N ’ 3d L

Physic 231 Lecture 7

• Newton’s 3d Law:– “When a body exerts a force on another, the second body exerts

an equal oppositely directed force on the first body.”• Frictional forces:

– kinetic friction:– static friction

Nf kk μ=Nf ss μ<

• Examples.f ss μ

Example

T f d t th 5 00 k bl k h i th d iF

F

• Two forces, and , act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1=60 N and F2 = 25 N. What is the horizontal acceleration (magnitude and direction) of the block?

1F 2F

1F

600

( )

Normal force cancels the y components of

1,y yF Normal W 0+ − =Normal

1F and W

2F

F F F 30N 25N 5N= − = − =

( )o1,xF 60cos 60 N 30N= =

W

F1 60 N

F 25 N

net 1,x 2F F F 30N 25N 5N= = =

2neta F / m 5N / 5kg 1m / s= = =

W

F2 25 N

m 5 kg

θ1 60o

Block accelerates to the right.

θ1 60

a ?

conceptual question

C id t di i l t th t i l ti• Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is

) l th– a) larger than– b) identical to– c) smaller than

the downward weight W of the person.

Example: with balance scale marked in Newtons, not the usual scale, but a quite reasonable one. A 100 k t d b th l i l t St ti f• A 100 kg man stands on a bathroom scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.80 s. It travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1 5 s and comes to rest What does the scalenegative acceleration for 1.5 s and comes to rest. What does the scale register (a) before the elevator starts to move? (b) during the first 0.8 s? (c) while the elevator is traveling at constant speed? (d) during the negative acceleration? Scale reading =Normal forcenegative acceleration? g

a) : a=0; N mg : scale N 980 Newtons= = =

b) N l N

m 100 kg

N ? N mg ma : scale N (in Newtons)− = =

b) : N mg ma : scale N − = =2a v / t (1.2m / s) / 0.8s 1.5m / s= Δ Δ = =

N m(g a);scale m(g a)= + = +

vf 1.2 m/sΔt 0.8 s

N

c) : N mg : scale N 980Newtons= = =

2

vi 1.2 m/s

2 2scale (100kg)(9.8m / s 1.5m / s ) 1130Newtons= + =

N 2f id) : a (v v ) / t 1.2m / s / (1.5s) 0.8m / s= − Δ = − = −

N m(g a);scale m(g a) 900Newtons= + = + =vf 0Δt 1.5 s

No Acceleration: Static Equilibrium

All bj t t t d i Th t f bj t t• All objects are at rest and remain so. The net force on any object must vanish. I.e. on an object:

iF 0=

• Example: Three ropes are arranged so as to support a 4 kg mass as shown below. Determine the tension in each rope.

ii

T

( )0i,x 1 2

iF 0 T T cos 60= = − +

60oA1T

2T

1 2 1 20 T T / 2; T T / 2= − + =

( )0i,y 2 3

iF 0 T sin 60 T= = −

4 kg

3T i

( )02 3 3T T / sin 60 1.16T= =

T mg 39 1N;T 45 3N= = =4 kg

3T

Solve by considering forces at point A

3 2

1

T mg 39.1N;T 45.3NT 22.6N

= = ==

mg

Atwood’s machine• Consider the Atwood machine to the

right. The massless string passes over a massless and frictionless pulley. It is under tension T, which we define to be the magnitude of the tension force. By this definition it is a positive number.

• Choosing up to be positive , what is the net force on mass 1?– a) T-m1g– b) T+m1g– c) m1g-T– d) none of the above m1d) none of the above

m2

Atwood’s machine• Choosing up to be positive , what is g p p ,

the net force on mass 2?– a) m2g-T– b) T+m2gb) T+m2g– c) T-m2g– d) none of the above

m1

m2

Atwood’s machine• From Newton’s 2nd law:

gmamTamgmT

111

111

+==−

• Also

amgmT 222 =−m1

m2

gmamTg

222

222

+=

• If m2 exceeds m1, m2 goes down and 22 1, 2 gm1 goes up. If m1 exceeds m2, m1goes down and m2 goes up. In either case

• Putting it together:

222111 gmamgmam +=+

12

12

vvyy

−=Δ−=Δ

( )121211

212

222111

gmgmamamgmam

gg

−=++−=

f12 aa −= ( )21

121 mm

gmma+−=

⇐ net forcepulling m1 upand m2 down.total mass

Reading Quiz4. An action/reaction pair of forces4. An action/reaction pair of forces

A. point in the same direction.B act on the same objectB. act on the same object.C. are always long-range forces.D. act on two different objects.

Slide 4-13

Newton’s Third Law

Wh b d f h h d b d• When a body exerts a force on another, the second body exerts an equal oppositely directed force on the first body.– Note: the two forces act on different bodiesNote: the two forces act on different bodies

• Force on body 2 due to body 1:due to body 1:F21

• Force on body 1 due to body 2:F12

3d l

body 1 body 2

• 3d law:F21=- F12

Example

T k t 82 k d 48 k t di i• Two skaters, an 82 kg man and a 48 kg woman, are standing on ice. Neglect any friction between the skate blades and the ice. The woman pushes the man with a force of 45 N due east. Determine the accelerations (magnitude and direction) of the man and the womanaccelerations (magnitude and direction) of the man and the woman.

2man

x components (West is positive)45Na 0.55m / s east

82kg−= =82kg

2woman

45Na 0.94m / s west48kg

= =

East

quiz

T k t 100 k d 50 k t di i• Two skaters, an 100 kg man and a 50 kg woman, are standing on ice. Neglect any friction between the skate blades and the ice. By pushing the man, the woman is accelerated at 2 m/s2 in the direction of due west What is the corresponding acceleration of the man?west. What is the corresponding acceleration of the man?– a) 4 m/s2 due east– b) 1 m/s2 due east– c) 2 m/s2 due east– d) 1.5 m/s2 due east

woman _ on _ man man _ on _ womanF F= −

man man woman womanm a m a= −

East

2 2woman

choosing wm 50a a ( 2m / s ) 1m / s east

est to be negative (just to be contra )ry

wo aman,x woman,x

man

a a ( 2m / s ) 1m / s eastm 100

= − = − − =

Note: the direction would still be east even if you chose west to be positive

Example21N

12N

A bl k ith 5 k d d bl k ith 10 k

m1=10 kg m2= 5 kg60 N 2112F =

• A block with mass 5 kg and a second block with mass 10 kg are supported by a frictionless surface. A force of 60 N is applied to the 10 kg mass. What is the force of the 5 kg block on the 10 kg block?

tx components:d

21,x 2 x 21,x 12,xbody 2: N m a ; N N from Newtons 3 law= = −x 12,x 1 x x 12,x 1 xbody 1: F N m a ; F N m a+ = = − +

( )x 21,x 1 x 2 x 1 x 2 1 xF N m a m a m a m m a= + = + = +( )x x 2 1a F / m m= +

( )N N F / +( )12,x 21,x 2 x 2 x 2 1

12,x

N N m a m F / m mN 5kg 60N / (15kg) 20NN 20N to the left

= − = − = − += − = −=

If objects move together, the acceleration is governed by the total mass

12,xN 20N to the left=

Friction

F i ti i d th ti f bj t l th f f th• Friction impedes the motion of one object along the surfaces of another. It occurs because the surfaces of the two objects temporarily stick together via “microwelds”. The frictional force can be larger if the two surfaces are at rest with respect to each othersurfaces are at rest with respect to each other.

• Experimentally we have two cases:– kinetic friction: v

N

kf

– static friction

Nf kk μ=

FN

sf

• The coefficient of static friction generally exceeds that for kinetic

sfNf ss μ<

• The coefficient of static friction generally exceeds that for kinetic friction:

F i ti l f l th ti f f ith t

ks μμ >

• Frictional forces always oppose the motion of one surface with respect to the other.

Example with static friction

C id th fi b l ith M 105 k d M 44 1 k Wh t i• Consider the figure below, with M1=105 kg and M2=44.1 kg. What is the minimum static coefficient of friction necessary to keep the block from slipping.

2T M g=

sT f=

If M1 doesn’t move

s s 1N M gμ μ≤ =s

2 1M g M gμ≤Putting it together

s s 1

2 s 1M g M gμ≤2

s1

MM

μ≤

If this isn’t true, M1 will slip