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