vectors in mechanics

Vectors

in mechanics

Photo: Wikimedia.org

Quantities

Vectors in mechanics 2

Scalar quantities Vector quantities

mass m (kg)

energy E (J)

power P (W)

displacement 𝑠 (𝑚)

velocity 𝑣 (𝑚𝑠−1)

acceleration 𝑎 (𝑚𝑠−2)

force 𝐹 (𝑁)

momentum 𝑝 (𝑘𝑔𝑚𝑠−1) torque 𝑇 (𝑁𝑚)

• Magnitude • Magnitude• Direction• Point of application


Vector notation

𝐹 arrow above orF bold face

6.0 𝑐𝑚

30 𝑁

6.0𝑐𝑚 ≜ 30𝑁

1.0𝑐𝑚 ≜ 5.0𝑁 Scale

Head

Tail

Point of application matters,…


𝐹

𝐹

but…is often compromised…


𝐹𝑔

𝐹𝑎𝑖𝑟

𝐹𝑟𝑜𝑎𝑑

𝐹𝑛

𝐹𝑔


𝐹𝑛

𝐹𝑎𝑖𝑟

𝐹𝑔


𝐹𝑛

𝐹𝑎𝑖𝑟

…to make life easier

Basic operations with vectors


Adding vectors 𝐹𝑛𝑒𝑤 = 𝐹𝑜𝑙𝑑1 + 𝐹𝑜𝑙𝑑2

Subtracting vectors 𝐹𝑛𝑒𝑤 = 𝐹𝑜𝑙𝑑1 − 𝐹𝑜𝑙𝑑2

Multiplying a vector with a scalar 𝐹𝑛𝑒𝑤 = 𝑎 ∙ 𝐹𝑜𝑙𝑑

Resolving a vector into components 𝐹𝑛𝑒𝑤1 + 𝐹𝑛𝑒𝑤2 = 𝐹𝑜𝑙𝑑

a = 2

𝜑𝐹 ∙ 𝐶𝑜𝑠 𝜑

𝐹 ∙ 𝑆𝑖𝑛 𝜑 𝐹

Resolve a vector along 2 working lines


𝐹𝑛𝑒𝑤1

𝐹𝑛𝑒𝑤2

𝐹𝑜𝑙𝑑

copy

copy

𝐹𝑜𝑙𝑑 = 𝐹𝑛𝑒𝑤1+ 𝐹𝑛𝑒𝑤2

𝐹𝑛𝑒𝑤1 =?

𝐹𝑛𝑒𝑤2 =?

𝐹𝑜𝑙𝑑

2 given working lines

Displacement vs. Distance


Type equation here.

𝑠1

𝑠2 = 𝑠1 + ∆ 𝑠

∆ 𝑠 = 𝑠2 − 𝑠1

vector scalar

Do not interpret ‘s’as the distance, because it represents the magnitude of the displacement vector 𝑠

Velocity vs. Speed


𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑡𝑖𝑚𝑒𝑠𝑝𝑒𝑒𝑑 =

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

vector scalar

Average vs. Instantaneous


t(s)

s(m)

∆𝑠

∆𝑡

𝑣𝑎𝑣𝑔 5𝑠 → 10𝑠 =𝑠 10 − 𝑠 5

10 − 5=∆𝑠

∆𝑡

= gradient of the secant line= difference quotient

= gradient of the tangent= differential quotient

𝑣 5𝑠 = lim𝑑𝑡→0

𝑠 5 + 𝑑𝑡 − 𝑠(5)

5 + 𝑑𝑡 − 5=𝑑𝑠

𝑑𝑡= 𝑠′

Acceleration – notation issues


𝑎 =∆𝑣

∆𝑡=𝑣 − 𝑢

∆𝑡

𝑎 =∆𝑣

∆𝑡=𝑣2 − 𝑣1∆𝑡

Initial velocity = 𝑢

Initial velocity = 𝑣0

Movement along a straight line:• 𝑠 → 𝑠 𝑜𝑟 𝑥• 𝑣 → 𝑣• 𝑎 → 𝑎• Choose an origin• Choose a + direction!

𝑣

𝑠 +

me

𝑎 =∆𝑣

∆𝑡⇒ ∆𝑣 = 𝑎 ∙ ∆𝑡 ⇒ 𝑣 − 𝑣0 = 𝑎 ∙ 𝑡 − 𝑡0 ⇒ 𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0

Uniformly accelerated motion - I


𝑡

𝑣

𝑡𝑜

𝑣𝑜

𝑎 > 0

𝑡

𝑣

𝑡𝑜

𝑣𝑜

𝑎 < 0

Uniformly accelerated motion - II


𝑡

𝑣

𝑡1

𝑣1

𝑡2

𝑣2

𝑣𝑎𝑣𝑔

𝑡

𝑣

𝑡1

𝑣1

𝑡2

𝑣2

𝑣𝑎𝑣𝑔

𝑣𝑎𝑣𝑔 =𝑣1 + 𝑣22

𝑣𝑎𝑣𝑔 =∆𝑠

∆𝑡⇒ ∆𝑠 = 𝑣𝑎𝑣𝑔 ∙ ∆𝑡

Uniformly accelerated motion - III


𝑣𝑎𝑣𝑔 =𝑣 + 𝑣02

∆𝑠 =𝑣 + 𝑣02

∙ ∆𝑡

𝑣 = 𝑣0 + 𝑎 ∙ ∆𝑡

∆𝑠 =𝑣0 + 𝑎 ∙ ∆𝑡 + 𝑣0

2∙ ∆𝑡 = 𝑣0 + 1 2𝑎 ∙ ∆𝑡 ∙ ∆𝑡 ⇒ 𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2 𝑎 ∙ 𝑡 − 𝑡0

2

Uniformly accelerated motion - IV


𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2𝑎 ∙ 𝑡 − 𝑡02

𝑡

𝑣

𝑡𝑜

𝑣𝑜

𝑎 > 0

𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0

𝑡

𝑠

𝑡𝑜

𝑠𝑜

Gradient of tangent = 𝑣0

Uniformly accelerated motion - V


𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2𝑎 ∙ 𝑡 − 𝑡02

𝑡

𝑣

𝑡𝑜

𝑣𝑜

𝑎 < 0

𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0

𝑡

𝑠

𝑡𝑜

𝑠𝑜

Uniformly accelerated motion - VI


𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 − 𝑡0 + 1 2𝑎 ∙ 𝑡 − 𝑡02

𝑣 = 𝑣0 + 𝑎 𝑡 − 𝑡0General equations

𝑡0 = 0 Generally

𝑠0 = 0 Accelerationfrom rest𝑣0 = 0

𝑠 = 𝑠0 + 𝑣0 ∙ 𝑡 + 1 2𝑎 ∙ 𝑡2

𝑣 = 𝑣0 + 𝑎𝑡

𝑠 = 1 2𝑎 ∙ 𝑡2

𝑣 = 𝑎𝑡

The math behind kinematics


𝑎 𝑡 =𝑑𝑣

𝑑𝑡= 𝑣′ 𝑡 = 𝑠′′ 𝑡

𝑣 𝑡 =𝑑𝑠

𝑑𝑡= 𝑠′ 𝑡

𝑠0

𝑠

𝑑𝑠 = 𝑡0

𝑡

𝑣 ∙ 𝑑𝑡

𝑣0

𝑣

𝑑𝑣 = 𝑡0

𝑡

𝑎 ∙ 𝑑𝑡

𝑠 𝑡 Displacement

Velocity

Acceleration 𝑎 𝑡

Differentiation

Differentiation

Integration

Integration

Displacement diagrams


𝑡

𝑠

𝑡

𝑠

∆𝑡

∆𝑡

∆𝑠

∆𝑠

Determine 𝑡 read 𝑡-axis

Determine 𝑠 read 𝑠-axis

Determine 𝑣 gradient of the tangent

Determine 𝑎

Velocity diagrams



Determine 𝑠 area under the graph

Determine 𝑣 read 𝑣-axis

Determine 𝑎 gradient of the graph

𝑡

𝑣1

𝑡

𝑣

𝑡1 𝑡2

𝑡1 𝑡2

𝑣2

∆𝑣

∆𝑡

Acceleration diagrams


𝑎

𝑎


Determine 𝑠

Determine 𝑣 area under the graph

Determine 𝑎 read 𝑎-axis

𝑡1 𝑡2

𝑡1 𝑡2

Relative velocity - I


How much time does it take A to catch up on B?

𝑣𝐴𝐵 = 𝑣𝐴 − 𝑣𝐵 = 15 − 10 = 5𝑚𝑠−1 ⇒ ∆𝑡 = ∆𝑠𝐴𝐵

𝑣𝐴𝐵 = 1005 = 20 𝑠

𝑣𝐴 = 15 𝑚𝑠−1 𝑣𝐵 = 10 𝑚𝑠

−1

100 𝑚

How much time does it take A to ‘meet’ B?

𝑣𝐴𝐵 = 𝑣𝐴 − 𝑣𝐵 = 15 − −10 = 25𝑚𝑠−1 ⇒ ∆𝑡 = ∆𝑠𝐴𝐵𝑣𝐴𝐵 = 150

25 = 6.0 𝑠

𝑣𝐴 = 15 𝑚𝑠−1 𝑣𝐵 = 10 𝑚𝑠

−1

150 𝑚

A

A

B

B

Relative velocity - II


How fast does A approach B? 𝑣𝐴 = 15 𝑚𝑠−1

𝑣𝐵 = 10 𝑚𝑠−1

B

30°

30°

𝑣𝐵

𝑣𝐴

𝑣𝐴𝐵

𝑣𝐴𝐵𝑥

𝑣𝐴𝐵𝑦

𝑣𝐴𝐵𝑥 = 15𝐶𝑜𝑠 30° − 10 = 3.0 m𝑠−1

𝑣𝐴𝐵𝑦 = 15𝑆𝑖𝑛 30° = 7.5 m𝑠−1

𝑣𝐴𝐵 = 3.02 + 7.52 = 8.1 𝑚𝑠−1

Forces


Name Symbol Value Direction Point of application

Weight 𝐹𝑊 𝐹𝑊 = m ∙ 𝑔 To the centre of the Earth Center of mass

Friction 𝐹𝐹 𝐹𝐹 = 𝑓 ∙ 𝐹𝑁 ∥ to surface Contact surface

Air resistance 𝐹𝐴𝑖𝑟 𝐹𝐴 = 1 2𝜌𝐶𝐷𝐴𝑣2 − 𝑣 Front of moving object

Rolling resistance 𝐹𝑅 𝐹𝑅 = 𝐶𝑅 ∙ 𝐹𝑁 − 𝑣 Contact surface

Buoyancy, Upthrust 𝐹𝑈 𝐹𝑈 = 𝜌 ∙ 𝑉 ∙ 𝑔 Along pressure gradient 𝛻𝑝 Lowest point of object

Tension 𝐹𝑇 Reaction force ∥ the cord End of the cord

Contact force 𝐹𝑃𝑢𝑠ℎ 𝐹𝑁 Reaction force ⊥ surface Contact surface

Tension, Compression, Shear


Compression

Tension

Shear

Resultant, Net, Overall force


𝐹𝑊

𝐹𝐴𝑖𝑟

Σ 𝐹

𝐹𝑁𝐸𝑇𝑇

𝐹𝑅𝐸𝑆

Σ 𝐹1st choice

Linear momentum


𝑝 = 𝑚 𝑣 Momentum is conserved in an isolated system Σ 𝑝 = constant

𝑚𝐴 = 5.0 𝑔

𝑣𝐴 = 300 𝑚𝑠−1

𝑚𝐵 = 1.0 𝑘𝑔

𝑣𝐵 = 0

𝑣𝑒𝑛𝑑 ?

Σ𝑝𝑏𝑒𝑓𝑜𝑟𝑒 = Σ𝑝𝑎𝑓𝑡𝑒𝑟 ⇒ 𝑚𝐴 ∙ 𝑣𝐴 +𝑚𝐵 ∙ 𝑣𝐵 = 𝑚𝐴 +𝑚𝐵 ∙ 𝑣𝑒𝑛𝑑

5.0 ∙ 10−3 ∙ 300 + 1.0 ∙ 0 = 5.0 ∙ 10−3 + 1.0 ∙ 𝑣𝑒𝑛𝑑 ⇒ 𝑣𝑒𝑛𝑑 = 1.5 𝑚𝑠−1

Newton’s laws


For each separate object:

Between objects:

Σ 𝐹 =Δ 𝑝

Δ𝑡Σ 𝐹 = 𝑚

Δ 𝑣

Δ𝑡⇒ Σ 𝐹 = 𝑚 𝑎

Constant mass2

If the resultant force on an object is zero, the object is in “translational equilibrium”1

3 ∆ 𝑝𝐴 = −∆ 𝑝𝐵 ⇒∆ 𝑝𝐴∆𝑡=∆ 𝑝𝐵∆𝑡

⇒ 𝐹𝐴𝐵 = − 𝐹𝐵𝐴

𝐹𝐵𝐴 𝐹𝐴𝐵

A B

“Force by A on B”

Wikimedia.org

Newton’s laws – Example (1/5)


Neglect air resistance in this problem.

A caravan (820 kg) is pulled by a car. The combination accelerates with 𝑎 = 0.60 𝑚𝑠−2.The caravan is subject to a rolling resistance of 𝐹𝑟𝑜𝑙𝑙 = 1.2 ∙ 10

2𝑁

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 ∙ 102𝑁 during the acceleration.

Question 2: Calculate the forward force by the road on the car.



A

B

𝑚𝐴 = 1580 𝑘𝑔

𝑚𝐵 = 820 𝑘𝑔

𝑎 = 0.60𝑚𝑠−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!



A

B

𝑚𝐴 = 1580 𝑘𝑔

𝑚𝐵 = 820 𝑘𝑔

𝑎 = 0.60𝑚𝑠−2

+

𝐹𝐴𝐵

𝐹𝑟𝑜𝑙𝑙,𝐵 = −1.2 ∙ 102𝑁

1: Draw the relevant force vectors on the caravan

Question 1

2: Apply 2nd law on B

Σ 𝐹𝐵 = 𝑚𝐵 𝑎 ⇒ 𝐹𝐴𝐵 + 𝐹𝑟𝑜𝑙𝑙,𝐵 = 𝑚𝐵 𝑎 ⇒ 𝐹𝐴𝐵 + −1.2 ∙ 102 = 820 ∙ +0.60 ⇒ 𝐹𝐴𝐵 = 612𝑁

Round result to 6.1 ∙ 102𝑁



A

B

𝑚𝐴 = 1580 𝑘𝑔

𝑚𝐵 = 820 𝑘𝑔

𝑎 = 0.60𝑚𝑠−2

+

𝐹𝐴𝐵 = 612𝑁

𝐹𝑟𝑜𝑙𝑙,𝐵 = −1.2 ∙ 102𝑁

1: Add the relevant force vectors on the car

Question 2

3: Apply 2nd law on A


𝐹𝐵𝐴

𝐹𝑟𝑜𝑙𝑙,𝐴 = −1.6 ∙ 102𝑁

𝐹𝑟𝑜𝑎𝑑,𝐴

2: Apply 3rd law on A: 𝐹𝐵𝐴 = − 𝐹𝐵𝐴 = −612𝑁

Σ 𝐹𝐴 = 𝑚𝐴 𝑎 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 + 𝐹𝐵𝐴 + 𝐹𝑟𝑜𝑙𝑙,𝐴 = 𝑚𝐵 𝑎 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 − 612 − 1.6 ∙ 102 = 1580 ∙ 0.60 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 = 1720𝑁



A

B

𝑚𝐴 = 1580 𝑘𝑔

𝑚𝐵 = 820 𝑘𝑔

𝑎 = 0.60𝑚𝑠−2

+

𝐹𝐴𝐵

𝐹𝑟𝑜𝑙𝑙,𝐵 = −1.2 ∙ 102𝑁

1: Add the relevant force vectors on the car

Question 2: Alternative solution: treat A & B as combination AB

3: Apply 2nd law on A and B together


𝐹𝐵𝐴

𝐹𝑟𝑜𝑙𝑙,𝐴 = −1.6 ∙ 102𝑁

𝐹𝑟𝑜𝑎𝑑,𝐴

2: Regard the combination as a whole. Now 𝐹𝐴𝐵 and 𝐹𝐵𝐴 are internal forces and cancel out !

Σ 𝐹𝐴𝐵 = 𝑚𝐴 +𝑚𝐵 𝑎 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 + 𝐹𝑟𝑜𝑙𝑙,𝐴 + 𝐹𝑟𝑜𝑙𝑙,𝐵 = 𝑚𝐴 +𝑚𝐵 𝑎 ⇒

⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 − 1.2 ∙ 102 − 1.6 ∙ 102 = 1580 + 820 ∙ 0.60 ⇒ 𝐹𝑟𝑜𝑎𝑑,𝐴 = 1720𝑁

AB

Impulse


𝐹 =∆ 𝑝

∆𝑡⇒ ∆ 𝑝 = 𝐹 ∙ ∆𝑡

𝑚 ∙ ∆ 𝑣 = 𝐹 ∙ ∆𝑡

𝑚 ∙ 𝑣2 − 𝑣1 = 𝑡0

𝑡1 𝐹 𝑡 ∙ 𝑑𝑡

Constant mass

𝐹

𝑡𝑡1 𝑡2

Constant force

Photo: Wikimedia.org

Impulse – Example (1/2)


𝐹 𝑁

𝑡

1000

A baseball (145 g) is thrown and flies with 155 m/s. The hitter’s club hits the ball frontally. Assume that the ballFlies back in the opposite direction. The force by the club on the ball is drawn in the diagram below.

Question: Determine the speed after the hit.

105𝑚𝑠

Impulse – Example (2/2)


𝐹 𝑁

𝑡

1000

95𝑚𝑠

3: ∆ 𝑝 = 𝐹 ∙ ∆𝑡 ⇒ 𝑚 𝑣2 − 𝑣1 = −42.3

2: The impulse 𝐹 ∙ ∆𝑡 equals the area under the graph. It can be approximated by the dashed triangle

𝐹 ∙ ∆𝑡 =1

2× 0.095 × −900 = −42.3𝑁𝑠

1: make sketch and choose PLUS direction 𝑣1 = +155𝑚𝑠

−1

𝑣2

∆ 𝑝 = 𝐹 ∙ ∆𝑡

Tock!

+ 𝐹

0.145 ∙ 𝑣2 − 155 = −42.3

𝑣2 = −140𝑚𝑠−1

4: Round to −1.4 ∙ 102𝑚𝑠−1

(graph area reading uncertainty)

END


DisclaimerThis document is meant to be apprehended through professional teacher mediation (‘live in class’) together with a physics text book, preferably on IB level.

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