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