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Equilibrium Forces and Unbalanced Forces
Topic Overview
•A force is a push or a pull applied to an object.
• A net Force (Fnet) is the sum of all the forces on an object (direction determines + or -)
Fnet = 6N to the right
•Isaac Newton has 3 laws that describe the motion of object
1st Law: Law of inertia
▫An object at rest will stay at rest unless acted on by an outside force▫Inertia: The amount of mass an object has▫More inertia = more mass = Harder to move
Topic Overview
•When an object is in equilibrium, the net force equals zero
•Equilibrium Fnet = 0
• Objects in equilibrium can either be at rest or be moving with constant velocity
•Up = Down Forces•Left = Right Forces
3rd Law: Equal and Opposite ▫For every action, there is an equal and opposite reaction.▫“Things push back”
Topic Overview
•When an object has unbalanced forces acting on it, the object will accelerate in the direction of that excess force:
• Fnet = ma
• This is called “Newton's Second Law”
Example Problem – Balanced Forces
Example Problem – Balanced Forces Solution
Example Problem – Unbalanced Forces
Example Problem – Unbalanced Forces - Solution
Common Mistakes
•Make sure you know if the object is in equilibrium or not
Circular Motion
Topic Overview
•An object in circular motion has a changing velocity but constant speed.
•This is possible because the objects speed does not change (same m/s) but the direction of its motion does change
Topic Overview
•The velocity of the object is always “tangent” to the path of the object.
•The circular force (Fc ) is always directed toward the center
• The acceleration is always toward the center of the circle
Force
Velocity
Equations
• r is the radius of the circle
Example Problem
Example Problem - Solution
Common Mistakes
•Be sure to square the velocity
•Cross multiply when solving for “r”
Momentum/Impulse
Momentum Recap
• Momentum: The product of the mass and velocity of an object
• Equation: p = mv
• Units: p = kilograms meters per second (kgm/s)
• Momentum is a vector: When describing the momentum of an object, the direction matters.
Momentum Recap
• Collision: When 2 or more objects interact they can transfer momentum to each other.
• Conservation of Momentum: The sum of the total momentum BEFORE a collision, is the same as the sum of the total momentum AFTER a collision
Momentum Before = Momentum After
p1i +p2i + p3i = p1f+p2f + p3f
Momentum Recap
• To Solve Collision Problems:
Step 1: Find the total momentum of each object before they interact
Step 2: Set it equal to the total momentum after they collide
Remember momentum is a vector, so you have to consider if the momentum is (+) or(-) when finding the total!!!
Initial = Final 0 = -1.2 (v) + (1.8)(2)
Initial = Final (1)(6)+ 0 = (1 + 3.0) v
Impulse
•An outside force will cause a change in the momentum of an object. This is called an impulse.
•IMPULSE: A change in momentum
Fnett = p = mv
Units = Ns
ImpulseTo find the impulse under a force vs. time graph, you
would find the area under the line.
I = Ft
Example Problem - 1
Example Problem 1- Solution
Kinematics
Topic Overview
•Kinematics is what we use to describe the motion of an object.
•We use terms such as displacement, distance, velocity, speed, acceleration, and time to describe the movement of objects.
Topic Overview
• Distance (m): The total meters covered by an object (odometer)
• Displacement (m): The difference between the start and end points.
Topic Overview
• Velocity (m/s): • How fast the car is moving is a • (+) and (–) indicate direction
• Acceleration (m/s2)• Acceleration is a change in velocity• If an object is accelerating, it is either
speeding up or slowing down
Topic Overview
•Some objects have constant velocity
•Again this means that acceleration = 0
•The equation for constant velocity is: x = vt
•If you are given an average velocity (vave), it is the same thing as constant velocity:
x = vavet
Topic Overview
•Some objects have constant acceleration
•This means they are speeding up or slowing down
•The equations for constant acceleration:a = (vf-vi) / t
x = vit + ½ at2
vf2 = vi
2 + 2ax
Topic Overview
•To solve problems with constant acceleration, make a table!
Vi Initial velocity
Vf Final velocity
a Accelerationx Distancet time
Reminder:If an object is “at rest”
Velocity = 0
Topic Overview
•One example of a “constant acceleration problem” is a falling object (an object traveling through the air)
•All falling objects accelerate at 9.8m/s2 due to gravity
•They also start withZero initial velocity
Vi 0m/s
Vf
a 9.8m/s2
xt
Example Problem
A stone is dropped from a bridge approximately 45 meters above the surface of a river. Approximately how many seconds does the stone take to reach the water's surface?
Vi 0m/s
Vf
a 9.8m/s2
x 45t ?
Reminder:Choose the equation
that does not have the “blocked off” variable.
In this case, choose the equation that does not have vfa = (vf-vi) / t
x = vit + ½ at2
vf2 = vi
2 + 2ax
Example Problem - Solution
Graphs
Topic Overview
•The motion of an object can be represented by three types of graphs (x, v, a)
1) Displacement vs. Time graphs
•Tells you where the object is•The slope (steepness) is the velocity•In the graph above A is faster than B
X(m)
Time (s)
A
B
Topic Overview
1) Types of Motion for x vs. t graphs
X(m)
Time (s)
X(m)
Time (s)
X(m)
Time (s)
Not moving because the position does not change
Constant velocity because the slope does not change (linear)
Accelerating because it is a curve
Topic Overview
2) Velocity vs Time graphs
•Tells you how fast the object is moving
•Slope of the line = Acceleration•Area under curve = Displacement
v(m/s)
(s)0
Topic Overview
2) Velocity vs. Time graphs
Constant speed because the value for velocity does not change
Speeding up because the value of the velocity is moving away from zero
v(m/s)
(s)0
v(m/s)
(s)0
v(m/s)
(s)0
Slowing down because the value of the velocity is moving toward zero
Topic Overview
3) Acceleration vs. time
•Tells you the acceleration of the object
•Area under curve = Velocity
a(m/s2)
(s)0
Topic Overview
3) Acceleration vs. Time graphs
This object has positive acceleration
No acceleration. This means the object has a constant speed
a(m/s2)
(s)0
a(m/s2)
(s)0
This object has negative acceleration
a(m/s2)
(s)0
Example Question-2
Example Question-2 Solution
Calculation Examples:
Example Question-1
Displacement: Difference between y value
Average Speed: Slope between 2 time points
Instantaneous speed: Slope at 1 time point (TANGENT)
Example Question-1 Solution
Example Question-2
What is the average speed from 0-6 seconds?
Example Question-2 Solution
What is the average speed from 0-6 seconds?
Average is the slope between 2 points
Slope = (4 – 0) = 0.66m/s (6 – 0)
Example Question-3
What is the displacement from 1-4 seconds?
Example Question-3 Solution
Position increased (+) from 2m to 4m
Answer:
4 – 2 = 2m
Projectile Motion
Topic OverviewObjects that travel in both the horizontal and vertical direction are called “projectiles”.
These problems involve cars rolling off cliffs, objects flying through the air, and other things like that.
Topic Overview
▫An object traveling through the air will:
▫ACCELERATE in the VERTICAL DIRECTION Because it is pulled down by gravity
▫ Have CONSTANT VELOCITY in the HORIZONTAL Because there is no gravity
▫Because they are different, we do calculations in the horizontal (x) and vertical (y)_ SEPARATELY.
Acceleration
VelocityHorizontal
VelocityVertical
-9.8m/s2
Constant
Increasing
Zero Launch Angle Projectile Motion
The time for an object to fall is determined by drop height ONLY
(horizontal velocity has no effect)
Projectile Motion
HorizontalConstant Velocity
x = vt
VerticalAccelerating
a = (vf-vi) / tx = vit + ½ at2
vf2 = vi
2 + 2ax
TIME = TIME
The key to solving projectile motion problems is to solve the horizontal and vertical parts SEPARATELY.
Time is the only thing that is the same!
Example 1
A bullet is shot at 200m/s from a rifle that is 2.5m above the ground. How far downrange will the bullet reach before hitting the ground?
vi0
vf
a -9.8
x 2.5
t ??
Step 1: Find the time.
In this case we have to use the vertical height to find the time. (acceleration)
t = 0.72s
Step 2: Use the time to find out the distance in the other direction (in this case the horizontal direction)
x = vt x = 200 (0.72)x = 150m
x = vit + ½ at2
Energy and Power
Energy is the ability to CHANGE an object. These types of energy are a result of a CHANGE in…….
Work: FORCEKinetic Energy: VELOCITYPotential Energy: HEIGHTElastic Energy: SHAPE
Conservation of Energy
• In a system, the TOTALMECHANICAL ENERGY never changes
• Energy can switch forms, but it cannot be created or destroyed.
KINETIC POTENTIAL ELASTIC
WORK
Conservation of Energy
• Mathematically speaking that looks like this
TME Initial = TME Final
KE + PE + EE + W = KE + PE + EE + W
Energy: Joules (J)
Work: W = FdKinetic Energy: KE = ½mv2
Potential Energy: PE = mghElastic Energy: EE = ½kx2
Examples
• A ball is dropped from a height of 12m, what is the velocity of the ball when it hits the ground?
Potential Energy Kinetic Energy
Since all of the energy is transferred, we cans set them equal to each other
PE = KEm(9.8)(12) = ½(m)v2 mass cancels
15.33 m/s = v
Examples
• A ball is dropped from a height of 12m, what is the velocity of the ball when it hits the ground?
Potential Energy Kinetic Energy
Since all of the energy is transferred, we can set them equal to each other
PE = KEm(9.8)(12) = ½(m)v2 mass cancels
15.33 m/s = v
Examples
• A force of 50N pushes horizontally on a 5kg object for a distance of 2m. What is the final velocity of the object?
Work Kinetic Energy
Since all of the energy is transferred, we can set them equal to each other
Work = KE(50)(2) = ½(5)v2
6.32 m/s = v
Examples
No matter type of energy transfer, the set up is the same. Even if there is more than one type of energy
present
TME Initial = TME Final
KE + PE + EE + W = KE + PE + EE + W
POWER
Power = Energy/ time
P = E/t
On your equation sheet it lists “Energy” as Work for the top of the fraction. But you can put any type of energy on the top part of this
equation.
Regular Physics ONLY:Electrostatics
Electrostatics:
•Atoms have protons, neutron, and electrons.
•Only electrons can move!
1 e- = 1.6E-19 Coulomb (C)
Electrostatics:
•An object becomes charged when its electrons are shifted or transferred
•Extra electrons = (-)•Fewer electrons = (+)
Electrostatics:
•How many more electrons than protons are there on an object with a -1.76E-18 C charge?
•What is the total charge of an object with a deficiency of 4.0 x108 electrons?
•Extra electrons = (-)•Fewer electrons = (+)
Electrostatics:
•Two charged objects will have an Electric Field between them.
Field Lines (+) (-)“Start Positive”
Electrostatics:
•Two charged objects will feel an Electric Force
•Opposite charges attract (+/-)•Same charges repel (+/+) (-/-)
Electrostatics:
•Calculating the Electric Force
•k = 9E9•q=charge • r= distance between the charges
F e k
q1q2
r2
Electrostatics:
•Calculating the Electric Force
•k = 9E9
What is the magnitude of the electrostatic force between two electrons separated by a distance of 1.00 × 10–8 meter?
221k
r
qqFe
Regular Physics ONLY:Circuits
•A circuit provides a COMPLETE path for electrons to move.
•The flow of electrons is called the current (I).
•Electrons flow because a voltage (V) provides an energy difference
•In order to get energy out of a circuit, there has to be resistors (R).
IRV
Circuits:
IRV Circuits:
Series Circuits•A circuit in which there is only one
current path
Circuits:
Series Circuit•Current is the same in all resistors
I = I1 = I2 = I3 = I4
•Voltage is distributed among the resistors
V = V1 + V2 + V3
•Total Resistance is the sum of all resistors.
RT= R1 + R2 + R3
Circuits:
Parallel CircuitsA circuit in which there are several
current paths
Circuits:
Series or Parallel?????
Circuits:
Parallel Circuit•Current is the added in all resistors
IT = I1 + I2 + I3
•Voltage is equal among the resistors
VT = V1 = V2 = V3
•Total Resistance is the reciprocal of all resistors.31/RT = 1/R1 + 1/R2 + 1/R3
Circuits:
RIVP TABLES!!!
Circuits:
AP ONLY:Torque
Torque Introduction
torque = distance from axis of rotation x force x sin (angle between r and F)
Units for torque: Nm (Newton-meters)
sinrF
Torque Introduction• Torque is analogous to a force.
Force TorqueLinear Acceleration Angular Acceleration
Torque
Torque Introduction
• The “distance” of the force also matters.
•Lever Arm (r): The distance between the pivot point and the force.
Torque Introduction
• The “angle (θ)” of the force also matters.
10N10N
Torque Introduction
• The “angle” of the force also matters.• Use the sin component of the force
•Measure angle from the LEVER ARM FORCE
10N10N
Torque – Direction and equilibrium
•Net Force = Zero No acceleration
•Net Torque = Zero No rotation
• Clockwise = (+) Counterclockwise = (-)
Is the object moving?
Ex 1
To weigh a fish a person hangs a tackle box of mass 6 kilograms and a cooler of mass 10 kilograms from the ends of a uniform rigid pole that is suspended by a rope attached to its center. The system balances when the fish hangs at a point 1/4 of the rod’s length from the tackle box. What is the mass of the fish?
6kg 10kg
•A store sign, with a mass of 20.0kg and 3.00m long, has its center of gravity at the center of the sign. It is supported by a loose bolt attached to the wall at one end and by a wire at the other end. The wire makes an angle of 25° with the horizontal. What is the tension in the wire?
Ex 2
AP ONLY:Oscillations
Oscillation:
Simple Harmonic Motion (SHM)•An object that “cycles” between having
maximum kinetic and maximum potential energy.
Simple Harmonic Motion (SHM) - Notes•This cycle will look like a sine or cosine wave•The graph will look like this for displacement,
potential energy, and kinetic energy as a function of time (It will just be shifted)
1 2 3 4
Simple Harmonic Motion (SHM)•Determine these for our system:
•Amplitude = Maximum displacement from equilibrium
• Period (T) = Time for 1 complete motion• Seconds per cycle • Units = seconds
• Frequency = cycles per second• Inverse of period• Units = s-1 or Hertz (Hz)
Equations
) ( cosA x t
) (sin A x t
T
2
Position Function for cosine – Class example
•If the object represented in the graph experiences a maximum displacement of 0.50m and completes a cycle in 2.0s, what is an appropriate expression for displacement as a function of time?
) ( cosA x t) (sin A x t
T
2
Position Function for cosine - Example•A 0.50kg object on a spring (k=494)
undergoes Simple Harmonic Motion. Its motion can be described by the following equation
x = (0.3 m) cos (10 t) π
•What is the amplitude of the vibration? •What is the period of the vibration? •What is the frequency? •What is the maximum elastic PE?•What is the maximum velocity?
AP ONLY:Gravitation
Gravitation
•The force of gravity depends on 2 things:•The mass of the two objects.•The distance between them.
2
21
r
mmGF
Gravitation - Example
Mars has a mass 1/10 that of Earth and a diameter 1/2 that of Earth. The acceleration of a falling body near the surface of Mars is most nearly
2
21
r
mmGF
Satellites•Satellites in orbit are in centripetal motion. •Fc is provided by the gravitational force.
•Combining Newton’s Law of Gravitation with the formula for centripetal force, we find:
•Rearrange for speed of satellite:
r
mv
r
mmG
2
221
r
GMv planet
Note that the mass of the satellite does not matter!
Example•How long would it take a satellite to orbit the
Mars if it were a distance of 6.79 x 106m from the center of the planet?
r
GMv Planet
Gravitation
•The force of gravity depends on 2 things:•The mass of the two objects.•The distance between them.
2
21
r
mmGF
Gravitation - Example
Mars has a mass 1/10 that of Earth and a diameter 1/2 that of Earth. The acceleration of a falling body near the surface of Mars is most nearly
2
21
r
mmGF
Satellites•Satellites in orbit are in centripetal motion. •Fc is provided by the gravitational force.
•Combining Newton’s Law of Gravitation with the formula for centripetal force, we find:
•Rearrange for speed of satellite:
r
mv
r
mmG
2
221
r
GMv planet
Note that the mass of the satellite does not matter!
Example•How long would it take a satellite to orbit the
Mars if it were a distance of 6.79 x 106m from the center of the planet?
r
GMv Planet
