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Applications & Examples of Newton’s 2 nd Law

Applications & Examples of Newton’s 2 nd Law. Example 4-11. A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force exerted is F P = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration of the box.

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Applications & Examples of Newton’s 2 nd Law

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  1. Applications & Examples of Newton’s 2nd Law

  2. Example 4-11 A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force exerted is FP = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration of the box. b. The magnitude of the upward normal force FN exerted by the table on the box. Free Body Diagram The normal force, FN is NOT always equal & opposite to the weight!!

  3. Example 4-12 Two boxes are connected by a lightweight (massless!) cord & are resting on a smooth (frictionless!) table. The masses are mA = 10 kg & mB = 12 kg. A horizontal force FP = 40 N is applied to mA. Calculate: a. Acceleration of the boxes. b. Tension in the cord connecting the boxes. Free Body Diagrams

  4. Example 4-13 (“Atwood’s Machine”) Two masses suspended over a (massless frictionless) pulley by a flexible (massless) cable is an “Atwood’s machine” . Example: elevator & counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = 1150 kg. Calculate a. The elevator’s acceleration. b. The tension in the cable. a   a aE = - a Free Body Diagrams aC = a

  5. Conceptual Example 4-14 Advantage of a Pulley A mover is trying to lift a piano (slowly) up to a second-story apartment. He uses a rope looped over 2 pulleys. What force must he exert on the rope to slowly lift the piano’s mg = 2000-N weight? mg = 2000 N Free Body Diagram

  6. Example 4-15 = 300 N Newton’s 3rd Law FBR = -FRB, FCR = -FRC FRBx = -FRBcosθ FRBy = -FRBsinθ FRCx = FRCcosθ FRCy = -FRCsinθ

  7. Example: Accelerometer A small mass m hangs from a thin string & can swing like a pendulum. You attach it above the window of your car as shown. What angle does the string make a. When the car accelerates at a constant a = 1.20 m/s2. b. When the car moves at constant velocity, v = 90 km/h? Free Body Diagram

  8. General Approach to Problem Solving • Read the problem carefully; then read it again. • Draw a sketch, then a free-body diagram. • Choose a convenient coordinate system. • List the known & unknown quantities; find relationships between the knowns & the unknowns. • Estimate the answer. • Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. • Keep track of dimensions. • Make sure your answer is REASONABLE!

  9. Problem 25  FT1 Newton’s 2nd Law ∑F = ma (y direction) for EACH bucket separately!!!  a Take up as positive. m1 = m2 = 3.2 kg m1g = m2g = 31.4 N Bucket 1: FT1 - FT2 - m1g = m1a Bucket 2: FT2 - m2g = m2a m1g  FT2 FT2   a  m2g

  10. Problem 29 Take up as positive. Newton’s 2nd Law:∑F = ma (y direction) on woman + bucket! m= 65 kg, mg = 637 N FT + FT - mg = ma 2FT - mg = ma Also, Newton’s 3rd Law:  FP = - FT FT   FT  FP  a  a  mg

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