Vectors in mechanics
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Newtonian mechanics in vector form: Elements of vectors, operations on vectors, kinematics, Laws of Newton, Momentum & Impulse
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Vectors in mechanics
- 1. Vectors in mechanics Photo: Wikimedia.org
- 2. Quantities Scalar quantities Vector quantities Vectors in mechanics 2 mass m (kg) energy E (J) power P (W) displacement 푠 (푚) velocity 푣 (푚푠−1) acceleration 푎 (푚푠−2) force 퐹 (푁) momentum 푝 (푘푔푚푠−1) torque 푇 (푁푚) • Magnitude • Magnitude • Direction • Point of application
- 3. Vectors in mechanics 3 Vector notation 퐹 arrow above or F bold face 6.0 푐푚 30 푁 6.0푐푚 ≜ 30푁 1.0푐푚 ≜ 5.0푁 Scale Head Tail
- 4. Point of application matters,… Vectors in mechanics 4 퐹 퐹
- 5. but…is often compromised… 퐹 푔 …to make life easier Vectors in mechanics 5 퐹 푔 퐹 푎푖푟 퐹 푟표푎푑 퐹 푛 퐹 푔 퐹 푟표푎푑 퐹 푛 퐹 푎푖푟 퐹 푟표푎푑 퐹 푛 퐹 푎푖푟
- 6. Basic operations with vectors Adding vectors 퐹 푛푒푤 = 퐹 표푙푑1 + 퐹 표푙푑2 Subtracting vectors 퐹 푛푒푤 = 퐹 표푙푑1 − 퐹 표푙푑2 Multiplying a vector with a scalar 퐹 푛푒푤 = 푎 ∙ 퐹 표푙푑 Resolving a vector into components 퐹 푛푒푤1 + 퐹 푛푒푤2 = 퐹 표푙푑 a = 2 퐹 ∙ 푆푖푛 휑 퐹 휑 퐹 ∙ 퐶표푠 휑 Vectors in mechanics 6
- 7. Resolve a vector along 2 working lines 퐹 푛푒푤1 퐹 표푙푑 퐹 표푙푑 = 퐹 푛푒푤1+퐹 푛푒푤2 퐹 푛푒푤2 copy copy Vectors in mechanics 7 퐹 푛푒푤1 =? 퐹 표푙푑 퐹 푛푒푤2 =? 2 given working lines
- 8. vector scalar Displacement vs. Distance Δ푠 = 푠 2 − 푠 1 Type equation here. 푠 1 Vectors in mechanics 8 푠 2 = 푠 1 + Δ푠 Do not interpret ‘s’ as the distance, because it represents the magnitude of the displacement vector 푠
- 9. Velocity vs. Speed Vectors in mechanics 9 푣푒푙표푐푖푡푦 = 푑푖푠푝푙푎푐푒푚푒푛푡 푡푖푚푒 푠푝푒푒푑 = 푑푖푠푡푎푛푐푒 푡푖푚푒 vector scalar
- 10. Average vs. Instantaneous 푣 5푠 = lim 푑푡→0 푠 5 + 푑푡 − 푠(5) 5 + 푑푡 − 5 Vectors in mechanics 10 t(s) s(m) Δ푠 Δ푡 푣푎푣푔 5푠 → 10푠 = 푠 10 − 푠 5 10 − 5 = Δ푠 Δ푡 = gradient of the secant line = difference quotient = gradient of the tangent = differential quotient = 푑푠 푑푡 = 푠′
- 11. Acceleration – notation issues Vectors in mechanics 11 푎 = Δ푣 Δ푡 = 푣 − 푢 Δ푡 Initial velocity = 푢 푎 = Δ푣 Δ푡 = 푣2 − 푣1 Δ푡 Initial velocity = 푣0 Movement along a straight line: • 푠 → 푠 표푟 푥 • 푣 → 푣 • 푎 → 푎 • Choose an origin • Choose a + direction! 푣 푠 + me
- 12. Uniformly accelerated motion - I 푎 = Δ푣 Δ푡 ⇒ Δ푣 = 푎 ∙ Δ푡 ⇒ 푣 − 푣0 = 푎 ∙ 푡 − 푡0 ⇒ 푣 = 푣0 + 푎 푡 − 푡0 Vectors in mechanics 12 푡 푣 푡표 푣표 푎 > 0 푡 푣 푡표 푣표 푎 < 0
- 13. Uniformly accelerated motion - II Vectors in mechanics 13 푡 푣 푡1 푣1 푡2 푣2 푣푎푣푔 푡 푣 푡1 푣1 푡2 푣2 푣푎푣푔 푣푎푣푔 = 푣1 + 푣2 2
- 14. Uniformly accelerated motion - III 푣푎푣푔 = Δ푠 Δ푡 ⇒ Δ푠 = 푣푎푣푔 ∙ Δ푡 Vectors in mechanics 14 푣푎푣푔 = 푣 + 푣0 2 Δ푠 = 푣 + 푣0 2 ∙ Δ푡 푣 = 푣0 + 푎 ∙ Δ푡 Δ푠 = 푣0 + 푎 ∙ Δ푡 + 푣0 2 ∙ Δ푡 = 푣0 + 1 2 푎 ∙ Δ푡 ∙ Δ푡 ⇒ 푠 = 푠0 + 푣0 ∙ 푡 − 푡0 + 1 2 푎 ∙ 푡 − 푡0 2
- 15. Uniformly accelerated motion - IV Vectors in mechanics 15 푠 = 푠0 + 푣0 ∙ 푡 − 푡0 + 1 2 푎 ∙ 푡 − 푡0 2 푡 푣 푡표 푣표 푎 > 0 푣 = 푣0 + 푎 푡 − 푡0 푡 푠 푡표 푠표 Gradient of tangent = 푣0
- 16. Uniformly accelerated motion - V Vectors in mechanics 16 푠 = 푠0 + 푣0 ∙ 푡 − 푡0 + 1 2 푎 ∙ 푡 − 푡0 2 푡 푣 푡표 푣표 푎 < 0 푣 = 푣0 + 푎 푡 − 푡0 푡 푠 푡표 푠표
- 17. Uniformly accelerated motion - VI 푣 = 푣0 + 푎 푡 − 푡0 General equations 푠 = 푠0 + 푣0 ∙ 푡 − 푡0 + 1 2 푎 ∙ 푡 − 푡0 2 Vectors in mechanics 17 푡0 = 0 Generally 푠0 = 0 Acceleration 푣0 = 0 from rest 푠 = 푠0 + 푣0 ∙ 푡 + 1 2 푎 ∙ 푡2 푣 = 푣0 + 푎푡 푠 = 1 2 푎 ∙ 푡2 푣 = 푎푡
- 18. The math behind kinematics 푠 푑푠 = 푡 푣 ∙ 푑푡 푣 푑푣 = 푡 푎 ∙ 푑푡 Integration Vectors in mechanics 18 푣 푡 = 푎 푡 = 푑푣 푑푡 푑푠 푑푡 = 푠′ 푡 = 푣′ 푡 = 푠′′ 푡 푠0 푡0 푣0 푡0 푠 푡 Displacement Velocity Acceleration 푎 푡 Differentiation Differentiation Integration
- 19. Displacement diagrams 푡 푠 푡 푠 Δ푡 Δ푡 Δ푠 Δ푠 Vectors in mechanics 19 Determine 푡 read 푡-axis Determine 푠 read 푠-axis Determine 푣 gradient of the tangent Determine 푎
- 20. Velocity diagrams 푡1 푡2 Vectors in mechanics 20 Determine 푡 read 푡-axis Determine 푠 area under the graph Determine 푣 read 푣-axis Determine 푎 gradient of the graph 푡 푣1 푡 푣 푡1 푡2 푣2 Δ푣 Δ푡
- 21. Acceleration diagrams 푎 푎 Vectors in mechanics 21 Determine 푡 read 푡-axis Determine 푠 Determine 푣 area under the graph Determine 푎 read 푎-axis 푡1 푡2 푡1 푡2
- 22. Relative velocity - I 푣퐴 = 15 푚푠−1 푣퐵 = 10 푚푠−1 B Vectors in mechanics 22 How much time does it take A to catch up on B? 푣퐴퐵 = 푣퐴 − 푣퐵 = 15 − 10 = 5푚푠−1 ⇒ Δ푡 = Δ푠퐴퐵 푣퐴퐵 = 100 5 = 20 푠 100 푚 How much time does it take A to ‘meet’ B? 푣퐴퐵 = 푣퐴 − 푣퐵 = 15 − −10 = 25푚푠−1 ⇒ Δ푡 = Δ푠퐴퐵 푣퐴퐵 = 150 25 = 6.0 푠 푣퐴 = 15 푚푠−1 푣퐵 = 10 푚푠−1 150 푚 A A B
- 23. Relative velocity - II 푣 퐵 = 10 푚푠−1 Vectors in mechanics 23 B How fast does A approach B? 푣 퐴 = 15 푚푠−1 30° 30° 푣 퐵 푣 퐴 푣 퐴퐵 푣퐴퐵푥 푣퐴퐵푦 푣퐴퐵푥 = 15퐶표푠 30° − 10 = 3.0 m푠−1 푣퐴퐵푦 = 15푆푖푛 30° = 7.5 m푠−1 푣퐴퐵 = 3.02 + 7.52 = 8.1 푚푠−1
- 24. Forces Name Symbol Value Direction Point of application Weight 퐹 푊 퐹푊 = m ∙ 푔 To the centre of the Earth Center of mass Friction 퐹 퐹 퐹퐹 = 푓 ∙ 퐹푁 ∥ to surface Contact surface Vectors in mechanics 24 Air resistance 퐹 퐴푖푟 퐹퐴 = 1 2 휌퐶퐷퐴푣2 −푣 Front of moving object Rolling resistance 퐹 푅 퐹푅 = 퐶푅 ∙ 퐹푁 −푣 Contact surface Buoyancy, Upthrust 퐹 푈 퐹푈 = 휌 ∙ 푉 ∙ 푔 Along pressure gradient 훻푝 Lowest point of object Tension 퐹푇 Reaction force ∥ the cord End of the cord Contact force 퐹푃푢푠ℎ 퐹 푁 Reaction force ⊥ surface Contact surface
- 25. Tension, Compression, Shear Tension Compression Shear Vectors in mechanics 25
- 26. Resultant, Net, Overall force Σ퐹 Vectors in mechanics 26 퐹 퐴푖푟 퐹 푊 Σ퐹 퐹 푁퐸푇푇 퐹 푅퐸푆 1st choice
- 27. Linear momentum 푝 = 푚푣 Momentum is conserved in an isolated system Σ푝 = constant Vectors in mechanics 27 푚퐴 = 5.0 푔 푣퐴 = 300 푚푠−1 푚퐵 = 1.0 푘푔 푣퐵 = 0 푣푒푛푑 ? Σ푝푏푒푓표푟푒 = Σ푝푎푓푡푒푟 ⇒ 푚퐴 ∙ 푣퐴 + 푚퐵 ∙ 푣퐵 = 푚퐴 + 푚퐵 ∙ 푣푒푛푑 5.0 ∙ 10−3 ∙ 300 + 1.0 ∙ 0 = 5.0 ∙ 10−3 + 1.0 ∙ 푣푒푛푑 ⇒ 푣푒푛푑 = 1.5 푚푠−1
- 28. Newton’s laws 1 If the resultant force on an object is zero, the object is in “translational equilibrium” Vectors in mechanics 28 For each separate object: Between objects: Σ퐹 = Δ푝 Δ푡 Σ퐹 = 푚 Δ푣 Δ푡 ⇒ Σ퐹 = 푚푎 2 Constant mass 3 Δ푝 퐴 = −Δ푝 퐵 ⇒ Δ푝 퐴 Δ푡 = Δ푝 퐵 Δ푡 ⇒ 퐹 퐴퐵 = −퐹 퐵퐴 퐹 퐵퐴 퐹 퐴퐵 A B “Force by A on B” Wikimedia.org
- 29. Newton’s laws – Example (1/5) Vectors in mechanics 29 Neglect air resistance in this problem. A caravan (820 kg) is pulled by a car. The combination accelerates with 푎 = 0.60 푚푠−2. The caravan is subject to a rolling resistance of 퐹푟표푙푙 = 1.2 ∙ 102푁 Question 1: Calculated the pulling force by the car on the caravan. The car with passengers (total 1580 kg) is subject to a rolling resistance of 1.6 ∙ 102푁 during the acceleration. Question 2: Calculate the forward force by the road on the car.
- 30. Newton’s laws – Example (2/5) A Vectors in mechanics 30 B 푚퐴 = 1580 푘푔 푚퐵 = 820 푘푔 푎 = 0.60푚푠−2 + Preparation: Make sketch of the connected car & caravan. Designate them ‘A’ & ‘B’. List the information given in the text: mass, acceleration and define a PLUS direction!
- 31. Newton’s laws – Example (3/5) A 1: Draw the relevant force vectors on the caravan Vectors in mechanics 31 B 푚퐴 = 1580 푘푔 푚퐵 = 820 푘푔 푎 = 0.60푚푠−2 + 퐹 퐴퐵 퐹 푟표푙푙,퐵 = −1.2 ∙ 102푁 Question 1 2: Apply 2nd law on B Σ퐹 퐵 = 푚퐵푎 ⇒ 퐹 퐴퐵 + 퐹 푟표푙푙,퐵 = 푚퐵푎 ⇒ 퐹 퐴퐵 + −1.2 ∙ 102 = 820 ∙ +0.60 ⇒ 퐹 퐴퐵 = 612푁 Round result to 6.1 ∙ 102푁
- 32. Newton’s laws – Example (4/5) A Vectors in mechanics 32 1: Add the relevant force vectors on the car B 푚퐴 = 1580 푘푔 푚퐵 = 820 푘푔 푎 = 0.60푚푠−2 + 퐹 퐴퐵 = 612푁 퐹 푟표푙푙,퐵 = −1.2 ∙ 102푁 Question 2 3: Apply 2nd law on A Round result to 1.7 ∙ 103푁 퐹 퐵퐴 퐹 푟표푙푙,퐴 = −1.6 ∙ 102푁 퐹 푟표푎푑,퐴 2: Apply 3rd law on A: 퐹 퐵퐴 = −퐹 퐵퐴 = −612푁 Σ퐹 퐴 = 푚퐴푎 ⇒ 퐹 푟표푎푑,퐴 + 퐹 퐵퐴 + 퐹 푟표푙푙,퐴 = 푚퐵푎 ⇒ 퐹 푟표푎푑,퐴 − 612 − 1.6 ∙ 102 = 1580 ∙ 0.60 ⇒ 퐹 푟표푎푑,퐴 = 1720푁
- 33. Newton’s laws – Example (5/5) Question 2: Alternative solution: treat A & B as combination AB AB A Vectors in mechanics 33 1: Add the relevant force vectors on the car B 푚퐴 = 1580 푘푔 푚퐵 = 820 푘푔 푎 = 0.60푚푠−2 + 퐹 퐴퐵 퐹 푟표푙푙,퐵 = −1.2 ∙ 102푁 3: Apply 2nd law on A and B together Round result to 1.7 ∙ 103푁 퐹 퐵퐴 퐹 푟표푙푙,퐴 = −1.6 ∙ 102푁 퐹 푟표푎푑,퐴 2: Regard the combination as a whole. Now 퐹 퐴퐵 and 퐹 퐵퐴 are internal forces and cancel out ! Σ퐹 퐴퐵 = 푚퐴 + 푚퐵 푎 ⇒ 퐹 푟표푎푑,퐴 + 퐹 푟표푙푙,퐴 + 퐹 푟표푙푙,퐵 = 푚퐴 + 푚퐵 푎 ⇒ ⇒ 퐹 푟표푎푑,퐴 − 1.2 ∙ 102 − 1.6 ∙ 102 = 1580 + 820 ∙ 0.60 ⇒ 퐹 푟표푎푑,퐴 = 1720푁
- 34. Impulse 푡1 퐹 푡 ∙ 푑푡 Vectors in mechanics 34 퐹 = Δ푝 Δ푡 ⇒ Δ푝 = 퐹 ∙ Δ푡 푚 ∙ Δ푣 = 퐹 ∙ Δ푡 푚 ∙ 푣 2 − 푣 1 = 푡0 Constant mass 퐹 푡1 푡 푡2 Constant force Photo: Wikimedia.org
- 35. Impulse – Example (1/2) A baseball (145 g) is thrown and flies with 155 m/s. The hitter’s club hits the ball frontally. Assume that the ball Flies back in the opposite direction. The force by the club on the ball is drawn in the diagram below. Vectors in mechanics 35 Question: Determine the speed after the hit. 퐹 푁 푡 1000 105푚푠
- 36. Impulse – Example (2/2) Tock! 2: The impulse 퐹 ∙ Δ푡 equals the area under the graph. It can be approximated by the dashed triangle 퐹 ∙ Δ푡 = 1 2 × 0.095 × −900 = −42.3푁푠 Vectors in mechanics 36 퐹 푁 푡 1000 95푚푠 3: Δ푝 = 퐹 ∙ Δ푡 ⇒ 푚 푣 2 − 푣 1 = −42.3 1: make sketch and choose PLUS direction 푣 1 = +155푚푠−1 푣 2 Δ푝 = 퐹 ∙ Δ푡 + 퐹 0.145 ∙ 푣 2 − 155 = −42.3 푣 2 = −140푚푠−1 4: Round to −1.4 ∙ 102푚푠−1 (graph area reading uncertainty)
- 37. END Disclaimer This document is meant to be apprehended through professional teacher mediation (‘live in class’) together with a physics text book, preferably on IB level. Vectors in mechanics 37