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Newton’s Laws

Newton’s Laws. Newton’s 1 st Law : The Law of Inertia. Copy this note!. “ Every object maintains a state of rest or uniform motion in a straight line unless acted upon by an unbalanced external force .”. Newton’s 1 st Law: The Law of Inertia. Copy this note!. Inertia

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Newton’s Laws

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  1. Newton’s Laws

  2. Newton’s 1st Law:The Law of Inertia Copy this note! “Every object maintains a state of rest or uniform motion in a straight line unless acted upon by an unbalanced external force.”

  3. Newton’s 1st Law: The Law of Inertia Copy this note! • Inertia • Depends on the mass of an object • Inertia is a resistance to a change in motion • Large mass = Large inertia • Small mass = Small inertia

  4. Copy this note! Newton’s 1st Law:The Law of Inertia • Simply put: • Objects like to stay where they are or to keep moving as they have been • The heavier an object is, the harder it is to change its state of motion. • Think of inertia as • Lazy • “I’m a keep on doing what I’m doing”

  5. Newton’s 1st LawThe Law of InertiaVideo

  6. Newton’s 1st Law: The Law of Inertia Copy this note! If the net forceacting on an object is zero, the object will stay at rest or continue to move at constant velocity Example: • A computer sitting on a desk (at rest) “At rest” means Velocity = 0

  7. Newton’s 1st Law: The Law of Inertia “At rest” means Velocity = 0 Copy this note! • A hockey puck moving across smooth ice at a constant velocity • A body in a vehicle keeps moving forward when the car stops suddenly • The seatbelt applies a force in the opposite direction to stop the body from going forward

  8. Note: Be safe and always wear your seatbelt when driving or as a passenger!

  9. Newton’s 2nd Law Copy this note! “An unbalanced force causes an object to accelerate” “If the net external force on an object is not zero, the object will accelerate in the direction of this net force”

  10. Newton’s 2nd Law Copy this note! • An object’s acceleration (a) is: • Directly proportional to the applied unbalanced force (F) • inversely proportional to the object’s mass (m) F = ma (N) (kg) (m/s2) F - NET FORCE in Newtons (N = kgm/s2) m- mass is in kg a- net acceleration is in m/s2 Highlight me!

  11. F = ma Newton’s 2nd Law • Newton’s 2nd Law tells us • The heavier an object is (i.e. more massive), the greater the force is required to accelerate it • Example: a pickup truck requires a much larger engine (larger force) compared to a small compact car. • To increase the rate of acceleration requires a greater force as well • Example: consider the acceleration of a Smart Car compared to a Lamborghini. • The Lamborghini’s bigger engine (bigger force) means it has bigger acceleration

  12. What do you think?

  13. Copy this note! Newton’s 2ndLaw Example 1 Find the net force acting on a 10 kg object that is accelerating at 5.0m/s2 south. Draw a FBD Label Directions +N G: = 5.0 m/s [S] m = 10 kg R: Net force A/S: FNET = ma FNET = (10)(-5) FNET = -50 N = 50 N [S] P: The net force is 50 N South. +E -W - S

  14. Copy this note! Newton’s 2nd Law Example 2 Find the acceleration of a 15 kg object that is experiencing a net force of 30 N to the right. Draw a FBD Label Directions +up G: = 30 N [R] m = 15 kg R: net acceleration A/S: FNET = ma a = FNET /m a = +30/15 a = +2.0 m/s2 ∴ = 2.0 m/s2 [R] P: The acceleration is 2 m/s to the right. +right -left - down

  15. Copy this note! Newton’s 2nd Law Example 3 Determine the net force on the following object. Label Directions +y There is only movement in the x direction. Perform the vector sum of the forces in the x direction = (+7.0) + (-5.0) = 7.0 – 5.0 = 2.0 N = 2.0 N [R] R means “right” +x -x 5.0 N 7.0 N - y

  16. Copy this note! Newton’s 2nd Law Example 4 Determine the net force on the following object. Label Directions +y There is only movement in the y direction. Perform the vector sum of the forces in the y direction = (+5.0) + (-1.0) = 5.0 – 1.0 = 4.0 N = 4.0 N [U] U means “up” +x 5.0 N -x - y If the mass is 0.50 kg, use Newton’s 2nd Law to find the acceleration. FNET = ma a = FNET /m a = 4.0/0.50 a = 8.0 m/s2 ∴ = 8.0 m/s2 [U] 1.0 N

  17. Copy this note! NET is assumed Newton’s 2nd Law Example 5 Determine the net force on the following object. Label Directions There is movement in both x and y!x direction = 7.0 – 5.0 = 2.0 N y direction = 4.0 – 1.0 = 3.0 N +y 5.0 N 7.0 N +x 4.0 N -x - y Use Pythagoras & SOH CAH TOA and are the components of the net force . Draw a right angle triangle: Fy points up (+), Fx points right (+). 1.0 N F2 = Fx2 + Fy2 F = F = F = 3.6 N tan Ѳ = Fy/Fx Ѳ = tan -1 (Fy/Fx) Ѳ = tan -1 (3/2)Ѳ= 56̊ Ѳ = 3.6 N [R56̊U]

  18. Important Announcements • Tuesday April 5th • Automotive Safety Report (due Monday April 11th) • Wednesday April 6th • Review • Thursday April 7th • Forces Unit Test • Friday April 8th • PA Day!

  19. Newton’s Laws

  20. Coefficient of Friction Tables

  21. From Yesterday Newton’s 1st Law:The Law of Inertia “Every object maintains a state of rest or uniform motion in a straight line unless acted upon by an unbalanced external force.”

  22. From Yesterday Newton’s 2nd Law • An object’s acceleration (a) is: • Directly proportional to the applied unbalanced force (F) • inversely proportional to the object’s mass (m) F = ma (N) (kg) (m/s2) F - NET FORCE in Newtons (N = kgm/s2) m- mass is in kg a- net acceleration is in m/s2 Highlight me!

  23. Copy this note! Newton’s 2nd Law Example 6 x direction = 7.0 – 7.0 – 3.0 = -3.0 N y direction = 3.0 + 4.0 - 4.0 = 3.0 N positive y direction Determine the net force on the following object. Label Directions 3.0 N +y 4.0 N +x 7.0 N 7.0 N -x 3.0 N - y Draw a right angle triangle: Fy points +y. Fx points -x. 4.0 N F2 = Fx2 + Fy2 F = F = F = 4.2 N tan Ѳ = Fy/Fx Ѳ = tan -1 (Fy/Fx) Ѳ = tan -1 (3.0/3.0)Ѳ= 45̊ Ѳ For the tanѲ calculation, use positive FX, FY numbers and then work out [L45̊U] = 4.2 N [L45̊U]

  24. Newton’s 3rd Law:Action-Reaction Copy this note! “Every action has an equal and opposite reaction” • If object A exerts a force on object B, then object B exerts the exact same force back on object A but in the opposite direction. A= B

  25. Newton’s 3rd Law:Action-Reaction Copy this note! If you exert 10N of force on an object, the object will exert 10N of force back on you! • You punch a wall, it punches back with an equal but opposite force • A shotgun recoils. The gun exerts a force on the bullet to move it forward. The bullet exerts and equal but opposite force against the gun, causing it to recoil madly!

  26. Newton’s 3rd LawAction-ReactionVideo

  27. Copy this note! Problem SolvingWith Newton’s Laws A box of mass 40 kg is pushed horizontally to the right across a floor with a force of 185N. μK = 0.30. a) Determine the weight of the box? +U +R Weight = G Given Newton’s 2nd Law, force isF = ma We have:G = ma “a” is acceleration On Earth for “a”, physics uses g = 9.81 m/s2 G = m = (40)(-9.81) = - 392N = 392N[Down] N -L - D FBD! F A = 185N G

  28. Copy this note! Problem SolvingWith Newton’s Laws b) Determine the size of the normal force. Normal force is in the y direction. = mY from Newton’s 2nd Law As there is no movement in the y direction: Y = 0 = mY = N + G 0 = N + (-392) N = + 392 N N = 392 N [Up] Notice that FN = FG as we have previously discussed. As shown, Newton’s Law supports this. +U +R N -L - D FBD! F A = 185N G

  29. Copy this note! Problem SolvingWith Newton’s Laws c) Determine the size of the frictional force. FF = µKFN FF = (0.3) (392) FF = 118N ∴F = 118N [LEFT] +U +R N -L - D FBD! F A = 185N G

  30. Copy this note! Problem SolvingWith Newton’s Laws d) Determine the resulting acceleration of the box. From the problem’s description, acceleration is along the x axis = mX from Newton’s 2nd Law = mX= F + A (40)X = (-118) + (185) X = X = + 1.675 m/s2 X = 1.675 m/s2 [right] +U +R N -L - D F A = 185N G

  31. Copy this note! Newton’s 3rd Law Example Standing on perfectly smooth ice, Eric pushes Andrew with a force of 20N to the right. Eric’s mass is 75kg. Andrew’s mass is 65kg. Find: • The reaction force that Andrew applies against Eric • Andrew’s acceleration • Eric’s Acceleration G: EonA = 20 N [R] mE = 75 kg mA= 65 kg R: a) AonEb) A c) E A/S: a) Since Eric pushes Andrew with a force of 20 N to the right, then Andrew produces a reaction force of 20N to the left (Newton’s 3rd Law) AonE = - EonA = - 20 N [R] = 20 N [L]

  32. Copy this note! Newton’s 3rd Law Example EonA = 20 N [R] b) aA = F/mA aA = (+20)/(65) aA= + 0.31 m/s2 = 0.31 m/s2 [R] c) aE= F/mA aE= (-20)/(75) aA= - 0.267 m/s2 = 0.267 m/s2 [L] Andrew and Eric both experienced the same force; however, their different masses mean they will have different accelerations. AonE= 20 N [L] Label Directions +N +E -W - S

  33. Copy this note! Forces at Different AnglesNewton’s 2nd Law Two people are pushing on a boat of mass 200 kg. They both push with a force of 300 N at an angle of 25̊ to each side of the boat. Sketch a FBD and calculate the net force and net acceleration of the boat. + We will use the Trigonometric Method to solve this = Net Force = the vector sum of all forces acting on the object We add the given force vectors head to tail Redraw the Force vectors so they make a triangle + 300 N 25̊ 25̊ 300 N 25̊ 25̊ 300 N 300 N 300 N Ѳ 300 N Ѳ = 180̊ - 25̊ - 25̊ Ѳ= 130̊ 25̊ 25̊ 25̊ 25̊ 25̊ uses the Z Rule

  34. Copy this note! Forces at Different AnglesNewton’s 2nd Law (cont’d) To Find the “Net Force” we will use the cosine law c2 = a2 + b2 – 2 abcos(C) or c = 25̊ 25̊ b = 300 N a = 300 N 130̊ 300 N 300 N C =130̊ 25̊ 25̊ FNET c = FNET c = FNET = FNET = 543 N FNET = maNET aNET = FNET/m aNET= 543 N/200 kg aNET= 2.72 m/s2

  35. Forces at Different Angles & Newton’s 2nd Law Copy this note! A person of mass 70 kg is sitting on a 20 kg toboggan. If two people are pulling her with two different ropes, find the person`s net force and net acceleration. The force of person 1 is 50N [E40̊N] and the force of person 2 is 60N[E25̊S] as seen from the top. N+ E+ 50 N 40̊ 25̊ 60 N

  36. Forces at Different Angles Newton’s 2nd Law Use the Trigonometric Method to solve this = Net Force Redraw the Force vectors so they make a triangle Vector addition is head to tail Copy this note! N+ E+ 50 N 40 ̊ is due to z rule C=180̊-40̊-25̊ C=115̊ 40̊ 25̊ C 60 N 25̊

  37. Copy this note! = b a= c = c = FNET = FNET = 92.9 N F= ma a = F/m a = 92.9 N/(70 kg + 20 kg) a = 1.03 m/s2 C = 115̊

  38. Copy this note! 2nd Law WorksheetGarfield #3 G: Weight = G = 5106 N [DOWN] R: Y=? ENGINE = 2.5107 N [UP] Just lifting off means no Normal Force A/S: Using Newton’s 2nd Law, consider X and Y directions XX= mX =0 no x direction movement. Y Y= mY = +G vector sum of all Y forces Show the known vectors using the correct signs given the directions mY = 2.5107 - 5106 mY = 2107 Y = (EQU I) m is unknown! This is where we must look at the given data and think about all the force relationships we have discussed in this unit. ENGINE G

  39. Copy this note! 2nd Law WorksheetGarfield #3 (cont`d) We were given G = 5106 N We know G= mg Thus: 5106= mg 5106= m (9.81) m = 5.1105kg Substitute m = 5.1105 kg into Y = (EQU I) Y = Y =39 m/s2 or Y =39 m/s2[up] ENGINE G

  40. Copy this note! 2nd Law WorksheetGarfield #4 G: Weight = G = 19600 N [DOWN] R: AIR = ? Y = 8 m/s2 MOTOR = 37600 N [UP] A/S: Using Newton’s 2nd Law, consider X and Y directions XX = mX =0 no x direction movement Y Y = mY = MOTOR +G + AIR vector sum of all Y forces Show the known vectors using the correct signs given the directions m(8)= + AIR 8m= + AIR m is unknown! Once again, this is where we must look at the given data and think about all the force relationships we have discussed in this unit… MOTOR AIR G

  41. Copy this note! 2nd Law WorksheetGarfield #4 (cont`d) We were given G = 19600 N We know G= mg Thus: 19600 = mg 19600 = m (9.81) m = 1997.96 kg Substitute m = 1997.96 kg into 8m= 18000+AIR (8)(1997.96) = 18000+AIR = 18000 + AIR AIR= -2016 N or AIR = 2016 [DOWN] MOTOR AIR G

  42. Questions

  43. Solving Newton Problems • Read the problem and identify what’s given: FN, FG, FF, FA,m, aX, aY, µ etc. Then identify what’s required and missing. • Draw a FBD of all Forces. Label conventional directions. • Apply Newton’s 2nd Law • Consider X direction • = mX = vector sum of all the x forces • Solve for all possible values: FX or aXor m, Ff, Fa etc. • Consider Y direction • = mY= vector sum of all the yforces • Solve for all possible values: FYor aY,m, or Fn, Fa etc. • Note! If there is no x or y movement in the problem, then aXor aYare 0! • This method usually yields 2 equations and 2 unknowns which we can solve (ie. Find Ff, m or u etc.). • If still values are missing, try relating them to FF=uFN, FG=mg etc. Look for clues in the problem (ie. “starting to move horizontally” then we know aX=0) • IF! there is both non zero netFxandFy forces • We find the net force with: • F2 = Fx2 + Fy2tan Ѳ = Fy/Fx(use absolute values) F=ma Often one of Fy or Fx is 0 so we don’t’ need to do this. Ѳ

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