Objectives - Lesson 3 : Forces in Two Dimensions
11/12 do now on a new sheetSketch a set of graphs that relate
the variables shown on the axes for an object that is thrown off a
building horizontally.
Due today - Extra creditHomework - Due Monday: Another Angle on
F-m-a packet Due MondayProject (projectile simulator)Packet
(Projectile Motion) correctionPacket (Regents Physics Projectile
Practice) correctionsHomework is posted on google classroom - code:
nsdnltForces in Two Dimensions - Objectivesknow the procedures of
Addition of Forces Resolution of Forces Understand the conditions
ofEquilibrium and StaticBe able to determineThe individual force
acting on an object in equilibriumIf equilibrium is achieved in a
systemForce is a vector quantity, it has ____________ and
___________.
The vector sum of all forces acting on an object is known as
_____________.
According to Newtons 2nd Law: Fnet = ________
Force Vector reviewmagnitudedirectionNet ForcemaAdding forcesTwo
ways to add vectors are ____________ and _______________.
Graphical method: _____________________.
Mathematical method (only if the vectors are perpendicular to
each other)To determine magnitude, we use _____________.To
determine direction, we use ______________.Graphical
methodMathematical methodHead-tail or tail-tail (parallelogram)c2 =
a2 + b2SOH CAH TOA4Resolving Force -Finding componentsTo determine
the magnitudes of the components of a force , we use
____________xy
AxAythe trig. functions.Ax = _______________Ay =
_______________AcosAsinEquilibrium and StaticWhen all the forces
that act upon an object are balanced, then the object is said to be
in a state of _______________F = ___, a = ____An object at
equilibrium is either ...at rest and staying at rest - "static
equilibrium.in motion and continuing in motion with the same speed
and direction dynamic equilibrium
equilibrium. 00Force at an angleWhen ever you see a force at an
angle acting on an object, you must resolve it into its x and y
components. Then you can compare all the x components and all the y
components to determine if an object is in equilibrium or
accelerating.Example - determine If a system is in
equilibriumForces A, B, C acting on a point. Determine if they
produce equilibrium. 36.9oxA = 5 NB = 4 NC = 3 NResolve A into x
and y components.Compare all x components and compare all y
components. If Fx = 0 and Fy = 0, then the system is in
equilibrium.To determine if a system is in equilibrium is to check
if F = 0Example determine individual force in an equilibriumWhen a
sign is hung at equilibrium symmetrically, the downward pull of
gravity must be balanced by the upward pull of the wires. The
tension in each wire must be the same. The vertical component of
the tension in each wire must support half the weight of the
frame.
Fx = 0Fy = 0What is the weight of the frame?The weight of the
frame is 50 NA vector is quantity which has _____________ and
______________.Two ways to add vectors are ______________ and
________________________.Vector resolution is use trigonometric
functions to resolve a vector into
________________________.Equilibrium is the state of an object in
which all the forces acting upon it are _____________. In such
cases, the net force is ____ Newton. Knowing the forces acting upon
an object, trigonometric functions can be utilized to determine the
_______________ and ______________ components of each force. At
equilibrium, all the vertical components must _______________ and
all the horizontal components must _________________.In
conclusionNet Force Problems RevisitedA force directed an angle can
be resolved into two components - a horizontal and a vertical
component.
To determine Net Force, add all the forces by components, then
use Pythagorean Theorem to solve for the magnitude and tangent
function to determine directionThe acceleration of an object can be
determined by using Newton's second law. Example - Determine the
net force and acceleration
Fnet = 69.9 N, right m = (Fgrav / g) = 20 kg a = (69.9 N) / (20
kg) =3.50 m/s/s, rightObject moves in horizontal direction
Fnet = 30.7 N, right a = 1.23 m/s/s, right.Example - Determine
the net force and accelerationObject moves in horizontal
directionProcedures for adding vectors at an angle with
horizontalResolve the vectors at an angle into x and y
components.Add all the x components togetherAdd all the y
components togetherUse Pythagorean Theorem to find the resultant
(hypotenuse)Resultant2 = x2 + y2Use trigonometric function to
determine the direction: tan = opp / adj Use Newtons 2nd Law to
determine acceleration.Practice 130FgFNThis applied force (FA)can
be broken intoCOMPONENTSFAFAXFAYA block of 10 kg mass is pushed
along a frictionless, horizontal surface with a force of 100 N at
an angle of 30 above horizontal.The total vertical force mustbe 0,
soRy = FN + FAY Fg = 0FN = Fg FAY R = Rx = Fax Acceleration depends
only onFAX
X Y
FAX FAY Fg FN Total = FAX Total = 0
FAx = 100cos(30o) = 87 NFAy = 100sin(30o) = 50 NLab shoot for
your gradeNAMES, DATE, TITLEPURPOSEHow far will a horizontally
launched ball land?MATERIALSlauncher, meter stick, target
paperPROCEDUREWrite a paragraph to describe how the lab is done so
that anyone reading your procedure can duplicate this lab. Include
the following with your paragraph: Draw a diagram of your
experiment List quantity you need to measure and the tools you use
to make the measurement. Indicate these quantities in your
diagramState equations you need to use to solve for the impact
point distance velocity
11/13 do nowSketch a set of graphs that relate the variables
shown on the axes for an object that is launched at an
angle.axtvxtaytvytHomework - Due Monday: Another Angle on F-m-a
packet Due MondayProject (projectile simulator)Packet (Projectile
Motion) correctionPacket (Regents Physics Projectile Practice)
correctionsDue TuesdayPacket (worksheet 1.2.3)Homework is posted on
google classroom - code: nsdnltPractice 2A man pulls a 40 kilogram
crate across a smooth, frictionless floor with a force of 20 N that
is 45 above horizontal.What is the net force on the sled?What is
the crates acceleration?Fnet = FA cos Fnet = (20 N)(cos 45)Fnet =
14.14 Na = Fnet / ma = (14.14 N) / (40 kg)a = 0.35 m/s2How could
the acceleration be increased?Pushing at a smaller angle will make
Fnet greater andtherefore increase acceleration.Pushing on an
Angle-30FgFNFAFAXFAYThis applied force (FA)can be broken
intoCOMPONENTSA block is pushed along a frictionless, horizontal
surface with a force of 100 newtons at an angle of 30 below
horizontal.The total vertical force mustbe 0, soFN = Fg + FAY
Acceleration depends only onFAX
X Y
FAX FAY Fg FN Total = FAX Total = 0
Practice 3A girl pushes a 30 kilogram lawnmower with a force of
15 N at an angle of 60 below horizontal.Assuming there is no
friction, what is the acceleration of the lawnmower?What could she
do to reduce her acceleration?Fnet = FA cos Fnet = (15 N)(cos
60)Fnet = 7.5 Na = Fnet / ma = (7.5 N) / (30 kg)a = 0.25 m/s2Push
at an greater anglePractice 4
Practice 5A student moves a box of books by attaching a rope to
the box and pulling with a force of 90.0 N at an angle of 30.0
degrees. The box of books has a mass of 20.0 kg, and the
coefficient of kinetic friction between the bottom of the box and
the sidewalk is 0.50. find the acceleration of the box.determine
the net force and acceleration
Fnet2 = (Fx)2 + (Fy)2 Fnet = 39 N = tan-1(-23/32) = -36o = 324o
CCWa = Fnet / m = 39 N / 5 kg = 8.0 m/s2, same direction as the
force11/16 do nowA projectile passes through points A and B as it
follows the path shown below.
Compared to the horizontal speed at point A in the projectiles
path, the horizontal speed at point B is (greater, smaller, the
same)Compared to the vertical speed at point A in the projectiles
path, the vertical speed at point B is (greater, smaller, the
same)Compared to the acceleration at point A in the projectiles
path, the acceleration at point B is (greater, smaller, the
same)Due today: Project (projectile simulator)Packet (Projectile
Motion) correctionPacket (Regents Physics Projectile Practice)
correctionsDue tomorrowAnother Angle on F-m-a packet Due
WednesdayWorksheet packet 2.1.2B packetDue FridayProjectile test
correctionworksheet 1.2.3 correctionHomework is posted on google
classroom - code: nsdnltInclined PlanesObjects accelerate down
inclined planes because of an unbalanced force. The unbalanced
force is caused by gravity.
Note: the normal force is not directed in the direction that we
are accustomed to. The normal forces are always directed
perpendicular to the surface that the object is on.
Fg on Inclined PlaneAn important ideaThe process of analyzing
the forces acting upon objects on inclined planes will involve
resolving the weight vector (Fgrav) into two perpendicular
components.
The perpendicular component of the force of gravity is directed
opposite the normal force and as such balances the normal force.
The parallel component of the force of gravity is part of the net
force that is responsible to objects motion along the incline.
27Forces on an Incline CalculationsConsider forces:Perpendicular
F = Fg cos Cancel out Normal (FN )
ParallelF// = Fg sin All the parallel components (including the
friction force) add together to yield the net force. Which should
directed along the incline. Tilt you head method28Essential
KnowledgeWhat happens to the component of weight that is
perpendicular to the plane as the angle is increased?Decreases Fg
perpendicular What happens to the component of weight that points
ALONG the plane as the angle is increased?Increases Fg parallelWhat
happens to the normal force as the angle is increased?Decreases
depends on Fg perpendicularWhat happens to the friction force as
the angle is increased?Decreases depends on normal forceExampleWhat
is the magnitude of the normal force?FN = Fg perpendicular = Fg cos
= 43.3 NIf the box is sliding with a constant velocity, what is the
magnitude of the friction force?Ff = Fg parallel = Fg sin = 25 NFg
= 50N30Example 1The free-body diagram shows the forces acting upon
a 100-kg crate that is sliding down an inclined plane. The plane is
inclined at an angle of 30 degrees. The coefficient of friction
between the crate and the incline is 0.3. Determine the net force
and acceleration of the crate.
In perpendicular direction:Fnorm = F = 850 NIn parallel
direction:Fnet = F// - FfFnet = 500 N - FnormFnet = 235 Na = Fnet /
m = 2.35 m/s2Fg = Fgravcos30o = 850 NFg// = Fgravsin30o = 500
NExample
practiceAn 8.0-newton block is accelerating down a frictionless
ramp inclined at 15 to the horizontal, as shown in the diagram
below. What is the magnitude of the net force causing the blocks
acceleration?
A child pulls a wagon at a constant velocity along a level
sidewalk. The child does this by applying a 22-newton force to the
wagon handle, which is inclined at 35 to the sidewalk as shown
below. What is the magnitude of the force of friction on the
wagon?
The diagram below shows a 1.0 105-newton truck at rest on a hill
that makes an angle of 8.0 with the horizontal. What is the
component of the trucks weight parallel to the hill?
A block weighing 10.0 newtons is on a ramp inclined at 30.0 to
the horizontal. A 3.0-newton force of friction, Ff , acts on the
block as it is pulled up the ramp at constant velocity with force
F, which is parallel to the ramp, as shown in the diagram below.
What is the magnitude of force F?
A force of 60. newtons is applied to a rope to pull a sled
across a horizontal surface at a constant velocity. The rope is at
an angle of 30. degrees above the horizontal. Determine the
magnitude of the frictional force acting on the sled.Calculate the
magnitude of the component of the 60.-newton force that is parallel
to the horizontal surface.
The diagram below represents a block at rest on an incline. Draw
arrows to indicate the directions of friction force (Ff), normal
force (FN), and gravity (Fg)
The diagram shows a sled and rider sliding down a snow-covered
hill that makes an angle of 30. with the horizontal. Draw an arrow
from the center of the rider to indicate the direction of normal
force.
A book weighing 20. newtons slides at a constant velocity down a
ramp inclined at 30. to the horizontal as shown in the diagram
below. What is the force of friction between the book and the
ramp?
11/17 do nowA box with mass m is at rest on an rough plane
inclined at angle with horizontal. Draw a free body diagram on the
box.In terms of m, g and Determine Fg and Fg// Determine
FNDetermine FfDetermine FnetobjectiveDue today: Another Angle on
F-m-a packet Due tomorrowworksheet 2.1.2B Forces on Angles
PacketCastle Learning Test Due tomorrow nightDue FridayProjectile
Projectile test correctionProject: projectile simulator
correctionWorksheet 1.2.3 ground launched projectile packet
correctionHomework is posted on google classroom - code:
nsdnltReviewFinish Lab11/7 do nowA ball is projected horizontally
with an initial velocity of 20. meters per second east, off a cliff
100. meters high. How many seconds does the ball take to reach the
ground? [Neglect air resistance.]
The diagram below shows a 4.0-kilogram object accelerating at
10. meters per second2 on a rough horizontal surface. What is the
magnitude of the frictional force Ff acting on the object?
3911/16 do nowWhat is the magnitude of the normal force?FN = Fg
perpendicular = Fg cos = 52 NIf the box is sliding with a constant
velocity, what is the magnitude of the friction force?Ff = Fg
parallel = Fg sin = 30NFg = 60N30Due today: Another Angle on F-m-a
packet Project (projectile simulator)Packet (Projectile Motion)
correctionPacket (Regents Physics Projectile Practice)
correctionsDue tomorrowPacket (worksheet 1.2.3)Homework is posted
on google classroom - code: nsdnlt