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Newton’s 3 rd Law Book page 48 - 49 14/9/2016 ©cgrahamphysics.com 2016

Transcript of Newton’s 3rd Law - WordPress.com · Newton’s 3rd Law •Valid for all of Physics •Forces...

Page 1: Newton’s 3rd Law - WordPress.com · Newton’s 3rd Law •Valid for all of Physics •Forces appear in equal and opposite pairs: •When a force acts on a body, an equal and opposite

Newton’s 3rd Law

Book page 48 - 49

14/9/2016 ©cgrahamphysics.com 2016

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Newton’s 2nd Law problem

Newton’s second law does not always work: - does not work when applied to atoms and molecules - does not work for particles moving close to the speed of light

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Newton’s 3rd Law

• Valid for all of Physics

• Forces appear in equal and opposite pairs:

• When a force acts on a body, an equal and opposite force acts on another body somewhere in the Universe

• In order to be a 3rd law pair, the following requirements must be met:

• Act for the same time

• Same line of action

• Same type

• Same magnitude

• Acts on / exerted by different bodies

• Opposite direction

14/9/2016

©cgrahamphysics.com 2016

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Confusion: what is a pair?

• Consider a book on a table • Write down the reaction – action pairs

• The reaction force to the weight is not the normal force - not the same type of force - both act on the same body - both can be drawn on the same FBD

• If N and W are not equal, they cannot be an action – reaction pair

• Action – reaction pairs are never drawn on the same FBD - using subscripts will help to avoid this trap

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𝐶𝑡𝑏

𝑊𝑏𝐸 𝑊𝑡𝐸

𝐶𝑏𝑡

𝐶𝐸𝑡

𝑊𝐸𝑏

𝑊𝐸𝑡

𝐶𝑡𝐸

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Two forces are equal and opposite if…

• x + y = 0 and A + B = 0

• X is the weight of the book 𝑊𝑏 • Y is the force the book exerts

on Earth

• Without table, the book would fall to Earth

• A is the force the table exerts on book 𝐶𝑡𝑏

• B is the force the book exerts on the table 𝐶𝑏𝑡

• This means for a 3rd law pair 𝑊𝑏𝑡 + 𝑊𝑡𝐸 + 𝑊𝐸𝑡 + 𝑊𝑡𝑏

𝑤ℎ𝑒𝑟𝑒 𝑊𝑡𝐸 = − 𝑊𝐸𝑡 • If a system of two isolated

particles that exert equal and opposite forces on each other, then the ratio of their acceleration will be in the ratio of their masses

©cgrahamphysics.com 2016

y

x

B

A

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Example • A truck is pulling two trailers starts from rest and

accelerates to a speed of 16.2𝑘𝑚ℎ−1 in 15s on a straight level section of the highway. The mass of the truck itself, T, is 5450kg, the mass of the first trailer, A, is 31500kg and the mass of the second trailer, B, is 19600kg. Assume friction is negligible. a) What magnitude of force must the truck generate in order to accelerate the entire vehicle? b) What magnitude of force must each of the trailer hitches withstand while the vehicle is accelerating?

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truck Trailer A Trailer B

Hitch between truck and trailer A Hitch between trailer A and trailer B

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Draw FBD for each part:

𝐹𝑝𝑢𝑙𝑙 𝑜𝑛 𝑇 𝐹𝐴𝑇 𝐹𝑇𝐴 𝐹𝐵𝐴 𝐹𝐴𝐵

𝑚𝐵𝑎

𝑚𝐵𝑎

𝑚𝐴𝑎 𝑚𝑇𝑎

𝐹𝐴𝑇

𝐹𝑝𝑢𝑙𝑙 𝑜𝑛 𝑇 𝐹𝑇𝐴

𝐹𝐵𝐴

𝐹𝐴𝐵

𝑚𝑇𝑎 𝑚𝐴𝑎

𝑚𝑇𝑎 + 𝐹𝐴𝑇=𝐹𝑝𝑢𝑙𝑙 𝑜𝑛 𝑇

𝑚𝐴𝑎+ 𝐹𝐵𝐴= 𝐹𝑇𝐴 𝑚𝐵𝑎 = 𝐹𝐴𝐵

31500kg 19600kg 5450kg

Given: 𝑢 = 0

v= 16.2𝑘𝑚ℎ−1 = 4.5𝑚𝑠−1

t = 15s

Truck: 𝐹𝑃𝑇 = 𝑚𝑇𝑜𝑡𝑎𝑙𝑎 = 31500 + 5450 + 19600 × 0.3 = 16965𝑁

a is the same for all

𝑎 =∆𝑣

𝑡=

4.5

15= 0.3𝑚𝑠−2

Solve for 𝐹𝐴𝑇 = 𝐹𝑃𝑇 − 𝑚𝑇𝑎 = 16965 − 5450 × 0.3 = 15330𝑁

𝐹𝐴𝑇

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• A- trailer: 𝑚𝐴𝑎+ 𝐹𝐵𝐴= 𝐹𝑇𝐴

• 𝐹𝐵𝐴 = 𝐹𝑇𝐴 − 𝑚𝐴𝑎=15330-31500 x 0.3= 5880N

• 2nd hitch 𝐹𝐵𝐴: 5.9 × 103N

• Check:

• 𝐹𝐵𝐴 should be equal to 𝐹𝐴𝐵

• 𝑚𝐵𝑎 = 𝐹𝐴𝐵

• 𝐹𝐴𝐵 = 19600 × 0.3 = 5880 = 5.9 × 103 N

14/9/2016 ©cgrahamphysics.com 2016

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Example with friction • A train is pulling two carriages. It is moving along a horizontal

track with an acceleration a. The engine has mass M and each carriage has mass m. The combined drag and frictional force on each carriage and on the engine is f. What is the thrust of the engine and the tension in each coupling? Identify any action – reaction pairs

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m m M F

𝑇21 𝑇12

𝑇23 𝑇32

f f f

ma + f = 𝑇32 ma + f + 𝑇23 = 𝑇12

Ma + f + 𝑇21 = F

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All together as one:

• (M + 2m)a + 3f = F

• The thrust in the engine is (M + 2m)a + 3f

Last carriage: ma + f = 𝑇32

2nd carriage: ma + f + 𝑇23 = 𝑇12

• ma + f +ma + f = 𝑇12

• 𝑇12= 2(f + ma)

14/9/2016 ©cgrahamphysics.com 2016

M + 2m F 3𝑓

(M + 2m)a

Check answer using first carriage

Ma + f + 𝑇21 = F F = Ma + f + 2(f + ma) = Ma + 2ma + 3f = (M + 2m)a + 3f

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5kg

Example • Two boxes, one of mass 5kg and the other of unknown mass M are

pushed along a rough floor with a force of 40N. If the frictional force is 10N and the boxes are accelerated at 2𝑚𝑠−2, what is the value of M?

• Consider the system: (5+M)a + 𝐹𝑓 = 𝐹

• 𝑀 + 5 =𝐹−𝐹𝑓

𝑎=

40−10

2= 15

• M = 15 – 5 = 10kg 14/9/2016 ©cgrahamphysics.com 2016

M

𝐹𝑓

F

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Example • A block of mass 𝑚1, lying on an inclined plane, is connected to mass

𝑚2 by a massless cord passing over a pulley.

• a) Determine the general formula for the acceleration of the system in terms of 𝑚1, 𝑚2, g and θ.

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𝑚2

𝑚1

𝑚2g

T

𝑚2a

𝑚1𝑔

N

𝑚1𝑔 cos 𝜃

𝑚1𝑔 sin 𝜃

𝐹𝑓

T

𝑚1𝑎 If 𝑚1sin 𝜃 > 𝑚2 the block will move down the plane

Incline: 𝑚1𝑎 + T + 𝐹𝑓 = 𝑚1𝑔 sin 𝜃

T = 𝑚1𝑔 sin 𝜃 − 𝑚1𝑎 − 𝐹𝑓

Hanging: T = 𝑚2g + 𝑚2a

𝑚2g + 𝑚2a = 𝑚1𝑔 sin 𝜃 − 𝑚1𝑎 − 𝐹𝑓

𝑚2a + 𝑚1𝑎 = 𝑚1𝑔 sin 𝜃 − 𝑚2g −𝐹𝑓

𝑎 =𝑚1sin 𝜃 − 𝑚2 𝑔 − 𝐹𝑓

𝑚1 + 𝑚2

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continued • b) Suppose the coefficient of

static friction between 𝑚1 and the plane is 𝜇𝑆 = 0.15 and that each block has an identical mass of 2.0kg. Determine the acceleration of the blocks if 𝜃 = 300

Remember: If 𝑚1 sin 𝜃 > 𝑚2 the block will move down the plane: 2sin 30 = 1, not >2 block will move up the plane

• 𝐹𝑠 = 𝜇𝑁 = 𝜇𝑚1𝑔𝑐𝑜𝑠𝜃

• 𝑚1𝑔𝑠𝑖𝑛 𝜃 + 𝐹𝑓 + 𝑚1𝑎 = 𝑇 and T + 𝑚2a = 𝑚2g T = 𝑚2g − 𝑚2a

• 𝑚1𝑔𝑠𝑖𝑛 𝜃 + 𝐹𝑓 + 𝑚1𝑎 = 𝑚2g − 𝑚2a

• 𝑚1𝑎 + 𝑚2a = 𝑚2g - 𝑚1𝑔𝑠𝑖𝑛 𝜃 − 𝐹𝑓 • 𝑎 𝑚1 + 𝑚2 = 𝑚2g - 𝑚1𝑔𝑠𝑖𝑛 𝜃 − 𝜇𝑚1𝑔𝑐𝑜𝑠𝜃

• 𝑎 =𝑚𝑔 1−𝑠𝑖𝑛𝜃−𝜇𝑐𝑜𝑠𝜃

2𝑚=

1

2× 10 1 − sin 30 − 0.15 cos 30 = 1.85𝑚𝑠−2

14/9/2016 ©cgrahamphysics.com 2016

𝐹𝑓

T 𝑚1𝑎

𝑚2g

T

𝑚2a

𝑚1 = 𝑚2

Up the ramp


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