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Newton’s Laws of Motion

Newton’s Laws of Motion. Basic Concepts. Force – push or pull on an object - Vector quantity Mass – amount of matter in a body. - Measured with a balance - Scalar quantity. First Law. Also known as the law of inertia

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Newton’s Laws of Motion

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  1. Newton’s Laws of Motion

  2. Basic Concepts • Force – push or pull on an object - Vector quantity • Mass – amount of matter in a body. - Measured with a balance - Scalar quantity HFinks '07

  3. First Law • Also known as the law of inertia • An object will remain at rest or continue to move at a constant speed unless acted upon by a net force. • Net force – vector sum of all forces acting on an object. HFinks '07

  4. Second Law • When a net force acts on a body, the acceleration is directly proportional to the force and indirectly proportional to the mass. • Equation: ∑F = ma or Fnet = ma Net force on an object = mass x acceleration Net = sum of the forces • Units of measurement for force metric system - Newton (N) metric system - Dyne (dyn) english system - Pound (lb) HFinks '07

  5. Third Law For every action, there is an equal and opposite reaction. Forces occur in pairs. Net force and acceleration will always be in the same direction. HFinks '07

  6. Examples and Labeling of Forces HFinks '07

  7. Applied Force (Fa) • Pushing, pulling or lifting an object. • Applied force is a vector quantity. HFinks '07

  8. Applied Force (Fa) Example 1: Lifting an object Fa HFinks '07

  9. Applied Force (Fa) Example 2: Pulling an Object Fa HFinks '07

  10. Applied Force (Fa) Example 3: Pushing an Object Fa HFinks '07

  11. Applied Force (Fa) Example 4: Pulling an object up an inclined plane Fa HFinks '07

  12. Normal Force (FN) • Normal force, FN , is a component of the force the surface exerts on an object. • Normal force is always perpendicular to the surface. • Normal force is a vector quantity. HFinks '07

  13. Normal Force (FN) Example 1: Block on floor FN HFinks '07

  14. Normal Force (FN) Example 2: Block on wall FN HFinks '07

  15. Normal Force (FN) Example 3: Block on inclined plane FN HFinks '07

  16. Weight (W) Weight – depends on the acceleration due to gravity. - Measured with a spring scale. - Vector quantity - Is a force - Equation: W = mg - Units of measurement metric system - Newton metric system - Dyne english system - Pound HFinks '07

  17. Comparing Units F = ma W = mg = kg m/s2 = kg m/s2 = N = N N = Newton HFinks '07

  18. Weight (W) and Mass (m) • Weight and mass are not the same. • Weight changes with a change in acceleration due to gravity. • Mass does not change with a change in acceleration due to gravity. Example: • Your weight will change if you go from the earth to the moon. • Your mass will remain the same. HFinks '07

  19. Weight (W) Example 1: Block on floor W HFinks '07

  20. Weight (W) Example 2: Block on wall W HFinks '07

  21. Weight (W) • Example 3: Block on inclined plane W HFinks '07

  22. Friction • Occurs when objects are in contact • Acts opposite to the direction of motion • Two kinds of frictional forces Static (object at rest) = fs Kinetic (object is moving) = fk • Equations Static fsmax =µsFN Kinetic fk = µkFN HFinks '07

  23. Coefficient of Friction (µ) • Ratio of the frictional force to the normal force. That is, you are comparing one force to another. StaticKinetic (sliding) µs = fs µk = fk FN FN • No unit of measurement because you are dividing a force by a force. HFinks '07

  24. Kinetic Frictional Force (fk) • The frictional force always act opposite to the applied force. Think of these two forces as being x-components. One is positive and the other is negative. Fa fk Direction of motion HFinks '07

  25. Kinetic Frictional Force (fk) Example 1: Box pulled across a floor Fa fk HFinks '07

  26. Kinetic Frictional Force (fk) Example 2: Object pulled up an inclined plane Fa fk HFinks '07

  27. Kinetic Frictional Force (fk) • Example 3: Object sliding down an inclined plane • No applied force fk HFinks '07

  28. Static Frictional Force (fs) - Friction between atoms or molecules - No applied force HFinks '07

  29. Frictional Force (fs) Example 1: Box at rest on a floor fs HFinks '07

  30. Frictional Force (fs) • Example 2: Object at rest on an inclined plane fs HFinks '07

  31. Fnet = ma Applications of this equation in the x direction Applied Force (Fa) and Frictional Force (fk) HFinks '07

  32. Fnet = ma Applications of this equation in the y direction Normal Force (FN) and Weight (W) Applied Force (FN) and Weight (W) HFinks '07

  33. Applied Force (Fa) and Weight (W) Example 1: Lifting Fa W Fa – W = ma HFinks '07

  34. Normal Force (FN) and Weight (W) Example 2: Block on floor FN W FN - W = ma HFinks '07

  35. Solving Force Problems • Label all forces • Write an expression for the forces acting in the x direction. • Write an expression for the forces acting in the y direction. • Substitute the known values into the expressions. • Solve for the unknown. HFinks '07

  36. Example 1 What is the magnitude and direction of FN? FN • Label forces • (No forces in x direction) b. Write expression Mass = 20.0 kg FN - W = ma At rest ay = 0 m/s2 W = mg = 196 N c. FN - W = ma FN = W + ma =196 N + (20.0 kg)(0 m/s2) FN = 196 N, upward HFinks '07

  37. Example 2 A 12.0 kg object is pulled upward by a massless rope with an acceleration of 3.00 m/s2. What is the tension (T) in the rope? Fa = T a = 3.00 m/s2 m = 12.0 kg W = (12.0 kg)(9.80 m/s2) = 118 N T – W = ma T = 118 N + (12.0 kg)(3.00 m/s2) T = W + ma T = 154 N HFinks '07

  38. Example 3 A 12.0 kg object is accelerating downward at 3.00 m/s2. What is the tension (T) in the rope? Fa = T a = -3.00 m/s2 m = 12.0 kg W = (12.0 kg)(9.80 m/s2) = 118 N T – W = ma T = 118 N + (12.0 kg)(-3.00 m/s2) T = W + ma T = 82.0 N HFinks '07

  39. Example 4 A 100. N crate is pulled across the floor. The tension in the rope is 75.0 N. Calculate the maximum frictional force. (µk = 0.5) a FN a. Label forces Fa = T fk b. Write expressions W = 100 n Fk = µk FN = (0.5)(100 N) Fk = 50.0 N HFinks '07

  40. Example 4 Calculate the resultant force (Fnet) in Example 4. a a. Label forces FN=100. N Fa = T = 75.0 N Fk =50.0 N b. Write expressions W = 100 n Fy = FN – W Fx = T – fk FN = W = 75.0 N – 50.0 N Fy = 100. N – 100. N Fx = 25.0 N Fy = 0 N HFinks '07

  41. Example 4 Calculate the acceleration of the object in Example 4. a FN a. Label forces fk= 50.0 N b. Write expressions for direction of motion only. Fa = T= 75.0 N W = 100. N m = w = 100. N = 10.2 kg g 9.80 m/s2 T – fk = ma a = 75.0 N – 50.0 N a = T – fk10.2 kg m a = 2.45 m/s2 HFinks '07

  42. Inclined Plane HFinks '07

  43. Example 5 A 20.0 kg object is at rest on an inclined plane (µs = 0.10) What is the value of fs? Ө = 10.0º fs = µsFN FN ≠ W Reason: They aren’t opposite each other “W” is a vector quantity and is acting at an angle. Calculate x- and y-components, FN fs W Ө Next HFinks '07

  44. Example 5 W = mg = 196 N Wy = W cos Ө Wx = W sin Ө Note: Ө is the same for both the large and small right triangles. Wx will always act “down” the incline.. ( -Wx) FN Wy fs Ө W Wx Ө Next HFinks '07

  45. Example 5 FN = Wy Fs = µsFN = µsWy = µs ( W cos Ө) = (0.10)(196 N)(cos 10.0º) fs = 19.3 N FN Wy fs Ө W Wx Ө HFinks '07

  46. Example 6 A 20.0 kg object is sliding down an incline at a constant velocity under the influence of gravity. Find the value of fK. fk HFinks '07

  47. Example 6 Label Forces and write expressions Object isn’t moving in the y direction. So…..don’t use the y expression: FN - Wy = ma Object is moving along the incline. Use…. ∑Fx fk - Wx = ma a = 0. Object is moving at a constant velocity. So….. Fk = Wx FN fk Wy Ө W Wx Ө Next HFinks '07

  48. Example 6 fk = Wx = W sin Ө =196N sin 10.0º fk = 34.0 N FN fk Wy Ө W Wx Ө Next HFinks '07

  49. Example 7 A 20.0 kg object is pulled up a 10.0º incline and µk = 0.05. The object is accelerated at a rate of 1.25 m/s2. What is the value of the applied force? W = 196 N (i) Fa - fk - Wx = ma Fa = fk + Wx + ma *’a” is positive. The object is moving up the incline. fk = µkFN = µkWy fk = µkW cos Ө FN Fa Wy fk Ө W Wx Ө Next HFinks '07

  50. Example 7 W = 196 N • (i) Fa - fk - Wx = ma • Fa = fk + Wx + ma • Fa = (µkW cos Ө) +W sin Ө + ma • = (0.05)(196 N)(cos 10.0º) + (196 N)(sin 10.0º) + (20.0 kg)(1.25 m/s2) • = 9.65 N + 34.0 N + 25.0 N • Fa = 68.7 N FN Fa Wy fk Ө W Wx Ө HFinks '07

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