Physics Chapter 7 Forces and Motion in Two Dimensions

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


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


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 EquilibriumHanging 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 EquilibriumHanging a Sign Is 168 N the sign’s weight,

resultant force, and/or equilibrant?

168 N

Creating EquilibriumHanging a Sign So both ropes together must pull

up with a force of 168 N to put the sign into equilibrium

168 N

Creating EquilibriumHanging 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 EquilibriumHanging 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 EquilibriumTry 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)


If the object is in equilibrium, what can you say about the relationship between Fg, Fn, Fa and Ff?


If the object is in equilibrium on a sloping surface, which of the following forces doesn’t change; Fg, Fn, Fa and Ff?


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?


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).


Fn, Fa, Ff all change with the angle of the slope


So lets look how Fg, Fa, Fn, and Ff are related


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


Fg is always the hypotenuse of the right force triangle formed

Fg


The force with which the object is pulled into the plane is called the perpendicular force (F)

Fg

F


The force with which the object is pulled down the plane is called the parallel force (F)

Fg

F

F


These forces are perpendicular and parallel to what?

Fg

F

F


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


So for F; cos = a/h Since a = F and h = Fg F = Fg cos

Fg

F = Fg cos

F


So for F; sin = o/h Since o = F and h = Fg F = Fg sin

Fg

F = Fg cos

F = Fg sin


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


So F , is equal and opposite Fn

Fg

F = Fg cos

F = Fg sin

Fn


So if gravity pulls the object down the plane with a force F , which force counteracts it?

Fg

F = Fg cos

F = Fg sin


So F is equal and opposite to Ff If there is no motion (or

acceleration)!

Fg

F = Fg cos

F = Fg sin

Ff


So F , is equal and opposite Fn

and F is equal and opposite to Ff Only if there is no

________or_______.


Example A trunk weighing 562 N is resting on a plane

inclined 30°. Find the perpendicular and parallel components of its weight.

AnswersF = - 487 NF = - 281 N

Why are they negative? What other forces are they equal to? Would the force change if the object was moving?


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?


What is the Five Forces Equation? Fnet = Fa + Ff + Fg + Fn Lets see how it can be modified to

deal with inclined planes


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


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


Fnet = Fa + Ff + F + F

Does F determine if the object accelerates down the plane?

Get rid of it Fnet = Fa + Ff + F


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 planesFnet = Fa + Ff + F


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 NF = - 281 Na = -1.41 m/s2

Projectile MotionProjectile—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 MotionA 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 MotionWe 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 ObjectExample A 2.5 kg stone is dropped from a

cliff 44 m high. How long is it in the air? Answer3.0 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


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?


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


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? Answer15 m/s

What is its vertical velocity right before it hits the ground? Answer- 29.4 m/s


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


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


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


A projectile is launched at an angle with a velocity, vo.

vo


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


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


Initial vertical component of its velocity (vv)

vv = vo sin

vo

vv

vh


Initial horizontal component of its velocity (vh)?

vh = vo cos

vo

vv

vh


What can you tell me about these perpendicular components of initial velocity?

vo

vv = vo sin

vh = vo cos


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


Motion horizontal

vertical

vo vo cos vo sin v Same Zero-at

top of path

a None - 9.8 m/s2


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


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.



Find its hang time, range and maximum height.Step #1: draw a diagram

(vector)vo

vv = vo sin vh = vo cos



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/svh = vo cos = 4.47 cos 66 = 1.82 m/s



Find its hang time, range and maximum height.Step #3: Make a table of the

horizontal and vertical components of the projectile’s motionvo



Step #3: Make a table of the horizontal and vertical components of the projectile’s motion

vo


Motion horizontal verticalvo 1.82 m/s 4.08 m/sv 1.82 m/s 0 m/s at

topa None -9.8 m/s2


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)


topa None -9.8 m/s2


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


topa None -9.8 m/s2


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/sv 1.82 m/s 0 m/s at

topa None -9.8 m/s2


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 MotionSummary: 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 pathCentripetal 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 MotionExample: 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 MotionSummary: Uniform circular motion

What is it? What changes and what doesn’t?

Circular MotionSummary: Circular velocity

What is it? Direction? Equation?

Circular MotionSummary: Centripetal acceleration (ac)


Circular MotionSummary: Centripetal Force (Fc)


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 MotionTwo quantities describe simple

harmonic motion: Period Amplitude

Simple Harmonic MotionPeriod (T)—the amount of time that

is needed to complete one complete cycle of motion

Units: seconds (per cycle) Amplitude

Simple Harmonic MotionAmplitude—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 MotionLets 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 springEffects 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?

PendulumEffects on period Size of mass

No effect Size of pendulum

Longer string means larger period Amplitude of vibration

No effect on period

PendulumThere 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)

PendulumExample: What is the length of a pendulum

that has a period of 2.25 s?Answer:

1.26 m

Documents

Physics Chapter 7 Forces and Motion in Two Dimensions