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Applications of Newton’s Laws of Motion

Applications of Newton’s Laws of Motion. Remember that forces are VECTORS ! Newton’s 2 nd Law : ∑ F = ma ∑ F = VECTOR SUM of all forces on mass m  We need VECTOR addition to add forces in the 2 nd Law ! Forces add according to the rules of VECTOR ADDITION !

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Applications of Newton’s Laws of Motion

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  1. Applications of Newton’s Laws of Motion

  2. Remember that forces are VECTORS! Newton’s 2nd Law: ∑F = ma ∑F = VECTORSUMof all forceson mass m  We need VECTORaddition to add forces in the 2nd Law! • Forces add according to the rules of VECTOR ADDITION! • Get Force Components from trig, as discussed earlier.

  3. Problem Solving Procedures • Make a Sketch. For each object separately, sketch a free-body diagram, showing allthe forces acting on that object. Make the magnitudes & directions as accurate as you can. Label each force. • Resolve Vectors into Components. • Apply Newton’s 2nd Lawseparately to each object & for each vector component. • Solve for the unknowns. • Note: This often requires algebra, such as solving 2 linear equations in 2 unknowns!

  4. Example Square Root! FR = [(F1)2 + (F2)2](½) = 141 N tanθ = (F2/F1) = 1  θ = 45º

  5. Example If the boat moves with acceleration a, ∑F = = FR = ma FRx = max, FRy = may

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

  7. Example • A box of mass m = 10 kg is pulled by an attached cord • along a horizontal smooth (frictionless!) surface of a • table. The forc 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.

  8. 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 equal & opposite to the weight!!

  9. A box of mass m = 10 kg is pulled by an attached cord along a • horizontal smooth (frictionless!) surface of a table. The force is • FP = 40.0 N at a 30.0° angle as shown. • Calculate: a. The acceleration. b. The normal force FN Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N

  10. A box of mass m = 10 kg is pulled by an attached cord along a • horizontal smooth (frictionless!) surface of a table. The force is • FP = 40.0 N at a 30.0° angle as shown. • Calculate: a. The acceleration. b. The normal force FN Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N

  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 is • FP = 40.0 N at a 30.0° angle as shown. • Calculate: a. The acceleration. b. The normal force FN Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Horizontal (x): Fx = ma FPx = ma; a = (FPx/m) a = 3.46 m/s2

  12. Example: Pulling against friction • A box, mass m = 10 kg, is pulled along a horizontal surface • by a force FP = 40.0 N applied at a 30.0° angle above • horizontal. The coefficient of kinetic friction is k= 0.3. • Calculate a. the acceleration, b. the normal force FN.

  13. A box, mass m = 10 kg, is pulled along a horizontal surface • by a force FP = 40.0 N applied at a 30.0° angle above • horizontal. The coefficient of kinetic friction is k= 0.3. • Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N

  14. A box, mass m = 10 kg, is pulled along a horizontal surface • by a force FP = 40.0 N applied at a 30.0° angle above • horizontal. The coefficient of kinetic friction is k= 0.3. • Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N

  15. A box, mass m = 10 kg, is pulled along a horizontal surface • by a force FP = 40.0 N applied at a 30.0° angle above • horizontal. The coefficient of kinetic friction is k= 0.3. • Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Friction: Ffr = kFN = (0.3)(78) = 23.4 N

  16. A box, mass m = 10 kg, is pulled along a horizontal surface • by a force FP = 40.0 N applied at a 30.0° angle above • horizontal. The coefficient of kinetic friction is k= 0.3. • Calculate a. the acceleration, b. the normal force FN. Components of FP FPx = FP cos() = 40cos(30.0°) = 34.6 N FPy = FP sin() = 40sin(30.0°) = 20 N Newton’s 2nd Law Vertical (y): Fy = 0 FN + FPy – mg = 0; FN = mg - FPy FN = 98 – 20 = 78 N Friction: Ffr = kFN = (0.3)(78) = 23.4 N Horizontal (x): Fx = ma FPx - Ffr = ma; a = (FPx - Ffr)/m) = 1.14 m/s2

  17. Example: Two boxes and a pulley • 2 boxes are connected by a cord running • over a pulley. The coefficient of kinetic • friction between box A & the table is • k = 0.2. Ignore mass of cord & pulley & • friction in the pulley, which means a • force applied to one end of the cord has • the same magnitude at the other end. • Find the acceleration, a, of the system, • which has the same magnitude for both • boxes if the cord doesn’t stretch. As box • B moves down, box A moves to right. • Find the tension FT in the cord. a  a  ∑F = ma For EACH mass separately! x & y components plus friction Ffr = μkFN

  18. Example: Two boxes & a pulley • Coefficient of kinetic friction is k = 0.2. • Find the acceleration, a, of the system • & the tension FT in the cord. • ∑F = ma For EACH mass • separately! x & y components • plus friction Ffr = μkFN • mA: ∑Fx = mAa = FT – Ffr • ∑Fy = 0 = FN – mAg • FN = mg, Ffr = μkFN = μkmAg • So: mAa = FT – μkmAg (1) • mB: ∑Fy = mBa = mBg – FT • So: mBa = mBg – FT (2) • Solve (1) & (2) to find a & FT a  a 

  19. Example: Two boxes and a pulley • Solutions are: • a = (mBg – Ffr)/(mA + mB) • a = 1.4 m/s2 • FT = Ffr + mAa • FT = 17 N a  a  ∑F = ma For EACH mass separately! x & y components plus friction Ffr = μkFN

  20. Conceptual Example: To push or to pull a sled? Your little sister wants a ride on her sled. If you are on flat ground, will you exert less force if you push her or pull her? Assume the same angle θin each case. ∑F = ma Pulling Pushing y forces: ∑Fy = 0 FN – mg –Fy = 0 FN = mg + Fy Ffr (max) = μsFN x forces: ∑Fx = ma y forces: ∑Fy = 0 FN - mg + Fy = 0 FN = mg - Fy Ffr (max) = μsFN x forces:∑Fx = ma

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