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Physics. Chapter 7 Forces and Motion in Two Dimensions. Forces and Motion in Two Dimensions. 7.1 Forces in Two Dimensions 7.2 Projectile Motion 7.3 Circular Motion And…Simple Harmonic Motion from Chapter 6!. Forces in Two Dimensions. - PowerPoint PPT Presentation
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Physics
Chapter 7Forces and Motion in Two
Dimensions
Forces and Motion in Two Dimensions
7.1 Forces in Two Dimensions 7.2 Projectile Motion 7.3 Circular Motion And…Simple Harmonic Motion
from Chapter 6!
Forces in Two Dimensions
We have already studied a few forces in two dimensions
When dealing with friction acting on an object (parallel), the normal force on the object (perpendicular) also plays a role in friction
In this chapter we will be looking at forces on a object that are at angles other than 90°
Equilibrium and the Equilibrant
Equilibrium—when the net forces on an object is equal to zero
This means an object can be stationary or moving but it can’t be _________.
When graphically adding vectors together, if a closed geometric shape is formed, there is no resultant (?) so net force is zero
Equilibrium and the Equilibrant
If an object is not in equilibrium this means that:
There is a net force on it When the vectors are added
together, there is a resultant It is accelerating
Equilibrium and the Equilibrant
If add a vector to the resultant that is equal and opposite to it, there will be no net force and the object will be in equilibrium
This vector is called the equilibrant The equilibrant is the vector that puts
an object in equilibrium It is equal and opposite to the resultant
of the existing vectors
Creating Equilibrium
Hanging a Sign A 168 N sign is equally supported by
two ropes that make an angle between them of 135°. What is the tension in the rope?
168 N
Creating Equilibrium
Hanging a Sign Is 168 N the sign’s weight,
resultant force, and/or equilibrant?
168 N
Creating Equilibrium
Hanging a Sign So both ropes together must pull
up with a force of 168 N to put the sign into equilibrium
168 N
Creating Equilibrium
Hanging a Sign So each ropes together must pull up
with a force of 84 N (1682) to put the sign into equilibrium
168 N 84 N
Creating Equilibrium
Hanging a Sign So to find the tension in the rope
(hypotenuse of a right triangle) use trig!
cos = a/h so h = a/cos H = 84/cos 67.5 H = 219.5 N Why was = 67.5° and not 135°?
84 N
Creating Equilibrium
Try this…..Hanging a Sign A 150 N sign is equally supported by
two ropes that make an angle between them of 96°. What is the tension in the rope?
Answer: 112 N
Motion Along an Inclined Plane
When an object sits on a flat surface, there are four forces that determine if there is a net force acting on the object.
What are these forces? (five force equation)
Motion Along an Inclined Plane
If the object is in equilibrium, what can you say about the relationship between Fg, Fn, Fa and Ff?
Motion Along an Inclined Plane
If the object is in equilibrium on a sloping surface, which of the following forces doesn’t change; Fg, Fn, Fa and Ff?
Motion Along an Inclined Plane
If the object is in equilibrium on a sloping surface, which of the following forces doesn’t depend on the angle of the slope; Fg, Fn, Fa and Ff?
Motion Along an Inclined Plane
The object weight (Fg) is always the same (regardless of the slope of the surface) and it always directed straight down (toward center of the earth).
Motion Along an Inclined Plane
Fn, Fa, Ff all change with the angle of the slope
Motion Along an Inclined Plane
So lets look how Fg, Fa, Fn, and Ff are related
Motion Along an Inclined Plane
When an object sits on an inclined plane it is being pulled into the plane and down the plane by the force of gravity
Fg
Motion Along an Inclined Plane
Fg is always the hypotenuse of the right force triangle formed
Fg
Motion Along an Inclined Plane
The force with which the object is pulled into the plane is called the perpendicular force (F)
Fg
F
Motion Along an Inclined Plane
The force with which the object is pulled down the plane is called the parallel force (F)
Fg
F
F
Motion Along an Inclined Plane
These forces are perpendicular and parallel to what?
Fg
F
F
Motion Along an Inclined Plane
We could solve for F and F|| using Fg and angle if we knew what was.
Angle is always equal to the slope of the plane
Fg
F
F
Motion Along an Inclined Plane
So for F; cos = a/h
Since a = F and h = Fg
F = Fg cos
Fg
F = Fg cos
F
Motion Along an Inclined Plane
So for F; sin = o/h
Since o = F and h = Fg
F = Fg sin
Fg
F = Fg cos
F = Fg sin
Motion Along an Inclined Plane
So if gravity pulls the object down into the plane with a force F , which force counteracts it?
Fg
F = Fg cos
F = Fg sin
Motion Along an Inclined Plane
So F , is equal and opposite Fn
Fg
F = Fg cos
F = Fg sin
Fn
Motion Along an Inclined Plane
So if gravity pulls the object down the plane with a force F , which force counteracts it?
Fg
F = Fg cos
F = Fg sin
Motion Along an Inclined Plane
So F is equal and opposite to Ff
If there is no motion (or acceleration)!
Fg
F = Fg cos
F = Fg sin
Ff
Motion Along an Inclined Plane
So F , is equal and opposite Fn
and F is equal and opposite to Ff
Only if there is no ________or_______.
Motion Along an Inclined Plane
Example A trunk weighing 562 N is resting on a plane
inclined 30°. Find the perpendicular and parallel components of its weight.
AnswersF = - 487 N
F = - 281 N Why are they negative? What other forces are they equal to? Would the force change if the object was moving?
Motion Along an Inclined Plane
Under what set of circumstances could the object be accelerating down the plane?
F > Ff
Slope too slippery—not enough friction to hold it
We could be pushing or pulling the object What can we say about the net force on the
object?
Motion Along an Inclined Plane
What is the Five Forces Equation? Fnet = Fa + Ff + Fg + Fn
Lets see how it can be modified to deal with inclined planes
Motion Along an Inclined Plane
Fnet = Fa + Ff + Fg + Fn
Does Fn determine if the object will accelerate down the plane?
So we get rid of it Fnet = Fa + Ff + Fg
Motion Along an Inclined Plane
Fnet = Fa + Ff + Fg
What are the two components of weight for an object on the plane?
We replace weight with these two Fnet = Fa + Ff + F + F
Motion Along an Inclined Plane
Fnet = Fa + Ff + F + F
Does F determine if the object accelerates down the plane?
Get rid of it Fnet = Fa + Ff + F
Motion Along an Inclined Plane
Fnet = Fa + Ff + F
Can we still push or pull the object? Does friction still act on it? So this is the new Four Forces
Equation for inclined planes
Fnet = Fa + Ff + F
Motion Along an Inclined Plane
Example A trunk weighing 562 N is resting on a plane
inclined 30°. Find the perpendicular and parallel components of its weight. If the force of friction is 200 N, find the trunk’s acceleration rate.
AnswersF = - 487 N
F = - 281 Na = -1.41 m/s2
Projectile Motion
Projectile—a launched object that moves through air only under the force of gravity
Ignoring air resistance!Trajectory—the path of a projectile
through space
Projectile Motion
A projectile has both horizontal and vertical components of its velocity
These components are independent of each other
This is because the force of gravity acts on the vertical component (causing acceleration) but not on the horizontal component (constant velocity).
But one can equal zero!
Projectile Motion
We will be learning how to solve three types of projectile motion problems:
Dropped Objects Objects Thrown Horizontally Objects Launched at an Angle
A Dropped Object
When an object is dropped from a height it will fall
It has two components of its velocity, horizontal and vertical and both are equal to zero
A Dropped Object
As it falls, it picks up speed (accelerates) in the vertical direction due to the force of gravity
Acceleration due to gravity =? So when dropped any object will
pick up negative vertical velocity at the rate of – 9.8 m/s for each second it falls
A Dropped Object
As it falls, its horizontal speed stays constant (why?)
So as an object falls its vertical speed changes but its horizontal speed doesn’t
These two perpendicular components of speed are independent of each other, they have no effect on each other (this is true of all perpendicular vectors)
A Dropped Object
Example A 2.5 kg stone is dropped from a
cliff 44 m high. How long is it in the air? Answer
3.0 sec What is its velocity right before it
hits the ground? Answer
- 29.4 m/s
An Object Thrown Horizontally
When an object is thrown horizontally, it has horizontal and vertical components to its velocity
The horizontal velocity is constant during the entire time its in the air (why?)
The vertical velocity starts as zero but increases as it falls just as if it were dropped! (why?)
These two components are independent of each other
An Object Thrown Horizontally
Example A 2.5 kg stone thrown horizontally
at 15 m/s from a cliff 44 m high. How long is it in the air? What is its horizontal velocity right
before it hits the ground? What is its vertical velocity right before
it hits the ground? How far from the cliff does it land?
An Object Thrown Horizontally
Example A 2.5 kg stone thrown horizontally at
15 m/s from a cliff 44 m high. How long is it in the air?
Answer3.0 sec
An Object Thrown Horizontally
Example A 2.5 kg stone thrown horizontally at 15 m/s
from a cliff 44 m high. How long is it in the air? 3.0 sec
What is its horizontal velocity right before it hits the ground? Answer
15 m/s What is its vertical velocity right before it hits
the ground? Answer
- 29.4 m/s
An Object Thrown Horizontally
Example A 2.5 kg stone thrown horizontally at 15
m/s from a cliff 44 m high. How long is it in the air?3.0 sec
What is its horizontal velocity right before it hits the ground? 15 m/s
What is its vertical velocity right before it hits the ground? - 29.4 m/s
How far from the cliff does it land? Answer
45 m
Objects Launched at an Angle
When a projectile is launched at an angle, there is a nonzero vertical and horizontal component to the initial velocity
Objects Launched at an Angle
The trajectory of the projectile is called a parabola
The amount of time the projectile is in the air (prior to landing) is called its flight time or hang time
The horizontal distance it travels (prior to landing) is called it range
The maximum vertical height is called the maximum height
Objects Launched at an Angle
Flight time or hang time, range and maximum height all depend on the initial vertical and horizontal velocity
And the initial vertical and horizontal components of velocity depend on the initial velocity and the angle at which the object is launched
Objects Launched at an Angle
A projectile is launched at an angle with a velocity, vo.
vo
Objects Launched at an Angle
At the beginning of its trajectory, it has an initial vertical component of its velocity (vv) and an initial horizontal component of its velocity (vh)
vo
vv
vh
Objects Launched at an Angle
Can we use trig to solve for the initial vertical component of its velocity (vv) and initial horizontal component of its velocity (vh)?
vo
vv
vh
Objects Launched at an Angle
Initial vertical component of its velocity (vv)
vv = vo sin
vo
vv
vh
Objects Launched at an Angle
Initial horizontal component of its velocity (vh)?
vh = vo cos
vo
vv
vh
Objects Launched at an Angle
What can you tell me about these perpendicular components of initial velocity?
vo
vv = vo sin
vh = vo cos
Objects Launched at an Angle
Lets make a table of what we know of what we know about the horizontal and vertical components of the projectile's motion
vo
vv = vo sin
vh = vo cos
Objects Launched at an Angle
Motion horizontal
vertical
vo vo cos vo sin
v Same Zero-at top of path
a None - 9.8 m/s2
Objects Launched at an Angle
Lets see how we find the hang time, range and maximum height for this type of projectile motion problem
vo
vv = vo sin
vh = vo cos
Objects Launched at an Angle
A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°.
Find its hang time, range and maximum height.
Objects Launched at an Angle
A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°.
Find its hang time, range and maximum height.Step #1: draw a diagram
(vector)
vo
vv = vo sin
vh = vo cos
Objects Launched at an Angle
A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°.
Find its hang time, range and maximum height.Step #2: Find the initial
horizontal and vertical components of the initial velocityvo
vv = vo sin = 4.47 sin 66 = 4.08 m/s
vh = vo cos = 4.47 cos 66 = 1.82 m/s
Objects Launched at an Angle
A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°.
Find its hang time, range and maximum height.Step #3: Make a table of the
horizontal and vertical components of the projectile’s motionvo
vv = vo sin = 4.47 sin 66 = 4.08 m/s
vh = vo cos = 4.47 cos 66 = 1.82 m/s
Objects Launched at an Angle
Step #3: Make a table of the horizontal and vertical components of the projectile’s motion
vo
vv = vo sin = 4.47 sin 66 = 4.08 m/s
vh = vo cos = 4.47 cos 66 = 1.82 m/s
Motion horizontal vertical
vo 1.82 m/s 4.08 m/s
v 1.82 m/s 0 m/s at top
a None -9.8 m/s2
Objects Launched at an Angle
Step #4: Use the vertical components of the projectile’s motion to solve for its hang time
v = vo + at 0 = 4.08 + -9.8t t1/2 = 0.416 s (time to top)
tt = 0.833 s (hang time)
Motion horizontal vertical
vo 1.82 m/s 4.08 m/s
v 1.82 m/s 0 m/s at top
a None -9.8 m/s2
Objects Launched at an Angle
Step #5: Use the vertical components of the projectile’s motion to solve for its maximum height
d = do + ½(v + vo)t d = 0 + ½ (4.08 + 0)(0.416) d = 0.849 m
Motion horizontal vertical
vo 1.82 m/s 4.08 m/s
v 1.82 m/s 0 m/s at top
a None -9.8 m/s2
Objects Launched at an Angle
Step #6: Use the horizontal components of the projectile’s motion to solve for its range
d = do + ½(v + vo)t d = 0 + ½ (1.82 + 1.82)(0.833) d = 1.52 m
Motion horizontal vertical
vo 1.82 m/s 4.08 m/s
v 1.82 m/s 0 m/s at top
a None -9.8 m/s2
Objects Launched at an Angle
A football is kicked at an initial velocity of 27 m/s at a 30°. Find its hang time, range and maximum height Hang time
2.76 s Range
64.6 m Maximum height.
9.27 m
Projectile Motion
Summary: Dropped Objects Objects Thrown Horizontally Objects Launched at an Angle
Circular Motion
Uniform circular motion (UCM)—when an object is moving in a circle at constant speed
Is an object in uniform circular motion accelerating? (why?)
What is uniform in UCM? What is changing in UCM?
Circular Motion
During uniform circular motion an object is accelerating because the direction (not magnitude) of its velocity is changing.
Circular Motion
During uniform circular motion an object is accelerating because the direction (not magnitude) of its velocity is changing.
Circular Motion
So to accelerate an object we must apply a ________ to it and its direction is ______.
Circular Motion
During UCM, the velocity's direction is tangential to its circular path and its acceleration’s direction is toward the center
Circular Motion
“Centripetal”—center seekingCentripetal acceleration (ac)—
acceleration that causes an object to move in a circular path Its direction is toward the center of the
circular path
Centripetal force (Fc)—force that causes an object to move in a circular path Its direction is toward the center of the
circular path
Circular Motion
Velocity = rate of change in position during a time interval
V = d/t When an object travels a circular path, this
is just the circumference of the circle (2r)
So for UCM; v = (2r)/T T = period of revolution (sec/rev) Time it takes object to complete circular path
Circular Motion
acceleration = rate of change in velocity during a time interval
a = v/t
But for UCM; ac = v2/r r = radius of circular path Other possible equations?
Circular Motion
force = cause a change in velocity (acceleration)
F = ma
so for UCM; Fc = mac = mv2/r r = radius of circular path Other possible equations?
Circular Motion
Example: A 13-g rubber stopper is attached
to a 0.93 m string. If the stopper is swung in a circular path at a rate of one revolution in 1.18 s, find the tension in the string (Fc)
Answer0.343 N
Circular Motion
Summary: Uniform circular motion
What is it? What changes and what doesn’t?
Circular Motion
Summary: Circular velocity
What is it? Direction? Equation?
Circular Motion
Summary: Centripetal acceleration (ac)
What is it? Direction? Equation?
Circular Motion
Summary: Centripetal Force (Fc)
What is it? Direction? Equation?
Simple Harmonic Motion
A type of periodic motion Periodic motion—motion that
repeats itself over and over again over same path
Simple Harmonic Motion (SHM)—periodic motion in which the force that will restore the object to equilibrium is directly proportional to its displacement
Simple Harmonic Motion
Examples: playground swing, pendulum, vibrating spring, guitar string
Simple Harmonic Motion
Two quantities describe simple harmonic motion:
Period Amplitude
Simple Harmonic Motion
Period (T)—the amount of time that is needed to complete one complete cycle of motion
Units: seconds (per cycle) Amplitude
Simple Harmonic Motion
Amplitude—the maximum distance the object moves from equilibrium
The larger the amplitude of vibration (displacement) the more force will need to be applied
Simple Harmonic Motion
Lets look at some types of objects undergoing simple harmonic motion:
Mass on a spring Pendulum
Mass on a spring
When a mass oscillates on a spring we can change many things and see what effect it has on the oscillation rate (period)
Lets look at: Size of mass Size of spring Stiffness of spring Amplitude of vibration
Mass on a spring
Effects on period Size of mass
Bigger mass means larger period Size of spring
Bigger spring means larger period Stiffness of spring
Stiffer spring means smaller period Amplitude of vibration
No effect on period
Pendulum
When a pendulum swings we can change many things and see what effect it has on the oscillation rate (remember your first lab!)
Lets look at: Size of mass (bob) Length of pendulum Amplitude of vibration
Does anyone remember which one changes the period?
Pendulum
Effects on period Size of mass
No effect Size of pendulum
Longer string means larger period Amplitude of vibration
No effect on period
Pendulum
There is an equation that can be used to calculate the period of any pendulum
T = 2l/g
Where:l = length of pendulum (meters)g = acceleration due to gravity (-9.8 m/s2)
Pendulum
Example: What is the length of a pendulum
that has a period of 2.25 s?Answer:
1.26 m