mass*Acceleration: Units
Mass vs. weight: Do you have the same amount of mass on the Moon as
you do the Earth?
Mass- a measure in the amount of matter in an object.
Weight: a measure of the amount of gravitational force acting on the mass of an object.
EX:
Wf= the weight of an object due to a gravitational fieldm = mass of the objectg = acceleration due to gravity
mgWf
Wf = 195lbgE = 32.2ft/s2
m = ?
Wf = mg m = Wf
g
Wf = mg m = 195lb * s2
32.2ft= 6.055 but what are the units?
Weight
W = mg weight is a force weight = mass x acceleration due to gravity
Units N = kg x m/s2
weight in Newtons = mass in kg x 9.81 m/s2
a 1 kg mass weighs 9.81 N
Wf = mg m = 195lb * s2
32.2ft
Derived Units
m = 195slug * ft * s2
s2 32.2ft
: Pause for a Cause
Your Turn: Calculate your mass
NOTE: MASS and WEIGHT are NOT the same thing. MASS never changesWhen an object moves to a different planet.
Pause for a Cause
What is the weight of an 85.3-kg person on earth? On Mars=3.2 m/s2?
NW
NWmgW
MARS 96.272)2.3)(3.85(
94.835)8.9)(3.85(
Force Diagrams The effect of a force depends on both magnitude
and direction. Thus, force is a vector quantity.
Diagrams that show force vectors as arrows are called force diagrams, or Free Body Diagrams.
Force diagrams that show only the forces acting on a single object are called free-body diagrams.
Force Diagrams
In a force diagram, vector arrows represent all the forces acting in a situation.
A free-body diagram shows only the forces acting on the object of interest—in this case, the car.
Force Diagram Free-Body Diagram
The sum of all the forces is called the Net Force.
The key to analyzing problems:pictorial representation of forces complete with labels.
W1,Fg1 or m1g
•Weight(mg) – Always drawn from the center, straight down•Force Normal(FN) – A surface force always drawn perpendicular to a surface.•Tension(T or FT) – force in ropes and always drawn AWAY from object.•Friction(Ff)- Always drawn opposing the motion.
m2g
T
T
FN
Ff
A women is pulling on her suitcase with a force of 70.0 N directed at an angle of +30.0° to the horizontal. Find the x and y components of this force.
Given:F = 70 NΘ = 30.0°
Force in the X-axisFx = F(cos θ) =
Force in the Y-axisFy = F(sin θ) =
Force in the X-axisFx = 70.0 N(cos 30.0°) =
Force in the Y-axisFy = 70.0 N(sin 30.0°) =
60.6 N
35.0 N
soh cah toa
X-axis
Y-axisHyp
Essential Question EQ: A fan blows on two balls, a bowling ball and a balloon.
Describe what you think will happen.
bowling ball
balloon
Thought Experiment: a Pro Golfer tees up a bowling ball and strikes it with his driver. Describe what you think will happen?
I)Inertia:
II)F = ma
III)Action-reaction
An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force.
Force equals mass times acceleration.
For every action there is an equal and opposite reaction.
Inertia: Video Inertia is the tendency of an object to resist being
moved or, if the object is moving, to resist a change in speed or direction.
Newton’s first law is often referred to as the law of inertia because it states that in the absence of a net force, a body will preserve its state of motion.
Mass is a measure of inertia.
Why?
An object in motion remains in motion in a straight line and at a constant speed OR an object at rest remains at rest, UNLESS acted upon by an EXTERNAL (unbalanced) Force.
There are TWO conditions here and one constraint.
Condition #1 – The object CAN move but must be at a CONSTANT SPEEDCondition #2 – The object is at RESTConstraint – As long as the forces are BALANCED!!!!! And if all the forces are balanced the SUM of all the forces is ZERO.
The bottom line: There is NO ACCELERATION in this case AND the object must be at EQILIBRIUM ( All the forces cancel out). 00 Facc
Pause for a Cause: If acceleration = 0, then the sum of the forces must = ?
Newton’s 1st Law of Motion
W
Ry
if Ry = Wthen resultant force = 0if v = 0 and F = 0STATIC EQUILIBRIUM
W
Ry
FpFr
Fr = resistive forceFp = propulsive forceif v = 0 and F = 0DYNAMIC EQUILIBRIUM
V
Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at EQUILIBRIUM.
TIPS for solving problems• Draw a FBD• Resolve anything into COMPONENTS• Write equations of equilibrium• Solve for unknowns
A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N.
a) Calculate the Force of Friction
b) Calculate the Force Normal
mg
Fn
Fa
Ff
NFF fa 50
NFmg n 98)8.9)(10(
Suppose the same box is now pulled at an angle of 30 degrees above the horizontal.
a) Calculate the Force of Friction
b) Calculate the Force Normal
mg
Fn
Fa
Ff30
NFF
NFF
axf
aax
3.43
3.4330cos50cos
Fax
Fay
NF
FmgF
mgFF
mgF
N
ayN
ayN
N
73
30sin50)8.9)(10(
!
If an object is NOT at rest or moving at a constant speed, that means the FORCES are UNBALANCED. One force(s) in a certain direction over power the others.
THE OBJECT WILL THEN ACCELERATE.
Unbalanced Forces
Unequal opposing forces produce an unbalanced force
causing motion
Balanced ForceEqual forces in opposite
directions produce no motion
What is the net force of the truck on bottom?
Force 1 (F1) and force 2 (F2) are applied to the ball at the same time with the same magnitude. Draw a new arrow describing the combined direction of F1 & F2 on the ball. This is called the vector sum.
Graphing assignment
F1 = F2F1 = F2
Vector Sum of Force
Graphing Assignment Two horizontal ropes are attached to a post
that is stuck in the ground. The ropes pull the post producing the following vector forces
Determine the direction and magnitude the support cable must have to prevent pole failure.
F1 = (5, 5) F2 = (-3, 6)
F1 = (5Nx, 5Ny) F2 = (-3Nx, 6Ny)
(X, Y) (X, Y)
Graphing Assignment
If F1 + F2 = -F3 diagram the direction and magnitude that the pole pulls back (F3) on the ropes?
Hint: -F3XY = (F1X + F2X) (F1y + F2y)
Graphing Assignment
What is the angle between F2 & F3?
What is the angle between F1 & F3?
What is the angle between F1 & F2?
Hint: tan(θ) = Opposite / Adjacent •θ = tan -1 (Opposite/Adjacent) or •(Y/X) •(-1) means (1/tan) or inverse.
F1
F2
Fnet = F1 + F2
F1 = F1x + F1y
F2 = F2x + F2y
F1x = F1 cos θ1x = F2x = F2 cos θ2x =
F1y = F1 sin θ1y = F2y = F2 sin θ2y =
F1x = 6.00E2 cos 30.0˚ = F2x = 6.00E2 cos -45.0˚ =
Given:Fnet = ma
a = ?m = 2.00E3 KgFh = 6.00E2 Nθ1 = 30.0˚θ2 = 45.0˚
5.02E2 N4.24E2 N
ΣFx = 9.44E2 N
F1y = 6.00E2 sin 30.0˚ = F2y = 6.00E2 sin -45.0˚ =
3.00E2 N-4.24E2 N
ΣFy = -1.24E2 N
Net Force
ΣFx = F1x + F2x
ΣFy = F1y + F2y
F1
F2
Given:a = ?m = 2.00E3 KgFh = 6.00E2 Nθ1 = 30.0˚θ2 = 45.0˚
ΣFx = 9.44E2 N
ΣFy = -1.24E2 N
F = maax = Fx
m ay = Fy
m
ax = 9.44E2 N = 2.00E3 Kg
2D: Accelerationa2 = ax
2 = ay2
a = √ax2 = ay
2
ay = -1.24E2 N = 2.00E3 Kg
0.472 m/s2
-0.062 m/s2
a = √(0.472)2 + (-0.062)2 =
a = √ax2 = ay
2
0.476 m/s2
Magnitude of Acceleration
F1
F2
Given:a = 0.476 m/s2
m = 2.00E3 KgFh = 6.00E2 Nθ1 = 30.0˚θ2 = 45.0˚
ΣFx = 9.44E2 N
ΣFy = -1.24E2 N
-0.062 m/s2
0.472 m/s2= -7.48˚
Direction of AccelerationThe direction will be the ratio of the acceleration in the x-axis & the y-axis
ay sin θ tan θax cos θ
= ay ax
θ = tan-1 ay ax
θ = tan-1
The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass.
maFm
Fa
maFa
NETNET
NET
1 FFNET
Tips:•Draw an FBD•Resolve vectors into components•Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force.•Solve for any unknowns
zz
yy
x
maF
maF
F
xma
A 10-kg box is being pulled across a frictionless table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box.
mg
FNFa
2/5
0.100.50
sma
a
maFNet
In which direction, is this object accelerating?
The X direction!
So N.S.L. is worked out using the forces in the “x” direction only
A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it.
mg
FNFa
Ff
2/8.3
101250
sma
a
maFF
maF
fa
Net
In which
direction, is this object accelerating?
The X direction!
So N.S.L. is worked out using the forces in the “x” direction only
Slope
F
a
Slope = rise / run = F / a, the slope is equal to the mass. Or, think of y = m x + b, like in algebra class. y corresponds to force, m to mass, x to acceleration, and b (the
y-intercept) is zero.
F
a
Friction- A non-conservative force that opposes motion- Acts parallel to the surfaces in contact.- µ is the coefficient of friction.
- It is unique for each material- It is determined experimentally
What about surface area? Which would have a greater friction force?
Friction force does not depend on the area of contact. Same FN and µ!
stress = F a
TWO types of Friction Static – Friction that keeps an object at rest
and prevents it from moving Kinetic – Friction that acts during motion
-A force that opposes motion
-Acts parallel to the surfacesin contact.
Force of Friction
The Force of Friction is directly related to the Force Normal.
Mostly due to the fact that BOTH are surface forces
Nkkf
Nssf
Nf
FF
FF
FF
friction oft coefficien
alityproportion ofconstant
Note: Friction ONLY depends on the MATERIALS sliding against each other, NOT on surface area.
The coefficient of friction is a unitless constant that is specific to the material type and usually less than one.
Pause for a CauseA 1500 N crate is being pushed
across a level floor at a constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the figure.
a) What is the coefficient of kinetic friction between the crate and the floor?
mg
FNFa
20
Ff
Fay
Fax
NN
N
FF
F
FF
F
F
F
FF
k
Ny
x
Ny
f
N
fk
Nkf
150020sin600
20cos600
sin
cos
kk
= 0.33
Pause for a CauseIf the 600 N force is instead pulling the
block at an angle of 20° above the horizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a)
mg
FN
Ff
20
Fa
Fax
Fay
m
Fa
maF
Net
Net
a
a
9.152
57.4288.563
81.91500
)20sin6001500(331.020cos600
gF
FFF
Net
yNx )sin(cos
m
FFF yfx )(
= 0.884 m/s2
FN = mg
Newton’s Third Law“For every action there is an EQUAL and
OPPOSITE reaction. This law focuses on action/reaction pairs (forces) They NEVER cancel out
All you do is SWITCH the wording!•PERSON on WALL•WALL on PERSON
N.T.LThis figure shows the force during a collision between a truck and a train. You can clearly see the forces are EQUAL and OPPOSITE. To help you understand the law better, look at this situation from the point of view of Newton’s Second Law.
TrainTrainTruckTruck
TrainTruck
aMAm
FF
There is a balance between the mass and acceleration. One object usually has a LARGE MASS and a SMALL ACCELERATION, while the other has a SMALL MASS (comparatively) and a LARGE ACCELERATION.
N.T.L ExamplesAction: HAMMER HITS NAILReaction: NAIL HITS HAMMER
Action: Earth pulls on YOUReaction: YOU pull on the earth
TensionIn physics, tension is the pulling force exerted by a string, cable, chain, or similar solid object on another object.
m1g
m2g
T
T
FN
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable. amTgmm
amTm
maFNet
222
11
14
)8.9)(11(
)(
21
2
122
mm
gma
mmagm
amamgm
amamgm
122
212
Acceleration = 7.7 m/s2
A traffic light weighing 100 N hangs from a vertical cable tied to two other cables that are fastened to a support at the angle illustrated. Find the tension in each of the three cables.
T3 =>T3 - Fg
T3 – Fg = 0equilibrium
T3 = Fg
T3 = 100 N
ΣFx = T1x - T2x
ΣFy = T1y + T2y = T3
ΣFx = T1x cosθ - T2x cosθ = 0
ΣFy = T1y sinθ + T2y sinθ – T3= 0
Hints:ΣFx there are two forces in the x-axis and they are opposing each other
ΣFy there three forces in the y-axis T1 & T2 are opposing T3
1) ΣFx = T1x cosθ - T2x cosθ = 0
2) ΣFy = T1y sinθ + T2y sinθ – T3= 0
There two equations & two unknows
Solve equation 1 for T2
T2x = T1x * cos37˚
cos53˚T2x = 1.33T1x
T1 sinθ + T2 sinθ – T3= 0
T1 sinθ + 1.33T1 sinθ – T3= 0
T1 sinθ + 1.33T1 sinθ = T3
T1 (sinθ + 1.33sinθ) = T3
T1 = _____T3_______
(sinθ + 1.33sinθ)
T1 = _____100 N_____
(sin37˚ + 1.33sin53˚)T1 = 60.1 N
T2 = 79.9 N
T1x cos37˚ - T2x cos53˚ = 0T2x cos53˚ = T1x cos37˚
Inclines
cosmg
sinmg
mg
FNFf
Tips•Rotate Axis•Break weight into components•Write equations of motion or equilibrium•Solve
Pause for a Cause
m2
m1
Two packing crates of masses 10.0 kg and 5.0 kg are connected by a light string that passes over a frictionless pulley. The 5.00 kg crate lies on a smooth incline of angle 40.0˚. Find the acceleration of the 5.00 kg crate and the tension in the string.
m1g
m2g
FNT
T
Ff
40
40
m2gcos40
m2gsin40
amTgmm
amTgmm
222
111
sin)2
)1
maFNET
Pause for a Cause
m2
m1
m1g
m2g
FNT
T
Ff
40
40
m2gcos40
m2gsin40
amTgmm
amTgmm
222
111
sin)2
)1
maFNET
)(
)sin(
)()sin(
sin
sin)2
)1
12
12
1212
1212
21122
111
mm
mmga
mmammg
termslikefactor
amamgmgm
amamgmgmm
amgmTm
44.4)0.50.10(
)0.1040sin0.5(8.9
a
Solve 1) for T
Sub 1) for T into 2) for T
Pause for a Cause
gmamTamgmT
maFNET
1111)1
m2
m1
Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string.
m1g
m2g
FNT
T
Ff
40
40 amTFgm f 22 )(sin)2 m2gcos40
m2gsin40
cos
cos
2
2
gmF
gmF
FF
kf
N
Nkf
Example
cos
sin
cossin
cossin
sin
)(sin
sin
2
2112
22112
21122
2112
2112
22
gm
amgmamgm
gmamgmamgm
amgmamgmgm
amgmamFgm
amgmamFgm
amTFgm
k
k
k
Nk
f
f
2
2
2
/125.0
)4(2101
21
sma
a
attvx ox
gmamTamgmT
maFNET
1111
amTFgm f 22 )(sin
NT 7.39)8.9(4)125(.4
235.057.67
125.12.395.07.56
k