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Unit 4: Newton’s Laws of Motion Causes of Motion Aristotle (384-322 BC) believed that all objects had a “natural place” and that the tendency of an object was to reside in its “natural place.” All objects were classified into categories of earth, water, air, or fire. “Natural motion” occurred when an object sought to return to its “natural place” after being moved from it by some type of “violent motion.” The natural state of an object was to be “at rest” in its “natural place.” To keep an object moving would require a force. These views remained widely supported until the 1500s when Galileo Galilei (1564-1642) popularized experimentation. Isaac Newton (1642–1727) proposed that the tendency of an object was to maintain its current state of motion. Forces • A force is a push or a pull • A force can cause – a stationary object to move – a moving object to stop – an object to accelerate (change speed or direction) • Net force – the combination of all the forces acting on an object. – changes an object’s state of motion. • Balanced Force – Net force = 0 – object at rest – Or constant velocity • Unbalanced – Net force is greater than zero – Object moves – Or accelerates Newton’s Laws of Motion • 1st Law – (Law of Inertia) An object at rest will stay at rest, and an object in motion will stay in motion at constant velocity, unless acted upon by an unbalanced force. • 2nd Law – (F=ma)Force equals mass times acceleration. • 3rd Law – (action-reaction)For every action there is an equal and opposite reaction. INERTIA the tendency of an object to resist any change in its motion Inertia is a property of matter and does not depend on the position or location of the object. But it does depend on: MASS a quantitative measure of inertia FORCE “a push or pull” Mass vs. Weight • MASS • WEIGHT – How much and what – Force of gravity acting on material an object is a mass made of (what types of – Measured in Newtons atoms and how many of them) Fg=mag – Measured in grams or kilograms (kg) – 1N = 1kg· m/s2 – Is constant for an object independent of location Normal Force and Gravity • Gravity always pulls straight down • Normal force (FN) is perpendicular to surface and equal and opposite to component of gravitational force (Fg) FN FN Fg Fg • This may lead to an unbalanced ‘sliding’ force that is the component of the gravitational force The net force acting on an object is the vector sum of all the forces acting on it. Examples: 9 lb 6 lb 8 lb 4 lb 8 lb 7 lb 5 lb 12 lb 4 lb If an object is remaining at rest, it is incorrect to assume that there are no forces acting on the object. We can only conclude that the net force on the object is zero. What direction is normal force (FN). Example 1 Example 9 Fnet magnitude _______ direction _________ Balanced or Unbalanced? Fnet magnitude _______ direction _________ Fnet Balanced or Unbalanced? magnitude _______ direction _________ Balanced or Unbalanced? Example 12 2nd Law The net force of an object is equal to the product of its mass and acceleration F=ma. 2nd Law • Relates an object’s mass and acceleration to the net force (force causes acceleration) • Mass is inversely related to acceleration • Acceleration is directly related to net force Newton’s 2nd Law proves that different masses accelerate to the earth at the same rate, but with different forces. • We know that objects with different masses accelerate to the ground at the same rate. • However, because of the 2nd Law we know that they don’t hit the ground with the same force. F = ma F = ma 98 N = 10 kg x 9.8 m/s2 9.8 N = 1 kg x 9.8 m/s2 Calculating force and acceleration • Remember Force = mass x acceleration F=ma • And acceleration is change in velocity over time v f vi a t Example 2 • How much force must a 30,000kg jet plane develop to achieve an acceleration of 1.5m/s2? (neglecting air friction) • F=ma • F=(30,000 kg) (1.5 m/s2)= 45,000 N Example 3 • If a 900 kg car goes from 0 to 60 mph (27 m/s) in 5 seconds, how much force is applied to achieve this? Example 4 • If I throw a 0.145 kg baseball at 20 m/s baseball and my ‘windup distance’ is 0.6 meters, how much force am I applying? Example 5 • A 2.2 kg book is slid across a table. If Fnet = 2.6 N what is the book’s acceleration? • F=ma • 2.6 N = (2.2kg) a • a = 1.18 m/s2 Example 6 • If you drop a 20 kg object what is its acceleration? What is its weight? • acceleration = 9.8 m/s2 • Weight = force • Fg=ma • Fg= (20 kg) (9.8 m/s2) • Q: If a jet cruises with a constant velocity and the thrust from its engines is constant 80,000 N. What is the acceleration of the jet? – A: Zero acceleration because the velocity does not change. • Q: What is the force of air resistance acting on the jet? – A: 80,000 N in the opposite direction of the jet’s motion Example 7: After a birthday party, Bozo the clown went to dinner in his 250 kg car. To save room in the car, he let the left over balloons hang out the window. The engine of the car is exerting a force of 360 N. The balloons are creating drag in the air with a force of 35 Newtons in the opposite direction of the car’s motion. • Draw the vector arrows on the free body diagram • What is the net force (Fnet) acting on the car? • What is the direction that force is acting? • Use Newton’s 2nd Law to calculate the net acceleration of the car. 3rd Law • For every action, there is an equal and opposite reaction. 3rd Law • There are two forces resulting from this interaction • a force on the chair (action) • a force on your body (reaction) action reaction • If all forces have equal and opposing forces, how does anything move? – Action-Reaction pairs are forces of objects on different objects – F Net is sum of external forces acting on ONE object 3rd Law Flying gracefully through the air, birds depend on Newton’s third law of motion. As the birds push down on the air with their wings, the air pushes their wings up and gives them lift. Other examples of Newton’s Third Law • Action: baseball forces the bat to the left • Reaction: bat forces the ball to the right Friction & Tension • Friction (Ff) - the force that opposes motion • Tension (FT) - the pulling force exerted by a string, cable, chain on another object. Drawing Free Body Diagrams Example 10 Drawing Free Body Diagrams Example 11 Example 7 • Draw free body diagram for table • Applied force from pusher, normal force, gravitational force, friction force • If applied force is greater than friction, table moves Friction • The force of friction (Ff): 1. Is always opposite to the direction of motion or impending motion 2. Usually has a smaller value if the object is moving than if it is stationary - (static friction > kinetic friction); 3. Depends on the nature of the pair of surfaces involved (the value of μ); Friction • The force of friction (cont’d): 4. Is proportional to the force pressing the surfaces together (the normal force); - static friction: Ff ≤ μs FN - kinetic friction: Ff =μk FN 5. Is usually independent of the contact area and speed. Example 13 • If a 1 kg mass sits on a flat surface with a coefficient of static friction of 0.5, what is the force of friction (Ff) if: – A horizontal force of 1 N is applied? – A horizontal force of 10 N is applied? – A horizontal force of 100 N is applied? Finding Force With Angles FN • Horizontal Ff Fapp – FN = Fg Fg FN • Incline Ff – Fnet = Fgsinθ 20° – FN = Fgcos θ FN Fg Fnet 20° Statics • The study of forces in equilibrium – Balanced forces – No acceleration Statics FT1 FT2 • If hanging from a wire – Weight is shared equally between each wire – Weight is NOT equal to Tension Fg Example 14 • At an art auction, you acquired a painting that now hangs from a nail on the wall. If the painting has a mass of 12.6 kg, what is the tension in each side of the wire supporting the painting? Physics 1 Assessment 4E 1. Two forces are applied to a 2.0 kg block on a frictionless, horizontal surface, as shown in the diagram. The acceleration of the block is A. 5.0 m/s2 to the right B. 3.0 m/s2 to the right C. 5.0 m/s2 to the left D. 3.0 m/s2 to the left Physics 1 Assessment 4E 2. The vector diagram below represents two forces, F1 and F2, simultaneously acting on an object. Which vector best represents the resultant of the two forces? A. B. C. D. Physics 1 Assessment 4E 3. A horizontal force is used to pull a 5.0 kg cart at a constant speed of 5.0 m/s across the floor, as shown in the diagram. If the force of friction between the cart and the floor is 10 N, the magnitude of the horizontal force along the handle of the cart is A.5.0 N B.10 N C.25 N D.50 N Physics 1 Assessment 4E 4. The diagram below shows a sled and rider sliding down a snow-covered hill that makes an angle of 30° with the horizontal. Which vector best represents the direction of the normal force, FN, exerted by the hill on the sled? A. B. C. D. Physics 1 Assessment 4E 5. An electric model of a Boeing 757, has a mass of about 12 kg. If the owner adjusts the wing flaps to create 123 N of lift upwards, what is the net vertical force on the plane? A.0 N B.5.4 N C.10.3 N D.111 N E.241 N Applying 2D Forces & Free-Body Diagrams • Forces are not always acting in one direction (same or opposite). • The forces along the x-axis and y-axis may not be in equilibrium. • We use Pythagorean theorem to resolve the net force acting on an object. Example • An object is being pulled by a 3 kiloNewtons force towards the north and a 4 kiloNewtons force eastward on a frictionless surface. What is the net force that will accelerate this object? Example • What is the applied force acting against a frictional force of 10 N, if an object is pulled with a force of 200 N at angle of 60o from the ground? What is the net force? • Solve for the Fa applied force along the x –axis Fa(x) Fa(x) = 200cos60 Fa(x) = 100 N • The force applied opposite frictional force is 100 N and not 200 N. • We can solve for the net force Fnet then. Fnet = Fa(x) – Ff = 100 N – 10 N = 90N Example • What is the normal force Fn acting on a 180 N object on the ramp that made an angle of 60o from the ground? • We will solve this problem using similar triangles. • Take note Fn = Fg’ , but opposite in direction. • We will solve Fg’ using cosine. • Our hypotenuse is the weight Fg =180 N. • Fg’ is the adjacent side with respect to the angle 60o. • cos Ɵ = adjacent side / hyp. • cos 60o = Fg’ / Fg • Fg’ = 180cos60 • Fg’ = 90 N • Since Fg’ = Fn • Fn = 90 N Example • A crate is being pulled by cables along a frictionless surface with a force of 500 kN eastward and by another force of 400 KN @ 120o. What is the net force acting on the crate? Hint: must find magnitude and direction! Sin 30= Fx / 400 kN Fx = 400sin30 Fx = 200 kN Cos 30 = Fy / 400kN Fy = 400cos30 Fy = 346.41 kN Magnitude: • Add all the vector forces along the x-axis. Fxtotal = 500 kN - 200 kN Fxtotal = 300 kN • Add all the vector forces along the y-axis. • Fytotal = 346.41 kN • Use Pythagorean Theorem to solve for Fnet Fnet = √(Fx2 + Fy2 ) = √(3002 + 346.412) = 458.23 kN Direction • Tangent Ɵ = opposite side/adjancent side Ɵ = tan-1(Fytotal/Fx total ) = tan-1346.41 kN/300 kN Ɵ = 49.11o Fnet = 458.23 kN @ 49.11o

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posted: | 3/3/2013 |

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