Chapter 8: Newton’s Third law - smu.cadclarke/PHYS1100/documents/chapter8.pdf · 2008-10-21 ·...

SMU PHYS1100.1, fall 2008, Prof. Clarke 1

Chapter 8: Newton’s Third law

Restating Newton’s Third law from Chapter 4:

When two bodies interact, the forces exerted by each body on the other are always equal in magnitude and opposite in direction.

Less precisely: For every action there is an equal and opposite reaction.

Warning: My approach is a somewhat abbreviated and simplified version of what is in the text, yet just as complete.Both my treatment and the text’s will prepare you to solve the same problems.



A system and its environment.

Often, there will be more than one object of interest in a problem, such as these two masses being pushed by a single force.

mM

F

Each object of interest (e.g., m and M) is referred to as a system; everything else (e.g., earth, table) is referred to as the environment.Typically, systems will interact (exert forces) with each other and with their environment.



Internal and external forces

Forces exerted by one system of interest on another are internal forces.

e.g., mass m pushes on mass M: Pm on M

mass M pushes on mass m: PM on m

Forces exerted on systems of interest by agents in the environment are external forces.

e.g., the Earth exerts a gravitational force on m: wE on m

the table exerts a normal force on m: nT on m

the table exerts a frictional force on M: fk,T on M

the external force F exerted on m: FX on m

etc.

The notation: Pm on M

means the force Pexerted by m on M.



Free-body diagrams for m and M:

fk,T on m

PM on m

FX on m

wE on m

nT on m

x

y

m

a

fk,T on M Pm on M

wE on M

nT on M

x

y

Ma

Using the “FA on B” notation, it is easy assign the correct forces to each system. Only forces with “on m” are put on the FBD for m, and only forces with “on M” are put on the FBD for M.

In particular, the force driving the two masses, F, becomes FX on m in this notation (X being some unknown, external agent), and is applied only to the FBD for m!



Action/reaction pairs

The two internal forces constitute an action/reaction pair.

Newton’s 3rd Law states that every force has a corresponding reaction force, but only internal forces have their reaction forces labeled on the FBDs, and never on the same FBD.

fk,T on m

PM on m

FX on m

wE on m

nT on m

x

y

m

a

fk,T on M Pm on M

wE on M

nT on M

x

y

Ma

Note the “symmetry” between the two internal forces:

Pm on M

PM on m



External forces (e.g., nT on m) are also part of an action-reaction pair, but the reaction force (e.g., nm on T, the normal force mexerts on the table) is irrelevant to the dynamics of m or M.

Thus, external forces do not appear in action/reaction pairs on the FBDs, only the internal forces do.

Action/reaction: a bit of a misnomer?

This term seems to imply that one force, being the “reaction” to the “action”, somehow happens after the “action”. This is incorrect: one cannot exist without the other. The two forces co-exist simultaneously, no exceptions.



Identifying action/reaction pairs

Easy!!! In our notation “FA on B”, just swap the labels A and B!

Thus, if FA on B is the “action” force, the “reaction” force is FB on A

e.g.: You (Y) are pushing a box (B) across the floor. If the “action” force is PY on B, the “reaction” force is the box pushing back on you, PB on Y.

PY on B

PB on Y

PY on B

BPB on Y

YNote that PY on B and PB on Y

appear on different FBDs!!

BY BY

ended here, 16/10/08



A B

nT on A

wE on B

T

E

More examples. Two boxes, Aand B, sit on a table, T, which sits on the earth, E.

a) If the “action” force is the normal force of the table, T, on box A, (nT on A), what is the “reaction” force?

nA on T = the normal force of box A on the table, T.

b) If the “action” force is the gravitational force the earth, E, exerts on box B (wE on B, otherwise known B’s weight), what is the “reaction” force?

wB on E = the gravitational force box B exerts on the earth, E.



Alert: A very common misconception:

m

n

wT

E

m

n

w

Regardless of what you may have learned or thought you learned elsewhere, the reaction force to the weight of m is NOT the normal force from the table!

First, gravitational and normal forces are entirely different kinds of forces. Action/reaction pairs are ALWAYS the same type of force.

Second, both the weight and normal force are put on the same FBD. Action/reaction pairs are NEVER put on the same FBD.



Alert: A very common misconception:

m

nT on m

wE on m

T

E

m

nT on m

wE on m

Regardless of what you may have learned or thought you learned elsewhere, the reaction force to the weight of m is NOT the normal force from the table!

First, gravitational and normal forces are entirely different kinds of forces. Action/reaction pairs are always the same type of force.

Second, both the weight and normal force are put on the same FBD. Action/reaction pairs are NEVER put on the same FBD.

Using the notation “FA on B” will help you avoid this “trap”.



Newton’s Third Law does more than just pair up forces into action/reaction pairs. It also states that the two forces are equal in magnitude and opposite in direction.

Thus, if Pm on M and PM on m are an action/reaction pair, the two are related by:

Pm on M = –PM on m

example: If your weight is 150 pounds, this means the earth is pulling you down with a gravitational force of 150 pounds.

At the very same time, your body is pulling up on the earth with a gravitational force of the same magnitude: 150 pounds! Who knew!



Clicker question 8.1

Solve a high-schooler’s conundrum: If I try to push a box across the floor, doesn’t its reaction force on me cancel my action force on it? If so, why am I able to push boxes across the floor?

a) the box doesn’t push back with quite the same force, and so the difference in forces allows it to move.

b) The friction force on my feet is greater than the friction force on the bottom of the box, so it moves.

c) The two forces act on different bodies (me and the box), and thus don’t even get the chance to cancel out.

d) Newton’s 3rd Law doesn’t apply in this case.




Solve a high-schooler’s conundrum: If I try to push a box across the floor, doesn’t its reaction force on me cancel my action force on it? If so, why am I able to push boxes across the floor?

a) the box doesn’t push back with quite the same force, and so the difference in forces allows it to move.

b) The friction force on my feet is greater than the friction force on the bottom of the box, so it moves.

c) The two forces act on different bodies (me and the box), and thus don’t even get the chance to cancel out.

d) Newton’s 3rd Law doesn’t apply in this case.

Buffalo muffins!

may be correct, but doesn’t answer the question

Oh yes it does!



Car B is stopped for a red light. Car A, whose mass is greaterthan Car B, doesn’t see the red light and runs into the back of Car B. Which of the following statements is true?


a) B exerts a force on A, but A doesn’t exert a force on B.

b) B exerts a larger force on A than A exerts on B.

c) B exerts the same amount of force on A as A exerts on B.

d) A exerts a larger force on B than B exerts on A.

e) A exerts a force on B but B doesn’t exert a force on A.

v



Consider yourself sitting in your chair. If the action force is your weight (i.e., the Earth’s pull on you), what is the reaction force?

a) the normal force you exert against the chair (pointing down);

b) the normal force the chair exerts against you (pointing up);

c) your gravitational pull on the Earth;

d) the chair pushing against the floor.




Challenge example. a) In the system shown, mass m is pushed to

the right with a force FX. Given the masses m and M and the coefficient of kinetic friction between the table and M (µµµµk,TM), what conditions must exist on µµµµk,Mm (coefficient of kinetic friction between M and m) and FX if m is to slide across M and M is to slide across the table?

M

m

µµµµk,Mm

µµµµk,TM

fk,M on m FX on m

wE on m

nM on m

x

y

m

am

fk,T on M

wE on M

nT on M

x

y

M

fk,m on Mnm on M

aM

FX on m

notes 8.1

T

E



y

FX on m

wE on m

x

m

am

nM on m

fk,M on m

FBD for m

Two forces internal to the system m + M:

nM on m (normal force exerted by M on m)

fk,M on m (kinetic friction exerted by M on m)

The action/reaction counterparts to these forces must appear on the FBD for M. They do not appear here!!!

All other forces are external to the system m + M ⇒⇒⇒⇒ their action/ reaction counterparts do not appear on either FBD.

mam = FX – µµµµk,Mm mg

⇒⇒⇒⇒ am = – µµµµk,Mm g 3FX

m

x/ FX – fk,M = mam

y/ nM – mg = 0 ⇒⇒⇒⇒ nM = mg 1

fk,M = µµµµk,Mm nM = µµµµk,Mm mg 2



Newton’s 2nd Law ⇒⇒⇒⇒

x/ fk,m – fk,T = MaM

y/ nT – nm – Mg = 0 ⇒⇒⇒⇒ nT = nm + Mg

wE on M

x

y

M

aM

nT on M

fk,T on M

nm on M

fk,m on M

FBD for M

In the equations, I drop the “on m” and “on M” part of the notation to make them less awkward to write. They were needed to help us put the forces on the correct FBD but serve no purpose in the equations.

fk,T = µµµµk,TM nT = µµµµk,TM (nm + Mg) = µµµµk,TM (m + M)g

Thus, aM = ⇒⇒⇒⇒ aM = g µµµµk,Mm – µµµµk,TM + 1 4m

M

fk,m – fk,T

Mm

M[ ( )]

Newton’s 3rd Law ⇒⇒⇒⇒

nm = nM = mg (from 1 )

fk,m = fk,M = µµµµk,Mm mg (from 2 )



We can now answer the question asked: What are the constraints on FX and µµµµk,Mm so that m slips on M and M slips on the table, i.e., for am > aM > 0?

From 4 : aM = g µµµµk,Mm – µµµµk,TM + 1 > 0m

M[ ( )]m

M

⇒⇒⇒⇒ µµµµk,Mm > µµµµk,TM + 1 ⇒⇒⇒⇒ µµµµk,Mm > µµµµk,TM 1 + 5m

M( )m

M

M

m( )

Next, using 3 and 4 :

am > aM ⇒⇒⇒⇒ – µµµµk,Mm g > g µµµµk,Mm – µµµµk,TM + 1 FX

m

m

M[ ( )]m

M

Solving for FX, we get: FX > mg (µµµµk,Mm – µµµµk,TM) + 1 6m

M( )



Challenge example. b) Let m = 2.0 kg, M = 3.0 kg, FX = 20 N, µµµµk,TM = 0.2, and µµµµk,Mm = 0.8. Verify that both masses will slip over their respective surfaces.

From 3 , am = 2.16 ms–2 and from 4 , aM = 1.96 ms–2. Thus,

∆∆∆∆sm = am ∆∆∆∆t2; ∆∆∆∆sM = aM ∆∆∆∆t2 ⇒⇒⇒⇒ ∆∆∆∆sm – ∆∆∆∆sM = = (am – aM)∆∆∆∆t2

⇒⇒⇒⇒ ∆∆∆∆t2 = W/(am – aM) = 0.45/(0.2) = 2.25 ⇒⇒⇒⇒ ∆∆∆∆t = 1.5 s

W

212

12

12

from 6 : mg (µµµµk,Mm – µµµµk,TM) + 1 = (2.0)(9.8)(0.8–0.2)( +1)m

M( )= 19.6 N < 20 N = FX

23

from 5 : µµµµk,TM 1 + = (0.2)(1+ ) = 0.5 < 0.8 = µµµµk,Mm

M

m( ) 32

c) If M has width W = 0.45 m and m starts directly over the centre of M, how much time passes before m is pushed off M?



Acceleration Constraints

If objects are “connected together” as they move, their accelerations will be related to each other: perhaps but not necessarily equal. The relationships among the accelerations are called acceleration constraints.

So long as the rope is under tension, the accelerations of the truck and car will be equal. aC = aT

So long as the rope doesn’t stretch, the magnitudes of the accelerations of blocks A and B are equal, though their directions are different. aA = aB



Note: In the middle of page 216 where the text discusses this example, it gives the acceleration constraint as:

aAx = –aBy

which relates the x-component of aA with the y-component of aB: The constraint on the previous slide (aA = aB) relates the magnitudes of the accelerations, and thus no negative sign.



a2

a1

∆∆∆∆s1

∆∆∆∆s2

Acceleration constraints can often provide the “missing piece” of information needed to solve a problem, but sometimes they can be tricky. Consider the problem below.



Acceleration constraints can often provide the “missing piece” of information needed to solve a problem, but sometimes they can be tricky. Consider the problem below.

but ∆∆∆∆s1 = ½ a1 ∆∆∆∆t2; ∆∆∆∆s2 = ½ a2 ∆∆∆∆t2 ⇒⇒⇒⇒ a1 = 2a2.

and this is the acceleration constraint.

In time ∆∆∆∆t, the length of rope that disappears in front of m1, ∆∆∆∆s1, must be accounted for by the two extra lengths of rope, ∆∆∆∆s2, that appear above m2.

⇒⇒⇒⇒ ∆∆∆∆s1 = 2∆∆∆∆s2

a1

a2

∆∆∆∆s1

∆∆∆∆s2



Rope 1 is fixed to a wall at one end and pulled at the other end with a force of 100N (top).

Rope 2 is pulled at both ends by a force of 100 N each (bottom).

Which statement is correct?


a) The tension in rope 2 is the same as the tension in rope 1.

b) The tension in rope 2 is twice that of the tension in rope 1.

c) The tension in rope 2 is less than the tension in rope 1.



M

Thus 100 N = FS on L = FM on L = FL on M = FR on M = FM on R = FW on R, and this is equivalent to the situation where two people are pulling.

Explanation to “Clicker question 8.4”: “Break up” the rope into three

bits: left (L), middle (M), and right (R) (or as many bits as you like).

Wall

R

Sam

L

FS on L

100 N

FM on RFR on MFL on MFM on L

action/reaction pairs

Newton’s 3rd Law identifies two action/reaction pairs: FM on L = FL on M

and FR on M = FM on R.

FW on R

Nothing moves. Thus Newton’s 2nd Law requires that FS on L = FM on L, FL on M = FR on M, and FM on R = FW on R.



8.4 Ropes and Pulleys

1. A stationary rope “transmits” an action/reaction pair so that the pull at one end of the rope is equal in magnitude and opposite in direction to the pull at the other end.

If a rope of mass m is used to accelerate an object of massM, and m is not negligible, then T1 – T2 = ma, and T1 > T2.

Massless string approximation: If m = 0, we recover T1 = T2. Thus, the tension along a massless rope (string) is constant even if the rope is accelerating.

–T T

Mm T1T2T2

a a



If a mass M hangs at the end of a rope of mass m, the tension at the top of the rope is greater than at the bottom since the top has to support both m and M, while the bottom only needs to support M.

Mg

mg

M

m

Tbot

Tbot

Ttop

Ttop

Ttop = Tbot + mg; Tbot = Mg

⇒⇒⇒⇒ Ttop = Mg + mg

If we use the massless string approximation, m = 0 and Ttop = Tbot. Once again the tension in the string is constant. For a vertical system, we need to assume the string is massless even if the system is not accelerating.

a/r

a/r



2. An ideal pulley is massless and frictionless, so that its only effect on the problem is to redirect the tension in the string.

The approximations of ideal pulleys and masslessstrings allow us to write: TS on A = TS on B = T.

We’ll do this particular problem as our first example.



Example. Mass A (1.0 kg) is held in place on a frictionless table and is attached via a masslessstring that passes over an ideal pulley to mass B (0.5 kg) that dangles freely, as shown.

a) When A is released, find the acceleration of the masses and the tension in the string.

nT on A

wE on A

TS on AA

x

y

a

For mass A:

x/ T = mAa 1

y/ irrelevant

Substitute 1 into 2 : mAa – mB g = –mBa

Solve for a: a = g = 3.27 ms-2

Evaluate T from 1 : T = (1.0)(3.27) = 3.27 N

mB

mA + mB

For mass B:

x/ irrelevant

y/ T – mB g = –mBa 2

wE on B

B

x

y

a

TS on B



b) What was the tension in the string when mass A was being held in place?

wE on B

TS on B

B

x

y

Look at mass B again, but this time with a = 0:

x/ irrelevant

y/ T – mB g = 0 ⇒⇒⇒⇒ T = (0.5)(9.8) = 4.9 N

The tension in the string is less when the masses are acceler-ating than when they are being held still.



Blocks A and B are pulled with a force F across a frictionless table by massless strings. Let the tension in string 1 be T1 and the tension in string 2 be T2. Which of the following is true?


a) T2 < T1 < F

b) T2 = T1 = F

c) T2 > T1 > F

d) T2 < T1 = F

e) We need to know about the masses before we can tell.

F




In the diagram, consider the tension in the (massless) string when A is held in place (Theld)and the tension in the string when A is let go and B falls (Tfall). Which of the following statements is true?

a) Theld > Tfall b) Theld = Tfall c) Theld < Tfall



Clicker question 8.7T1

T2

T3In the system shown, the string is massless and all pulleys are ideal.

T1 is the tension in the string between m1 and the pulley on the table.

T2 is the tension in the string between the pulley on the table and m2.

T3 is the tension in the string between the ceiling and m2.

Which tension is the greatest?

a) T1 b) T2 c) T3 d) They are all the same.



Clicker question 8.7T1

T2

T3In the system shown, the string is massless and all pulleys are ideal.

T1 is the tension in the string between m1 and the pulley on the table.

T2 is the tension in the string between the pulley on the table and m2.

T3 is the tension in the string between the ceiling and m2.

Which tension is the greatest?

a) T1 b) T2 c) T3 d) They are all the same.

ended here, 21/10/08



a2

a1

Example: Find the tension in the

string and accelerations in terms of m1, m2, and g. Assume a massless string, ideal pulleys, and a frictionless table.

m1

m1g

nT on 1

T x

y

a1

m1: x/ T = m1a1

y/ irrelevant



a2

a1

m2

m2g

T

x

y

a2

T

m2: x/ no forces

y/ 2T – m2g = –m2a2



m1

m1g

nT on 1

T x

y

a1

m1: x/ T = m1a1

y/ irrelevant



a2

a1

m2

m2g

T

x

y

a2

T

m2: x/ no forces

y/ 2T – m2g = –m2a2

⇒⇒⇒⇒ 2m1a1 – m2g = –m2a2

acceleration constraint from slide 26: a1 = 2a2

⇒⇒⇒⇒ 4m1a2 – m2g = –m2a2 ⇒⇒⇒⇒ a2(m2 + 4m1) = m2g

⇒⇒⇒⇒ a2 = ⇒⇒⇒⇒ a1 = ⇒⇒⇒⇒ T = m2g

m2 + 4m1

2m2g

m2 + 4m1

2m1m2g

m2 + 4m1



m1

m1g

nT on 1

T x

y

a1

m1: x/ T = m1a1

y/ irrelevant



∆∆∆∆s1

∆∆∆∆sP

acceleration constraint: ∆∆∆∆sP = ∆∆∆∆s1

⇒⇒⇒⇒ aP = a1 = – 1 = 2.45 ms-21

2

m2

m1( ) g

2

1

2

Example: Atwood’s machine

m1 m2

F

a) If m1 = 1.0 kg and m2 = 1.5 kg, with what minimumforce F must the ideal pulley be pulled so that both masses lift off the ground?

b) What is the acceleration of the pulley?

T T

F

P aP

F – 2T = 0 (massless pulley)

⇒⇒⇒⇒ F = 2 m2g = 29.4 N

m1

m1g

T

a1

T – m1g = m1a1

⇒⇒⇒⇒ a1 = – g = – 1 gT

m1

m2

m1( )

P

m1

m2

F

P

T – m2g = 0 ⇒⇒⇒⇒ T = m2g

T

m2

m2g

a2 = 0nG on 2 = 0

Chapter 8: Newton’s Third law - smu.cadclarke/PHYS1100/documents/chapter8.pdf · 2008-10-21 ·...

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Transcript of Chapter 8: Newton’s Third law - smu.cadclarke/PHYS1100/documents/chapter8.pdf · 2008-10-21 ·...