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11/12 do now – on a new sheet Sketch 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 credit Homework - Due Monday: Another Angle on F-m-a packet Due Monday Project (projectile simulator) Packet (Projectile Motion) correction Packet (Regents Physics Projectile Practice) corrections Homework is posted on google classroom - code: nsdnlt
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Forces in Two Dimensions - Objectives
know the procedures of Addition of Forces Resolution of Forces Understand the conditions of Equilibrium and Static Be able to determine The individual force acting on an object in equilibrium If equilibrium is achieved in a system
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Force Vector review Force is a vector quantity, it has ____________ and ___________. The vector sum of all forces acting on an object is known as _____________. According to Newton’s 2nd Law: Fnet = ________ magnitude direction Net Force m∙a
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Adding forces Two 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 method Mathematical method Head-tail or tail-tail (parallelogram) c2 = a2 + b2 SOH CAH TOA
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Resolving Force - Finding components
To determine the magnitudes of the components of a force , we use ____________ the trig. functions. y Ax = _______________ A∙cosθ Ay A∙sinθ Ay = _______________ θ x Ax
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Equilibrium and Static
When 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.
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Force at an angle When 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.
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Example - determine If a system is in equilibrium
Forces A, B, C acting on a point. Determine if they produce equilibrium. To determine if a system is in equilibrium is to check if ∑F = 0 A = 5 N 36.9o x B = 4 N C = 3 N Resolve 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.
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Example – determine individual force in an equilibrium
When 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 = 0 ∑Fy = 0 What is the weight of the frame? The weight of the frame is 50 N
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In conclusion A 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 _________________.
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Net Force Problems Revisited
A 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 direction The acceleration of an object can be determined by using Newton's second law.
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Example - Determine the net force and acceleration
Object moves in horizontal direction Fnet = N, right m = (Fgrav / g) = 20 kg a = (69.9 N) / (20 kg) =3.50 m/s/s, right
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Example - Determine the net force and acceleration
Object moves in horizontal direction Fnet = 30.7 N, right a = 1.23 m/s/s, right.
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Procedures for adding vectors at an angle with horizontal
Resolve the vectors at an angle into x and y components. Add all the x components together Add all the y components together Use Pythagorean Theorem to find the resultant (hypotenuse) Resultant2 = x2 + y2 Use trigonometric function to determine the direction: tanθ = opp / adj Use Newton’s 2nd Law to determine acceleration.
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This applied force (FA) The total vertical force must
Practice 1 A 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. This applied force (FA) can be broken into COMPONENTS FAx = 100cos(30o) = 87 N FAy = 100sin(30o) = 50 N X Y FAX FAY Fg FN Total = FAX Total = 0 FN The total vertical force must be 0, so Ry = FN + FAY –Fg = 0 FN = Fg – FAY FA FAY 30˚ FAX R = Rx = Fax Acceleration depends only on FAX Fg
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Lab – shoot for your grade
NAMES, DATE, TITLE PURPOSE How far will a horizontally launched ball land? MATERIALS launcher, meter stick, target paper PROCEDURE Write 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 diagram State equations you need to use to solve for the impact point distance velocity
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11/13 do now Sketch a set of graphs that relate the variables shown on the axes for an object that is launched at an angle. Homework - Due Monday: Another Angle on F-m-a packet Due Monday Project (projectile simulator) Packet (Projectile Motion) correction Packet (Regents Physics Projectile Practice) corrections Due Tuesday Packet (worksheet 1.2.3) Homework is posted on google classroom - code: nsdnlt ax vx t t ay vy t t
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Practice 2 A 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? How could the acceleration be increased? Fnet = FA cos θ Fnet = (20 N)(cos 45°) Fnet = N Pushing at a smaller angle will make Fnet greater and therefore increase acceleration. What is the crate’s acceleration? a = Fnet / m a = (14.14 N) / (40 kg) a = 0.35 m/s2
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This applied force (FA) The total vertical force must
Pushing on an Angle A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of 30° below horizontal. This applied force (FA) can be broken into COMPONENTS X Y FAX FAY Fg FN Total = FAX Total = 0 FN The total vertical force must be 0, so FN = Fg + FAY FAX FAY -30˚ FA Acceleration depends only on FAX Fg
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Practice 3 A 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? Fnet = FA cos θ Fnet = (15 N)(cos 60°) Fnet = 7.5 N a = Fnet / m a = (7.5 N) / (30 kg) a = 0.25 m/s2 What could she do to reduce her acceleration? Push at an greater angle
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Practice 4
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Practice 5 A 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 find the acceleration of the box.
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determine the net force and acceleration
Fnet2 = (∑Fx)2 + (∑Fy)2 Fnet = 39 N θ = tan-1(-23/32) = -36o θ = 324o CCW a = Fnet / m = 39 N / 5 kg = 8.0 m/s2, same direction as the force
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11/16 do now Due today: Project (projectile simulator) Packet (Projectile Motion) correction Packet (Regents Physics Projectile Practice) corrections Due tomorrow Another Angle on F-m-a packet Due Wednesday Worksheet packet 2.1.2B packet Due Friday Projectile test correction worksheet correction Homework is posted on google classroom - code: nsdnlt A projectile passes through points A and B as it follows the path shown below. Compared to the horizontal speed at point A in the projectile’s path, the horizontal speed at point B is (greater, smaller, the same) Compared to the vertical speed at point A in the projectile’s path, the vertical speed at point B is (greater, smaller, the same) Compared to the acceleration at point A in the projectile’s path, the acceleration at point B is (greater, smaller, the same)
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Inclined Planes Objects 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.
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Fg on Inclined Plane
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An important idea The 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 object’s motion along the incline.
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Forces on an Incline Calculations
Consider forces: Perpendicular F┴ = Fg cos θ Cancel out Normal (FN ) Parallel F// = Fg sin θ All the parallel components (including the friction force) add together to yield the net force. Which should directed along the incline. θ Tilt you head method θ
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Essential Knowledge What 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 parallel What happens to the normal force as the angle is increased? Decreases – depends on Fg perpendicular What happens to the friction force as the angle is increased? Decreases – depends on normal force
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Example What is the magnitude of the normal force?
Fg = 50N 30° What is the magnitude of the normal force? FN = Fg perpendicular = Fg cos θ = 43.3 N If the box is sliding with a constant velocity, what is the magnitude of the friction force? Ff = Fg parallel = Fg sin θ = 25 N
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Example 1 The 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. Fg┴ = Fgrav∙cos30o = 850 N Fg// = Fgrav∙sin30o = 500 N In perpendicular direction: Fnorm = F┴ = 850 N In parallel direction: Fnet = F// - Ff Fnet = 500 N - µFnorm Fnet = 235 N a = Fnet / m = 2.35 m/s2
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Example
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practice An 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 block’s 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?
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The diagram below shows a 1
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 truck’s 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?
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Determine the magnitude of the frictional force acting on the sled.
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)
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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?
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11/17 do now A box with mass m is at rest on an rough plane inclined at angle θ with horizontal. Draw a free body diagram on the box. In terms of m, g and θ Determine Fg┴ and Fg// Determine FN Determine Ff Determine Fnet
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objective Review Finish Lab Due today: Another Angle on F-m-a packet
Due tomorrow worksheet 2.1.2B – Forces on Angles Packet Castle Learning Test – Due tomorrow night Due Friday Projectile Projectile test correction Project: projectile simulator correction Worksheet ground launched projectile packet correction Homework is posted on google classroom - code: nsdnlt
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11/7 do now A 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?
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11/16 do now What is the magnitude of the normal force?
Due today: Another Angle on F-m-a packet Project (projectile simulator) Packet (Projectile Motion) correction Packet (Regents Physics Projectile Practice) corrections Due tomorrow Packet (worksheet 1.2.3) Homework is posted on google classroom - code: nsdnlt Fg = 60N 30° What is the magnitude of the normal force? FN = Fg perpendicular = Fg cos θ = 52 N If the box is sliding with a constant velocity, what is the magnitude of the friction force? Ff = Fg parallel = Fg sin θ = 30N
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