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Transcript
NEWTON’S LAWS & FORCES 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. Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass. Newton’s Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs. Newton: The Unit of Force One newton is that resultant force which imparts 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 = m a F=? 3 kg F (3 kg)(4 m/s2 ) a = 4 m/s2 F = 12 N Acceleration and Force With Zero Friction Forces Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration. Example 2. 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? First we find the acceleration a of plane. F = 4.2 x 104 N F 4.2 x 104 N a m 3.2 x 104 kg + F = ma m = 3.2 x 104 kg a = 1.31 m/s2 To find F on 75-kg pilot, assume same acceleration: F = ma = (75 kg)(1.31 m/s2); F = 98.4 N Problem Solving Strategy (For the Simpler Problems.) • Read problem; draw and label sketch. • List all given quantities and state what is to be found. • Make sure all given units are consistent with Newton’s second law of motion (F = m a). • Determine two of the three parameters in Newton’s law, then solve for the unknown. Example 3. 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? First, draw sketch and list given quantities: Given: vo = 0; vf = 48 m/s d = 40 cm; m = 54 g a=? Consistent units require converting grams to kilograms and centimeters to meters: Given: vo = 0; vf = 48 m/s d = 0.40 m; Cont. . . m = 0.0540 kg; a = ? Example 3 (Cont). A 54-gm 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? Knowing that F = m a, we need first to find acceleration a: 0 2ax v v ; a 2 f (48 m/s)2 a ; 2(0.40 m) 2 0 v 2 f 2x a 2880 m/s 2 F= (0.054 kg)(2880 m/s2); F = ma F = 156 N Weight and Mass • Weight is the force due to gravity. It is directed downward and it varies from location to location. • Mass is a universal constant which is a measure of the inertia of a body. F = m a so that: Fw = mg and m = Fw g Weight and Mass: Examples • What is the weight of a 10-kg block? 10 kg m 9.8 m/s2 Fw Fw = mg = (10 kg)(9.8 m/s2) Fw = 98 N Newton’s Third Law (Reviewed): • Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs. Action Reaction Reaction Action Acting and Reacting Forces • Use the words by and on to study action/reaction forces below as they relate to the hand and the bar: Action The action force is exerted by bar the _____ hands on the _____. The reaction force is exerted bar on the _____. hands by the _____ Reaction Example 6: A 60-kg athlete exerts a force on a 10-kg skateboard. If she receives an acceleration of 4 m/s2, what is the acceleration of the skateboard? Force on runner = -(Force on board) mr ar = -mb ab Force on Board (60 kg)(4 m/s2) = -(10 kg) ab Force on Runner (60 kg)(4 m/s) 2 a 24 m/s -(10 kg) a = - 24 m/s2 Applying Newton’s Second Law • Read, draw, and label problem. • Draw free-body diagram for each body. • Choose x or y-axis along motion and choose direction of motion as positive. • Write Newton’s law for both axes: SFx = m ax SFy = m ay • Solve for unknown quantities. Example 7: 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. Diagram for Cart: FN FW Fapp + x 2. Draw a vector force diagram and label forces. 3. Choose x-axis along motion and indicate the right direction as positive (+). Example 7 (Cont.) What force F is required to give an acceleration of 6 m/s2? 4. Write Newton's Law equation for both axes. m = 120 kg Diagram for cart: FN F FW ay = 0 SFy = 0; FN - Fw= 0 The normal force FN is equal to weight FW + x SFx = max; F = ma F = (120 kg)(6 m/s2) F = 720 N Example 8: What is the tension T in the rope below if the block accelerates upward at 4 m/s2? (Draw sketch and free-body.) T a 10 kg a = +4 m/s2 T mg + SFx = m ax = 0 (No info) SFy = m ay = m a T - mg = m a mg = (10 kg)(9.8 m/s) = 98 N m a= (10 kg)(4 m/s) = 40 N T - 98 N = 40 N T = 138 N