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- 1. Newton’s Laws El Paso Independent High School
- 2. Newton’s First Law Newton’s First Law: An object at rest or an object in motion at constant speed will remain at rest or at constant speed in the absence of a resultant force. A glass is placed on a board and the board is jerked quickly to the right. The glass tends to remain at rest while the board is removed.
- 3. Newton’s First Law (Cont.) Newton’s First Law: An object at rest or an object in motion at constant speed will remain at rest or at constant speed in the absence of a resultant force. Assume glass and board move together at constant speed. If the board stops suddenly, the glass tends to maintain its constant speed.
- 4. Inertia • Newton's first law is often referred to as the law of inertia. • Inertia is the resistance or unwillingness of an object to accelerate (speed up, slow down, or change directions). • The more mass an object has, the harder it is to accelerate. Thus, more mass equals more inertia.
- 5. Understanding the First Law (a) The driver is forced to move forward. An object at rest tends to remain at rest. Discuss what the driver experiences when a car accelerates from rest and then applies the brakes. (b) Driver must resist the forward motion as brakes are applied. A moving object tends to remain in motion.
- 6. Newton’s Second Law • Second Law: Whenever a resultant force acts on an object, it produces an acceleration that is directly proportional to the force and inversely proportional to the mass. m F amaF Force Units: Newton’s or N
- 7. Acceleration and Force Pushing a cart with twice the force produces twice the acceleration. Three times the force triples the acceleration.
- 8. Newton: The Unit of Force One newton is the force required to give an acceleration of 1 m/s2 to a mass of 1 kg. F (N) = m (kg) a (m/s2) What resultant force will give a 3 kg mass an acceleration of 4 m/s2? F = 12 N F = ? a = 4 m/s2 3 kg )/4()3( 2 smkgmaF
- 9. Example A 40 N resultant force causes a block to accelerate at 5 m/s2. What is the mass? F = 40 Nm=? a = 5 m/s2 kg sm N a F m maF 8 /5 40 2
- 10. Example A net force of 4.2 x 104 N acts on a 3.2 x 104 kg airplane during takeoff. What is the force on the plane’s 75-kg pilot? F = 4.2 x 104 N m = 3.2 x 104 kg + F = ma a = 1.31 m/s2 To find the force on the pilot, assume same acceleration: First we find the acceleration of the plane. kgx Nx m F a 4 4 102.3 102.4 NsmkgmaF 4.98)/31.1)(75( 2
- 11. Example A 54-g tennis ball is in contact with the racket for a distance of 40 cm as it leaves with a velocity of 48 m/s. What is the average force on the ball? Given: vo = 0; vf = 48 m/s x = 0.40 m; m = 0.0540 kg; a = ? First, draw sketch and list given quantities: Given: vo = 0; vf = 48 m/s x = 40 cm; m = 54 g a = ? Consistent units require converting grams to kilograms and centimeters to meters: Cont. . .
- 12. Example (Cont). A 54-g tennis ball is in contact with the racket for a distance of 40 cm as it leaves with a velocity of 48 m/s. What is the average force on the ball? F= (0.054 kg)(2880 m/s2) F = 156 N 2 2 fv a x Knowing that F = m a, we need first to find acceleration a: 2 2222 22 /2880 )4(.2 048 2 2 sm x vv a axvv of of
- 13. Weight and Mass • Weight is the force an object applies as a result of gravity pulling it downward. It is directed downward and it varies with gravity. • Mass is a universal constant which is a measure of the matter that makes up an object. It is always constant regardless of location mgW maFBecause weight is a force created by the downward acceleration of gravity, we can make this substitution.
- 14. Weight and Mass: Examples What is the weight of a 10-kg block? 9.8 m/s2 W m10 kg W = mg = (10 kg)(9.8 m/s2) W = 98 N The weight of an object is the force it causes as a result of gravity. The weight of an object is also referred to as the force of gravity
- 15. Always Remember!! In Physics, the use of Newton’s second law and many other applications makes it absolutely necessary to distinguish between mass and weight. Use the correct units! Metric SI units: Mass is in kg; weight is in N. Always give preference to the SI units. Pounds should never be used!!!
- 16. Example A resultant force of 40 N gives a block an acceleration of 8 m/s2. What is the weight of the block near the surface of the Earth? W=? F = 40 Na 8 m/s2 To find weight, we must first find the mass of the block: Now find weight of a 5-kg mass on earth. W = mg = (5 kg)(9.8 m/s2) W = 49.0 N a F mmaF ; kg sm N a F m 5 /8 40 2
- 17. Newton’s Third Law • Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs. Action Reaction ActionReaction
- 18. Action and Reaction Forces Use the words by and on to study action/reaction forces below as they relate to the hand and the bar: The action force is exerted by the _____ on the _____. The reaction force is exerted by the _____ on the _____.bar hands bar hands Action Reaction
- 19. m= 1 kg Forces If a bar of gold that has a mass of 1 kg is sitting at rest on a table, what forces are acting on it and to what magnitude? FN= 9.8 N The normal force of an object pushes up with a force equal to the bar’s downward force. Because both forces are present, the bar remains at equilibrium. Gravity acts on the object in the downward direction. However, because the bar is not moving downward, there must be another force counteracting it. Fg= 9.8 N
- 20. Free-body Diagrams • A free body diagram is a diagram that shows all of the forces present on a given object. • When drawing a free body diagram, objects are always represented as points or dots and forces are drawn as arrows pointing away from the object. • For now the only forces we will consider are gravity, tension, normal, applied, and friction.
- 21. Free-body Diagrams • Gravitational Force: This is the only force that will always be present and will always point straight down. • Tension Force: This type of force is caused by a rope, chain, string, etc. and goes in the direction of the rope, chain, string, etc. • Normal Force: This type of force is present when the object is in contact with a surface. It always points perpendicular to the surface.
- 22. Free-body Diagrams • Applied Force: This type of force is used when an outside agent is acting on the object such as a person. • Frictional Force: This type of force is the result of two surfaces being drug across each other, creating friction. It always points in the direction opposite of the movement. The size of the arrow should be proportional to the size of the force. Larger forces should have larger arrows!!!
- 23. Examples Fa FN FF Fg Fa FT Fg FT Friction acts parallel to the surface but opposite the direction of travel. There is no normal force present because the object is not resting on a surface.
- 24. Example A cart and driver have a mass of 120 kg. What force F is required to give an acceleration of 6 m/s2 with no friction? 1. Read problem and draw a sketch. 2. Draw a free body diagram and label forces. Diagram for Cart: FN Fg Fa 3. Choose x-axis along motion and indicate the right direction as positive (+). x +
- 25. Example What force F is required to give an acceleration of 6 m/s2? Fy = 0; FN - Fg = 0 The normal force FN is equal to weight Fg Fx = max; Fa = ma Fa = (120 kg)(6 m/s2) Fa = 720 N m = 120 kg 4. Write Newton's Law equation for both axes. ay = 0 Diagram for Cart: FN Fg Fa
- 26. Example What is the tension FT in the rope below if the block accelerates upward at 4 m/s2? (Draw sketch and free-body) 10 kg a = +4 m/s2 FT a FT Fg + Fx = m ax = 0 Fy = m ay = m a FT- Fg = m a Fg=mg = (10 kg)(9.8 m/s) = 98 N m a= (10 kg)(4 m/s) = 40 N FT - 98 N = 40 N FT = 138 N
- 27. Example Find the acceleration of the blocks if there is no friction on the surfaces. 2 kg 4 kg 12 N First apply F = ma to entire system (both masses). 12 N FN Fg=(m2 + m4)g Fx = (m2 + m4) a 12 N = (6 kg) a 2 /2 6 12 sm kg N a Because the blocks are tied together and move together, we can treat them as one object
- 28. Now find the tension force in the rope connecting the two blocks. 2 kg 4 kg 12 N Now find tension T in connecting cord. Apply F = m a to the 2 kg mass where a = 2 m/s2. FT FN m2 g Fx = m2 a FT = (2 kg)(2 m/s2) FT= 4 N
- 29. Example (Cont.) The two-body problem. 2 kg 4 kg 12 N Same answer for FT results from focusing on 4-kg by itself. Apply F = m a to the 4 kg mass where a = 2 m/s2. Fx = m4 a 12 N - FT = (4 kg)(2 m/s2) T = 4 N Fa FT m4g FT

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