Physics IWelcome to:

I’m Dr Alex Pettitt, and I’ll be your guide!

EPhysics I:

Lecture 5: 13-10-2019

Forces and more ofNewton’s Laws

Review: Forces & circular motion A space station’s shape is a ring, 450 m in diameter.

How many revolutions (turns) per minute should it rotate in order

to simulate the Earth’s gravity, ?g = 9.81m/s2

450 m

(a) ~1.5 revs/min

(b) ~2.0 revs/min

(c) ~4.0 revs/min

(d) ~30 revs/min

A space station’s shape is a ring, 450 m in diameter.

How many revolutions (turns) per minute should it rotate in order

to simulate the Earth’s gravity, ?g = 9.81m/s2

Review: Forces & circular motion

(a) ~1.5 revs/min

(b) ~2.0 revs/min

(c) ~4.0 revs/min

(d) ~30 revs/min

450 m

a =v2

r= g = 9.81

v =2⇡r

T

= 30.1s

rev/min = 60.0/30.1 ⇠ 2

A space station’s shape is a ring, 450 m in diameter.

How many revolutions (turns) per minute should it rotate in order

to simulate the Earth’s gravity, ?g = 9.81m/s2

Review: Forces & circular motion

450 m

T =

s4⇡2r

g=

s4⇡2D

2g

If centripetal acceleration acts towards the centre, why does the person walk on the outside edge?

(a) Bad diagram!

(d) there is an extra force creating artificial gravity

(c) Newton’s laws are different on the space station

(b) There is another physical force (centrifugal) acting to balance the centripetal force

Review: Forces & circular motion

Review: Forces & circular motion If centripetal acceleration acts towards the centre, why does the person walk on the outside edge?

(a) Bad diagram!

(d) there is an extra force creating artificial gravity

(b) There is another physical force (centrifugal) acting to balance the centripetal force

(c) Newton’s laws are different on the space station

Person thinks he should walk in a straight line (Newton’s 1st)

But the space station is accelerating:

This is called the ‘centrifugal force’ but it is a fake force due to this not being an inertial frame (see lecture 4)

Review: Forces & circular motion

But to the person, the force seems to push him outwards

A force is acting

v

It pushes his path inwards (centripetal force)

This lecture:Calculate the forces from: Friction

Springs

Drag

Calculate the friction force for stationary ( ) and moving objects.

v = 0

Use Hooke’s Law to calculate the force for springs

Draw a diagram showing the forces acting on an object

different forces

The origin of forces

N0

Electromagnetism

WeakStrong

Electricity, magnetic fields

Keeps the nuclei of atoms together Governs radioactive decay

Attraction of massive bodies

Large scale

Small scale

Gravity“weakest”

“strongest”

“Wel

l” u

nder

stoo

d

P+ P+

N0

N0P+

e- νe

e-

e-

“Physics” as we know it can be derived from one of the four fundamental forces;

The origin of forces

Why is gravity the “weakest”?Because we have to pile up lots and lots to see its effect (it only attracts).

xkcd1489

Friction

.... or lack of it!

FrictionFriction is the force that exists between two objects in contact.

A surface might look smooth,

Friction

Direction of motion It always acts against the

direction of motion

but at microscopic level it is far more irregular.

It is proportional to the normal force.

FN

FN = N

FrictionFriction is the force that exists between two objects in contact.

A surface might look smooth,

Friction

Direction of motion It always acts against the

direction of motion

but at microscopic level it is far more irregular.

It is proportional to the normal force.

FN

FN = N

128 | C H A P T E R 5 Additional Applications of Newton’s Laws

5-1 FRICTION

If you shove a book that is resting on a desktop, the book will probably skid acrossthe desktop. If the desktop is long enough, the book will eventually skid to a stop.This happens because a frictional force is exerted by the desktop on the book in adirection opposite to the book’s velocity. This force, which acts on the surface of thebook in contact with the desktop, is known as a frictional force. Frictional forces are anecessary part of our lives. Without friction our ground-based transportationsystem, from walking to automobiles, could not function. Friction allows you to startwalking, and once you are already moving, friction allows you to change either yourspeed or direction. Friction allows you to start, steer, and stop a car. Friction holds anut on a screw, a nail in wood, and a knot in a piece of rope. However, as importantas friction is, it is often not desirable. Friction causeswear whenever moving pieces of machinery are incontact, and large amounts of time and money arespent trying to reduce such effects.

Friction is a complex, incompletely understoodphenomenon that arises from the attraction betweenthe molecules of one surface and the molecules on asecond surface in close contact. The nature of thisattraction is electromagnetic—the same as themolecular bonding that holds an object together.This short-ranged attractive force becomes negligi-ble at distances of only a few atomic diameters.

As shown in Figure 5-1, ordinary objects that look smooth and feel smooth arerough and bumpy at the microscopic (atomic) scale. This is the case even if the sur-faces are highly polished. When surfaces come into contact, they touch only atprominences, called asperities, shown in Figure 5-1. The normal force exerted by asurface is exerted at the tips of these asperities where the normal force per unit areais very large, large enough to flatten the tips of the asperities. If the two surfacesare pressed together more strongly, the normal force increases and so does this flat-tening, resulting in a larger microscopic contact area. Under a wide range of con-ditions the microscopic area of contact is proportional to the normal force. The fric-tional force is proportional to the microscopic contact area; so, like the microscopiccontact area, it is proportional to the normal force.

1 m

10 m

Magnified section of a polished steel surface showing surface irregularities.The irregularities are high, a height that corresponds toseveral thousand atomic diameters. (From F. P. Bowden and D. Tabor,Lubrication of Solids, Oxford University Press, 2000.)

about ! 5 " 10#7 m

F

F

f

f

F I G U R E 5 - 1 The microscopic area ofcontact between box and floor is only a smallfraction of the macroscopic area of the box’sbottom surface. The microscopic area ofcontact is proportional to the normal forceexerted between the surfaces. If the box restson its side, the macroscopic area is increased,but the force per unit area is decreased, so themicroscopic area of contact is unchanged.Whether the box is upright or on its side, thesame horizontal applied force F is required tokeep it sliding at constant speed.

The computer graphic shows gold atoms (bottom) adhering to thefine point of a nickel probe (top) that has been in contact with thegold surface. (Uzi Landman and David W. Leudtke/Georgia Institute ofTechnology.)

10microns

Polished steel surface layer. Bumps 1e-7 high (1000s of atoms)

FrictionThere are 2 kinds of friction:

Fs µsN

Static friction exists between 2 surfaces not moving relative to each other.

Range 0 to max value. The max value depends on the type of surface.

is the coefficient of static frictionµs

Kinetic friction exists between 2 surfaces that are moving relative to each other.

Fk = µkN

is the coefficient of kinetic frictionµk

v

Force

Fk

Force

Fs

Fn

N is the normal forceor: ~n

Frictionµs > µkstatic

frictionkinetic friction

Fric

tion

time

µsN

µkN

Pushing a heavy object:

It is very hard to start moving.

Once it is moving, it is easier to push.

at rest

constant speed

accelerating

increases as your force increasesFs

Fs Fkis replaced by

m1

Fs

Fapplied

N

Fg

Friction

material 1 material 2

ice ice 0.01 0.01

wood wood 0.25 0.129

leather wood 0.61 0.52

leather metal 0.61 0.25

glass glass 0.9 - 1.0 0.4

rubber concrete (wet) 0.3 0.25

rubber concrete (dry) 1 0.8

waxed ski snow 0.1 0.05

µs µk

These also depend on temperatures of the materials.

Friction Example

2 children are pulled on a sled on snow, with the sled initially at rest.

The children have a total mass of 45 kg and the sled mass is 5 kg.

The sled is pulled by a rope that makes an angle of with the horizontal.

40�

Find the frictional force exerted by the ground on the sled and the acceleration of the sled if the tension in the rope is:

(a) 100 N (b) 140 N.

✓

The coefficients of friction are and .µs = 0.2 µk = 0.15

Fg FN

FT

?

Fg,sled

Friction Example

✓

Fg FN

FT

?

Fg,sled

Rope tension components:

FT,x

= FT

cos 40

�

= (100N)(0.766) = 76.6N

FT,y = FT sin 40�

= (100N)(0.643) = 64.3N

Newton 2nd: F = mavertical

FN + FT,y + Fg = m⇥ 0 (ay = 0)

FN = mg � FT,y = (50 kg)(9.81m/s2)� 64.3N

= 426N

(a) FT = 100N

Friction Example

✓

Fg FN

FT

?

Fg,sled

Maximum static friction:

Fs,max

= µsFN

= 0.2(426N) = 85.2N

FT,x = 76.6N < F

s,max

since:

sled doesn’t move, a = 0.

Therefore friction force is:

Ffriction

= Fs

= FT,x = 76.6N

(less than maximum value)

Friction Example

✓

Fg FN

FT

?

Fg,sled

(b) FT = 140N

Rope tension components:

FT,x = 140N cos 40

�= 107N

FT,y = 140N sin 40� = 90N

Newton 2nd: F = mavertical

FN = mg � FT,y = 490N� 90N = 400N

Maximum static friction:

Fs,max

= µsFN

= 0.2(400N) = 80.0N

Fs,max

< FT,x sled will move.

frictional force will be kinetic friction

Friction Example

✓

Fg FN

FT

?

Fg,sled

Kinetic friction:

FK = µKFN

= 0.15(400N) = 60.0N

Newton 2nd: F = mahorizontal

FT,x � F

K

= ma

107N� 60N = (50 kg)a

a =107� 60

50= 0.94m/s2

Friction Example

A practical example:

A braking system in carscalled ABS (anti-lock).

If you hit the brakes very hard normally, you will likely skid/slip.

ABS uses the fact that static friction > kinetic friction by making several small brakes in turn rather than one big one, staying out of the kinematic friction regime.

µk < µs

Kinetic when slipping

Static when momentarily

at rest

Friction Example

A practical example:

A braking system in carscalled ABS (anti-lock).

µk < µs

If you hit the brakes very hard normally, you will likely skid/slip.

ABS uses the fact that static friction > kinetic friction by making several small brakes in turn rather than one big one, staying out of the kinematic friction regime.

Kinetic when slipping

Static when momentarily

at rest

Please see handout Q1

: spring constant, measure of spring stiffness, depends on material, size, etc.

k

Ideal spring:

Springs

Displacement from equilibrium is proportional to the force it produces:

equilibrium(rest point)

Hooke’s law

Fsp = �k�x

SpringsFsp = �k�x

equilibrium

Hooke’s law

SpringsFsp = �k�x

Hooke’s law

x

A helicopter rises vertically, carrying a

35 kg block on a spring scale. The spring

constant is k = 3.4 kN/m.

How much does the spring extend

(a) when the helicopter is at rest?

(b) when its accelerating upwards at 1.9 m/s2 ?

Springs Example

Springs Example

A helicopter rises vertically, carrying a

35 kg block on a spring scale. The spring

constant is k = 3.4 kN/m. v Fs

Fg

Newton 2nd: F = mavertical

m~a = ~Fs + ~Fg may = �k�y �mgy

�y = �m(g + ay)

kso extension is

down in y

(a) v = 0 (at rest) �y = � (35kg)(9.81m/s2 + 0)

3400N/m= �10cm

(b) a = 1.9m/s2 �y = � (35kg)(9.81m/s2 + 1.9m/s2)

3400N/m= �12cm

Please see handout Q2 i, ii

Drag

Not everything falling to the Earthwill fall at g.

Every object that falls through theatmosphere experiences some drag force from air resistance

Drag forces depend on several factors including: ⇢air

area

Fluid density

The object’s cross-sectional area

The object’s speed. v Go

ask

an e

ngin

eer!

DragAn object falling:

Speed is initially low, so drag is low. vFdragAs object gains speed, drag increases.

Fdrag

Eventually the drag force and gravity will be equal.

Fg

At this point , and the speed is constant, called the terminal speed.

Fnet = 0

Because drag depends on area and not on mass, a sheet of paper falls much slower than a golf ball.

DragGalileo Galilei has a famous experiment about gravity and drag.

Galileo Galilei [1564-1642]

He proposed that objects falling due to gravity will do soat the same rate, regardless of their mass.

This was hard to prove because heavier things tend to follow: large volume => large surface area => greater drag i.e. greater air resistance…

DragWithout drag they would fall at the same rate Fnet = mg = ma

DragTesting without air

Please see handout Q2 iii, iv

This lecture:Calculate the forces from: Friction

Springs

Drag

Calculate the friction force for stationary ( ) and moving objects.

v = 0

Use Hooke’s Law to calculate the force for springs

Draw a diagram showing the forces acting on an object

different forces