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Forces in Two Dimensions - Objectives 1. 2. 3. 4. 5. Addition of Forces Resolution of Forces Equilibrium and Static Net Force Problems Revisited Inclined Planes Addition of Forces • Newton’s 2nd Law: ∑F = ma (∑F is the net force) Two ways to add vectors 1.Graphical Method • Head and tail • parallelogram 2.Mathematical method: Using vector resolution and Pythagorean Theorem to determine magnitude and tangent function to determine direction 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 equilibrium. ∑F = 0, a = 0 • 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. example • Forces A, B, C acting on a point. Determine if they produce equilibrium on the point. A B C Force A Force B Force C Magnitude 3.4 N 9.2 N 9.8 N Direction 161 deg. 70 deg. 270 deg Head and tail method to determine resultant Parallelogram method to determine resultant Mathematical Method 1. Resolve vectors 2. Use Pythagorean Theorem to determine magnitude 3. Use Tangent function to determine direction Forces Fx Fy A (3.4 N)(cos161o) = -3.2 N (3.4N)(sin161o) = 1.11 N B (9.2N)(cos70o) = 3.2 N (9.2N)(sin70o) = 8.65 N C (9.8N)(cos270o) = 0 N (9.8 N)(sin270o) = -9.8 N R 0N 0N • The data in the table above show that the forces nearly balance. We could say it's "close enough." Example • A frame is shown with the given tension. Determine the weight of the frame. B A (50N)sin30 = ½ (50 N) (50N)cos120 = -43 N (50N)cos30 = 43 N Fg = 50 N C Cx = 0 Cy = - 50 N The weight is 50 N downward Forces Fx Fy A (50 N)(cos150o) = -43 N (50 N)(sin150o) = 25 N B (50 N)(cos30o) = 43 N (50 N)(sin30o) = 25 N C Cx= ? Cy = ? R 0N 0N Example • A sign is shown with the given mass of 5 kg. Determine the tension of each cable. Tsin40o + Tsin40o A=T B=T 40o Tcos140o 40o Tcos40o Fg = mg = 49 N Forces Fx Fy A T∙cos140o T∙sin140o Tsin40o + Tsin140o = 49 N B T∙cos40o T∙sin40o T = 38.1 N C 0N Cy = mg = 49 N R 0N 0N An important principle Fg = 10 N • As the angle with the horizontal increases, the amount of tensional force required to hold the sign at equilibrium decreases. In conclusion • Equilibrium is the state of an object in which all the forces acting upon it are balanced. In such cases, the net force is 0 Newton. Knowing the forces acting upon an object, trigonometric functions can be utilized to determine the horizontal and vertical components of each force. At equilibrium, all the vertical components must balance and all the horizontal components must balance. 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. Example - Determine the net force and acceleration Object moves in horizontal direction Fnet = 69.9 N, right m = (Fgrav / g) = 20 kg a = (69.9 N) / (20 kg) =3.50 m/s/s, right Example - Determine the net force and acceleration Object moves in horizontal direction Fnet = 30.7 N, right a = 1.23 m/s/s, right. Procedures for adding vectors at an angle with horizontal 1. 2. 3. 4. 5. 6. 7. 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. 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. FN FAY FA 30˚ This applied force (FA) o) = 87 N FAxcan = 100cos(30 be broken into F =COMPONENTS 100sin(30o) = 50 N Ay X verticalYforce must The total be 0, so FAY RyFAX = FN + FAY –Fg = 0 FN = Fg F–g FAY FN FAX Fg Total =R FAX = Rx Total = Fax= 0 Acceleration depends only on FAX 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 Fnetacceleration = FA cos θ be increased? Fnet = (20 N)(cos 45°) F greater and Pushing at a smaller angle will make net Fnetincrease = 14.14acceleration. N therefore What is the crate’s acceleration? a = Fnet / m a = (14.14 N) / (40 kg) a = 0.35 m/s2 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. FAX FAY The total verticalg force must N be 0, Fso = FgTotal + FAY Total F =N FAX =0 -30˚ FA Fg FAX F FN FAY This applied force (FA) canXbe broken Y into COMPONENTS Acceleration depends only on FAX 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 Practice 4 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 0.50. find the acceleration of the box. 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 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. Fg on Inclined Plane 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. Forces on an Incline Calculations • Consider forces: – Perpendicular • F┴ = Fg cos θ • Cancel out Normal (FN ) θ – 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. Tilt you head method θ 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 Example 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 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. F┴ = Fgrav∙cos30o = 850 N F// = 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 Example practice 1. 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? 2. 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? 3. 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? 4. 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? 5. a. b. 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. 6. 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) 7. The diagram shows a sled and rider sliding down a snowcovered 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. 8. 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?