-
Forces (AQA) SummaryTable of ContentsForces (AQA)
SUmmary......................................................................................................................1
Terms used for force
systems............................................................................................2Forces
and
Motion...........................................................................................................................3
The fundamental
forces...............................................................................................................3Newtons
laws of
motion......................................................................................................................5
Newtons first
law............................................................................................................................5Superposition
of
forces......................................................................................................6
Free body
diagrams.....................................................................................................................8Newton's
second
law......................................................................................................................10
Mass..........................................................................................................................................11Inertial
mass.....................................................................................................................11
Mass and
Weight..................................................................................................................13Free
fall............................................................................................................................13
Gravitation Field
Strength....................................................................................................13Statics........................................................................................................................................15
Newton's third
law.........................................................................................................................16Connected
Bodies.................................................................................................................18
Friction...........................................................................................................................................19Friction
on an inclined
plane...........................................................................................23
Density......................................................................................................................................24Pressure..........................................................................................................................................24
Pascal's law or Pascal's
principle..........................................................................................25Buoyancy/Upthrust........................................................................................................................26
Archimedes
Principle...............................................................................................................26Hydrometer...........................................................................................................................27
Viscosity....................................................................................................................................28Stokess
Law.........................................................................................................................28
Fluid resistance:
Drag...............................................................................................................29Terminal
velocity..................................................................................................................29
Moments.............................................................................................................................................31Levers...................................................................................................................................31The
principle of
moments.....................................................................................................32
Couple.......................................................................................................................................33Centre
of
mass/gravity...................................................................................................................34Conditions
for
Equilibrium............................................................................................................36
Ladder
Problems.......................................................................................................................37Stability.....................................................................................................................................38
Page 1 of 38 G.F.Farrelly
-
Kinematics is a description of motionDynamics is an explanation
for the causes of motion.Statics is the treatment of cases in which
there is no motion
Terms used for force systems
coplanar - in the same plane, e.g. two pens (vectors) on
desk,then on inclined surface.
line of action of a force: this is shown here by the dotted
line.Note that it is NOT the same as the displacement vector r.
intersection - show two intersecting lines. If the lines
represent force vectors, we can say that the forces are coincident
at the point of intersection.
In a concurrent force system, all forces pass through a common
point.
If forces are in equilibrium, there can be no moment
(turningeffect), thus the forces are concurrent.
Page 2 of 38 G.F.Farrelly
-
Forces and MotionAristotle (384-322 BC) held that some kind of
force was necessary to keep an object moving - this seems correct
in our experience (due to friction).
Galileo (1564-1642) found that a pendulum always returned to the
same height; he extended this to a ball rolling down onto a flat
surface, then back up another slope to the original height [do this
experiment with track]. He reasoned (in a thought experiment) that
the ball would roll indefinitely along an infinitely flat, smooth
surface, contradicting Aristotle, i.e. Galileo realised that a
force was NOT necessary for the ball to be in motion.
Isaac Newton (1642-1727) was a genius. He invented the
calculusto help his study of planetary motion, developed a theory
ofgravity and forces, and produced theories on light, etc. His laws
offorce and motion underpin classical physics, but need
correction(not rejection) for objects travelling near the speed of
light andthey need reinterpretation for subatomic particles.
The fundamental forces
In fact we now know that there are four fundamental forces or
interactions:
Interaction Current theory Mediators
RelativestrengthLong-distance
behaviourRange
(m)
Strong
Quantum chromodynamics(QCD) gluons 10
38 1015
ElectromagneticQuantum electrodynamics(QED)
photons 1036 1/r2
Weak Electroweak Theory W and Z bosons 1025 1
remW , Z r 1018
Gravitational General Relativity(GR)gravitons (hypothetical) 1
1/r
2
Page 3 of 38 G.F.Farrelly
-
Note that gravitational forces are the weakest of all!
The strong force occurs between the hadrons - and is responsible
for holding the nucleus together. Itmust be stronger than the EM
force since it overcomes the mutual repulsion of the protons.
The weak force is responsible for nuclear beta decay (and other
types of decay - it involves leptons)Both nuclear forces are very
short range.
The electromagnetic force occurs between charges or magnetic
poles (moving charges) - and is responsible for holding
atoms/molecules, etc. together. Like charges/poles repel, unlikes
attract.
The gravitational force occurs between masses. At a fundamental
level it is the 'odd one out'. The gravitational force on objects
on the earth is very large since the earth is very near ! and has a
very large mass.
Page 4 of 38 G.F.Farrelly
-
Newtons laws of motionThese laws are fundamental, i.e. not
otherwise derivable: they are the foundation of Classical
Mechanics. Philosophiae Naturalis Principia Mathematica (1687)
Newtons first lawN1: A body continues at rest or in uniform
linear motion (constant linear velocity) unless acted on by a
(resultant) force.
A body acted on by no net force moves with constant velocity
(which may be zero). The tendency to keep moving/at rest is the
inertia of the body. Inertia is the reluctance of a body to change
its motion, e.g. a car seat seems to push on you as car
accelerates/moves off and the door pushes on you as you turn a
corner; you go forwards, protected by a seatbelt, as the car
brakes; you seem to continue travelling up as a lift going up slows
down quickly.
The measure of inertia is the mass.
In free space, i.e. in the absence of gravitational forces and
with no friction, N1 is confirmed. Rocket engines do not continue
firing once the rocket has 'escaped' the influences of friction and
gravity.
On earth we can use a linear air track to make friction
negligible - even less friction than on ice.This can be used to
investigate forces.
A resultant force causes acceleration. NB. Force causes a change
in motion, e.g. a change in speed, direction or both (that is a
change in velocity); forces do not cause motion itself; an object
initially moving at 100 [ms-1] in a given direction in 'empty'
space, will continue on that trajectory unless acted on by some
external force (usually gravitational or a collision force).The
acceleration produced is inversely proportional to the mass (see N2
below).
When there is no resultant/net force the body is in equilibrium:
F=0 , hence Fx=0, Fy=0, etc.NB. No resultant force does not mean
that no forces act, e.g. the weight of the floor above in a
building.
Q. What balances this downward force of the floor above?A. The
normal contact force from walls.
Page 5 of 38 G.F.Farrelly
-
Superposition of forcesForces have to be added vectorially,
e.g.
The free body diagram (see below) is:
The vector diagram adds the vectors, tail-to-head, the resultant
being from the initial point to the last. The fact that these form
a closed polygon indicates that the forces are in equilibrium, i.e.
there is no net force.
Page 6 of 38 G.F.Farrelly
T
LW
D
-
Net/Resultant force: F i=F
This is for a particle; for a rigid body we can have rotation,
despite F 0 , and for a non-rigid body we can have deformation,
i.e. change in size/shape, e.g. spring.We can resolve forces into
perpendicular (orthogonal) components - independent of each
other,e.g. resolve the weight (force) into a component down the
slope and one normal, i.e. at 90 to it:
NB. Force causes a change in motion, not motion itself.
Parallel to slope (down the slope): W sin Perpendicular (normal)
to slope (downwards): W cos
ExcursusInertial Frames of Reference: Consider being on roller
skates in an accelerating train: a force seems to act even though,
in absolute terms, there is no force. Similarly, in a lift
accelerating upwards we feel as if we are heavier, i.e. greater
gravitational force. Einsteins Principle of Equivalence was used to
equate acceleration and gravity in his general theory of
relativity.
An inertial frame of reference is one in which N1 holds (i.e. a
non-accelerating frame of reference). N1 is also known as the law
of inertia. If frame A is inertial and frame B is moving with
constant velocity wrt. frame A then B is also an inertial frame of
reference.
A rotating frame of reference is non-inertial, e.g. cycling in a
circle, shown below. Note how there does seem to be a resultant
moment, yet the cyclist does not topple over this is because it is
a non-inertial frame, so we cannot use the normal Newtonian
analysis from this diagram.
Page 7 of 38 G.F.Farrelly
-
Free body diagrams
These show all the forces that act on the body, not bythe body,
[ask yourself: what other body applies thisforce?], free of its
surroundings. The length of thearrow indicates the magnitude of the
force.
Action-reaction pairs should never be shown (since they act on
different bodies). Several such diagrams may be required (for the
different bodies).
Page 8 of 38 G.F.Farrelly
-
Force is a vector: we resolve in two perpendicular directions:
horizontal and vertical; parallel and perpendicular; radial and
tangential.
The normal force is not always equal to the weight.
Q. If the mass of an object is 2.5 [kg], what is the component
of its weight down an inclined slope of 30? (Use g=9.8 ms-2)
A. 2.5g sin 30 = 12.25=12 [N] (2.s.f.)
Q. If the object moves with constant velocity, what is the
frictional force up the slope?
A. Constant velocity means no acceleration, thus no resultant
force (N1), so frictional force =12 [N]
Q. What is the normal force?
A. No resultant force perpendicular to the slope so N=2.5g cos
30=21 [N] (2 s.f.)NB. This is less than the weight.An astronaut in
orbiting shuttle is apparently weightless since he/she has same
acceleration. as shuttle, not because he/she is outside
gravitational field.
Page 9 of 38 G.F.Farrelly
-
Newton's second lawMomentum: p = mv. Units: kg m s-1
Q. Which has more momentum, a bullet, mass 20 [g] travelling at
500 [ms-1], or a lorry, mass 2 [t], travelling at 1 [ms-1]?A. the
lorry, with momentum 2000 [kg m s-1 ] compared to the bullet's 10
[kg m s-1 ].
Q. Which would be harder (i.e. require more force) to stop?A.
the lorry
N2: The rate of change of momentum is proportional to, and in
the same direction as, the resultant external force:
F d pdt
Defining the newton [N] such that the constant of
proportionality is 1 when mass is measured in [kg] and acceleration
in [ms-2], enables us to write:
N2: (valid only in inertial frames of reference).
NB. This involves the vector sum of all the forces.
The average force: F = p t
For constant mass, if a net external force acts on a body, the
rate of change of momentum is:
F=d pdt =d ( mv )
dt=m d v
dt=m a
N.B. N2 is more general than F=ma since it includes situations
involving a change of mass.N.B. The acceleration is always in the
same direction as the resultant force.
N2 enables us to define 1 [N] as the force required to give a
mass of 1 [kg] an acceleration of 1 [ms-2].
Q. A trolley of mass 500 [g] travelling at 2 [ms-1] is brought
to a stop by a constant frictional force in 3 [s]. Calculate the
frictional force.A. v=u+at 0=2 + a(3) a=-2/3; F=ma=0.5(-2/3)=-0.33
[N], i.e. 0.33 [N] in the opposite direction to the velocity.
Page 10 of 38 G.F.Farrelly
F=d pdt
-
The direction of acceleration is the same as the direction of
the net force. Strictly, F ma so F=kma, but in S.I. units: kg,
m/s2, N, k=1 so F=maComponents obey N2: F =ma F x =max , F y =ma y
etc.
N.B. ma is not a force itself !NB. Unlike the ticker-timer
trolleys, here the motion is constrained to the 'grooves', so it is
more accurate.
Impulse: This is F dt or t. Units: [Ns]N2 implies that this must
be p.So: impulse = change in momentum. Both impulse and momentum
are vector quantities.
MassThis is the amount of matter (atoms/molecules) in a
substance.There are three major states or phases of matter: solid,
liquid, gas. Solids have the strongest 'bonds' between their
constituent particles (atoms/molecules). Liquids and gases are
fluids, being able to flow.A gas will always fill a container - it
has very weak bonds; its particles move very quickly.In all three
states, the vibration of particles increases with temperature. When
at certain critical temperatures, the substances changes state.
These are melting/freezing point and boiling point.
We can distinguish inertial mass from gravitational mass:
Inertial mass
m= Fa
. This is a quantitative measure of inertia, the resistance of
an object to change its
motion. S.I. UNIT: the kilogram (kg)
cf. definition of the Newton: 1 [N] is the net force that gives
an acceleration of 1 [ms-2] to a body of mass 1 kg.
This can be compared with gravitational mass, linked with
weight: W=mg.
Weight is the force on a mass due to gravity, so N2: w=mg, where
g is the gravitational field strength, about 9.8 [N kg-1] on the
surface of the earth; on the Moon, it is 1.60 [N kg-1] and on
Jupiter 26.0 [N kg-1].e.g. a 5 [kg] mass weighs 5 x 9.8 = 49 [N] on
the Earth's surface. The mass here is gravitational mass. All
experiments (and theory) indicate that gravitational and inertial
mass are equivalent. Thus, since F=ma, W=mg means that g should
also be the gravitational acceleration at the surface of the earth:
9.81 [ms-2] (3 s.f.).
Typical forces: (on earth) weight of apple 1 [N], weight of
person 500 [N], rocket engines 1 [MN].
Simple accelerometer: e.g. a mass hanging from the roof of q
car, making an angle with the
Page 11 of 38 G.F.Farrelly
-
vertical, tan = ag
, so at 450, a=g.
Our experience of weight is actually based on forces that
balance it, e.g. bathroom scales:
Equilibrium: N=mg
Q. A person of weight 600 [N] stands on bathroom scales in a
lift moving up at constant 2 [ms-1]. Draw a free body diagram.What
is the weight shown on scales (the normal force)? A. 600 [N]
Q. The lift now accelerates upwards at 3ms-2. What is the weight
shown on scales? A. N-mg=ma, so N-600=60(3), so N=780 [N] (using
g=10 [N/kg])
Q. The lift now 'falls' because the cable breaks. What is the
weight shown on scales? A. 0 [N] (N-mg=ma=-mg)
This is apparent 'weightlessness'. This effect is used for
training astronauts - they 'float' in a plane that temporarily
undergoes near free-fall.
Page 12 of 38 G.F.Farrelly
N=Normal force from scales (molecular-electromagnetic)
w=weight=mg
-
MassandWeightMass is a scalar and is additive: the concept of
mass characterizes the quantity of matter in a body.mass is a
measure of inertia whereas weight, a vector, is a measure of force:
e.g. a large stone is justas hard to throw on the moon (F-ma), but
easier to lift (F=mg). Weight depends on location, mass does not.
Weight w=mg1 [kg] has a weight of 9.8 [N] at the surface of the
earth; here, m is the gravitational mass and g is the gravitational
field strength, the force exerted on a unit mass in a particular
region. g varies, increasing the mass of the earth (or moon /star
etc.) and decreasing with the distance from the the object. On the
moon, g=1.62 [N kg-1], so a 1 [kg] has a weight of 1.6 [N] at the
surface of the moon.
Experiment confirms the gravitational mass to be the same as
inertial mass.N.B. The terms "mass" and "weight" and their units
are often used loosely, e.g. a ship of weight 40 000 tonnes.
Q. Explain why a beam balance really measures relative mass
rather than weight (unlike a digital balance) - consider weighing
objects using these instruments on the moon.A. On e.g. the Moon,
the beam will still be balanced for equal masses, since the weights
are equal (though 1/6 that of their weight on Earth), whereas a
digital/spring balance will show the true (reduced) weight.
Free fallProof that acceleration is independent of mass in free
fall:
F=mg=maa=g , independent of the mass.
The final speed is given from the uniform motion equations by:
v2=2gh v=2gh
Task: Show that this can also be obtained by the Principle of
Conservation of Energy.
GravitationFieldStrength
This is defined as the force per unit mass: g=Fm [Nkg-1].
At the surface of the earth, g=9.8 [Nkg-1]. Also, gravitational
acceleration uses the same symbol but different units: g=9.8
[ms-2]. Show that these are equivalent:
N2: F=W=ma=mg
Page 13 of 38 G.F.Farrelly
-
Gravitational field: F=mg
So g is both the gravitational acceleration and the
gravitational field strength; in fact, the mathematics shows this
to be true for any mass at any distance above any object.
Field theories enable an analysis of action-at-a-distance in
which objects exert forces on each other without being in direct
contact. We envisage forces as arising from a force field around a
unit object(here unit mass, i.e. 1 [kg]).
At the earths surface this is approx. 9.8 N/kg. If inertial and
gravitational mass are deemed to be the same, F=ma=mg, so a=g=9.8
m/s2.
Field lines show direction of force on a test mass: lines of
force. Since gravity is always attractive between masses, the field
lines are directed towards the mass causing the field. Greater
field strength is indicated by a greater density of field lines.
The gravitational field around a planet is radial. Over small
distances, the field may be assumed to be uniform and parallel
(perpendicular to Earths surface).
The Earth is not of uniform density, becoming denser (solid
mantle) towards the centre. The Earth, however, is not a perfect
sphere, bulging somewhat at the equator, and at the equator thereis
also a centripetal reaction, further reducing g (see below) to
about 9.78 N/kg, compared with g=9.82 N/kg at the poles.
Page 14 of 38 G.F.Farrelly
-
StaticsHere the forces are in equilibrium AND there is no motion
(in the rest frame).The total (net) force is zero.
Use vectors to resolve forces into components.
Q. hanging picture; simplify (model) as two strings holding up
the weight (of the picture) at the centre of gravity of the
net.
If the mass of the picture is 500 [g] and the strings are
symmetrically placed at an angle of 20 above the horizontal,
calculate the tension in each string.
A.
Resolving vertically: 2Tsin200 =W= 0 .5gT= 0 . 5g
2sin200=7 . 2 [N ] .
Resolving horizontally is unnecessary but would give: T cos200
=T cos200 (trivial!)
Page 15 of 38 G.F.Farrelly
T T
W
-
Newton's third lawN3: F A onB=F B on AIf body A exerts a force
on body B (an action) then body B exerts a force on body A (a
reaction) equal in magnitude but opposite in direction. NB. These
two forces act on different bodies.
Action-reaction pair: the two forces in an action-reaction pair
never act on the same body. They are always forces of the same
type, e.g. the reaction of a persons weight is NOT the normal force
(don'tcall it normal 'reaction' as is often done in
Maths-Mechanics!) at the ground but the gravitational force pulling
the earth upwards towards the person.
N.B. The 'normal' upward force of the table on the book is a
'reaction' force to the downward electromagnetic force of the book
on the table. It is NOT a reaction to the weight of the book
(different types of force). The reaction to the weight is an equal
and opposite force on the earth:
Useful diagram Johnson p49 Example 5
N3: the force you exert on the table is equal in size to the
force the table exerts on you.
Page 16 of 38 G.F.Farrelly
-
Q. A car of mass 1 tonne pulls a trailer of mass 200 kg with
constant velocity. If the frictional force on the trailer is 30 N,
calculate a) the tension in the tow-bar, b) the tractive/motive
force of the car's engine if the frictional force on the car is
also 30 [N];
A. a) Free body diagram for trailer (only) with N1 and N2:T-30=0
so T=30 [N]
b) Free body diagram for car (only) with N1 and N2:F-30-T=0 so
F-60=0 so F=60[N]
Q. If the tractive force is now increased to 100 [N], but the
frictional forces remain unchanged, calculate the acceleration of
the car.
A. Consider the whole system (car + trailer): 100-60=1200a so
a=40/1200=0.033 [ms-2]
For equilibrium: F=0 ; In general, the sum of forces on the
Earth is zero.
Page 17 of 38 G.F.Farrelly
-
ConnectedBodies
(Connected bodies are not really part of syllabus yet the
situation describe below is often used with light gates and
trolleys).
Stevinus' proof of combining resolving forces see Adams &
Allday p.52
A rope/rod of negligible mass [or in a non-gravitational
(horizontal) equilibrium situation] may be considered to transmit
undiminished the force of the agent. At equilibrium, the tension in
a rope is equal at both ends and throughout the rope. (The tension
at the top of a 'heavy' rope is greater than that at the
bottom).
Q. Two masses are connected by a light, inextensible string over
a pulley with negligible friction.Find an expression for the
acceleration of the masses and the tension in the string.
A. For m1: T= m1a; For m2: m2g-T= m2a;
Solve by adding: m2 g=m1a+ m2a=(m1+ m2)aa=m2 g
m1+ m2; and sub. for a in the first equation:
T=m1 m2 gm1+ m2
Page 18 of 38 G.F.Farrelly
-
FrictionTribology is the study of friction. Normal and
frictional forces are both contact forces, unlike gravitational
force which involves action at-a-distance. These contact forces are
electromagnetic forces acting between the atoms/molecules of the
surfaces in contact.
Friction always opposes relative motion (but cannot cause
relative motion) Friction/resistance/drag forces cannot be greater
than the forces they oppose Frictional/resistive forces convert
kinetic energy into (waste) heat, i.e. energy is dissipated.
For tyres/wheels, it is the forward friction reaction to the
friction backwards (by the engine on the wheels) that provides the
motive force. Although then, there is a sense here that friction
causes motion, it causes no relative motion since tyre and road are
stationary as the wheels roll.
Friction is a resistive force. It is a contact force between
surfaces for solids, whereas for fluids (gases/liquids) it is more
complicated, depending on complex molecular forces. Friction
between surfaces is due to the microscopic roughness of the
surfaces (hence friction is a property of TWO surfaces - one cannot
talk of the friction coefficient for steel but for steel on steel,
steel on rubber, etc.). Surface friction can be reduced by
lubrication/lubricants; these liquids keep the surfaces from making
contact yet permit motion, e.g. oil.
For solids:
The frictional force is proportional to the normal force, f N
The frictional force is independent of the area of contact
In problems/questions 'smooth' means 'negligible friction' and
'rough' means friction has to be
Page 19 of 38 G.F.Farrelly
-
considered.
Static friction: here there is no relative motion (e.g. wheels,
normally). Friction increases with the applied force up to a
maximum: f Ss N , limiting friction, where s is the coefficient of
static friction. The maximum (limiting) frictional force is the
force just before it starts to slide.
Q. An object of mass 4 [kg] is just about to slide when a force
of 2 [N] is applied horizontally. Calculate the coefficient of
static friction.A. fs=N so 2=(4g) so =0.05
Q. A force of 1.5 [N] is now applied. What is the frictional
force?A. 1.5 [N] (friction is just sufficient to oppose motion) For
a body just about to slide down an inclined plane, angle
Parallel: mg sin s N=0mg sin= s NPerp: mg cos=N
Dividing: tan =s
Kinetic friction: Frictional force, fk, and normal force, N, are
always perpendicular.f k =k N , where k is the coefficient of
kinetic friction (scalar equation).
N.B. This is only an approximate representation of a complex
phenomenon: microscopic, intermolecular forces; actual contact area
is less than surface contact area, number of contact points varies
so is not really constant (it also depends on speed). Two smooth
surfaces (sic) can actually cold weld, microscopic bonding, hence
lubricants such as oils are used.
Page 20 of 38 G.F.Farrelly
-
Q. A 500 N crate is dragged across a floor with a horizontal
rope at a constant velocity of 4 m/s. The coefficient of kinetic
friction is 0.40. Find the tension in the rope.
A. FR=kN and N=W=500 so FR=0.40 x 500 = 200 [N]
The rope is now angled at 30 above the horizontal. Find the new
tension.
Parallel: T cos 30 - FR = 0 so T cos 30 - 0.40 N = 0
Perp: T sin 30 + N - 500 = 0
Solve simultaneously (e.g. by substitution): T=188 [N]. NB the
normal force here is less than the weight.
As soon as sliding starts, the frictional force usually
decreases.Stick-slip situations: static and kinetic friction
quickly alternate, e.g. violin bow.
Page 21 of 38 G.F.Farrelly
-
Materials s kSteel on Steel 0.74 0.57
Aluminum on Steel 0.61 0.47
Copper on Steel 0.53 0.36
Rubber on Concrete 1.0 0.8
Wood on Wood 0.25-0.5 0.2
Glass on Glass 0.94 0.4
Waxed wood on Wet snow 0.14 0.1
Metal on Metal (lubricated) 0.15 0.06
Ice on Ice 0.1 0.03
Teflon on Teflon 0.04 0.04
Synovial joints in humans 0.01 0.003
Rolling friction: Magnitude of frictional force for constant
speed with rolling: , where is the coefficient of rolling friction,
~ around 0.002 for steel wheels on rails; 0.02 for rubber tyres on
concrete, so less friction on trains than lorries.
Q. A tow truck of mass 4 [t] pulls a car of mass 1.5 [t] at a
uniform velocity of 340 [km h-1].The coefficient of kinetic
friction is 0.7 for both the car and the truck wheels on the
road
surface. Calculate the tension in the tow-bar and the tractive
force of the truck.
A.
NB. The tension is constant throughout the bar; the truck and
car move with the same velocity sincerigidly connected.
Vertically: N 1=1500g=15000[N ] , N 2=4000g=40000 [N]
N2 for the truck:FF 2T =4000 a FF 2T=0 F0.7 N 2T=0
F0.7(40000)T=0 (1)
N2 for the car: TF 1=1500 aTF 1=0T0.7 N 1=0T0.7 (15000)=0T=10
500[N ]
Substituting for T in Error: Reference source not found: F28
00010500=0 F=38500[N ]
Page 22 of 38 G.F.Farrelly
T4 t1.5 t
N1N2F2F1
340 km/h
F
1.5 g 4 g
-
Friction on an inclined planeQ. a) An object of mass 5 [kg]
slides with constant velocity down a slope inclined at 20 above the
horizontal. Calculate the frictional force.b) If the object is now
pulled UP the slope with a constant velocity by a constant force F,
the frictional force remaining the same as in part (a), determine
the size of F.
A.a) .Constant velocity means equilibrium (N1):
Parallel: W sin 20 = FR (1)Perp: W cos 20 = N (2)
and W=mg=5 x 9.8 = 49 [N]. Sub. in (1) gives FR = 17 [N] (2
s.f.)
b) Constant velocity means equilibrium (N1):Parallel: F-FR -W
sin 20 =0 (1)Perp: W cos 20 = N (2)Sub. W=49 [N] and FR = 16.75...
[N] in (1) gives F = 34 [N](2 s.f.)
Page 23 of 38 G.F.Farrelly
200
200
FR
N
W
200
200
FR
N
W
F
-
DensityThis is mass per unit volume: m
V (for a homogeneous material). It depends on the material,
not
on its shape (cf. resistivity and resistance).Density may depend
on temperature and pressure. The above equation represents average
density.
Specific gravity or relative density is the ratio of the density
to that of water at 4C (1000 [kg/m3]).NB. 1 [g cm-3 ]= 1000 [kg
m-3]
Recall that V(sphere)=4/3 r3
PressurePressure is defined as p
FA
. It is a scalar quantity.
Unit: Pascal: 1 [Pa]=1 [Nm-2]; 1 [bar] = 105 [Pa]; Atmospheric
pressure: pa , 1 atmosphere: 1 [atm] = 1.013 x 105 [Pa]=1.013
[bar]=1013 [mbars]
More force gives more pressure; Less area gives more
pressure
Soa small area, e.g. drawing pin/knife/stiletto exerts more
pressure than the same force over a larger area, e.g. slipper,
blunt knife, skis, snow shoes, caterpillar tracks.
Demo: girl in stilettos compared with same girl in flat shoes -
impression in clay.Demo: weight on drawing pin above soft wood
compared to weight on wood.
e.g. A young woman of 70 kg mass wears flat shoes with a total
heel area of 10 cm2 .a) What is the pressure she exerts in N/cm2
?b) She now wears stilettos with a heel area of only 1 cm2. What is
the pressure she now exerts in N/cm2?
Pressure is more generally used for fluids. In a fluid the force
acts perpendicularly to any
cross-section, so pFA
Pressure (under gravity) increases with depth: p=gh.
Proof:
p= p2p1=FA=mg
A=Vg
A= Ahg
A= gh
Note that this is the extra pressure due to the depth
Page 24 of 38 G.F.Farrelly
-
Demo. can with spouts at different depths.
A dam wall is thicker at base.
But don't forget the large air pressure acting, e.g. on the
surface of the sea:p p gh 0 , where p0 = pressure at surface
(atmospheric pressure), h=depthThis is for a fluid of uniform
density: a good approximation for liquids, less good for gases over
large height differences.
Pressure depends on depth, not shape
Q. What is the pressure at the bottom of a tube of water of
depth 10 [m]?A. p=gh=1000x9.81x10=9.81 x 104 , i.e. around 100 kPa,
i.e. 1 atm. So air pressure will support about 10m of water or 76
cm of Hg [Derive this from p=gh].
Pascal'slaworPascal'sprinciplethe pressure exerted anywhere in a
confined fluid is transmitted equally in all directions throughout
the fluid.i.e. the pressure applied to an enclosed fluid is
transmitted undiminished to every portion of the fluid and the
walls of the container.
Liquids are virtually incompressible. Use is made of this in
hydraulics in which a small force on a small area is transmitted to
a large area to cause a large force, e.g. car brakes.
Q. If an input force of 12 [N] acts on an area of 0.01 [m2],
what is the outputforce where the area is is 0.06 [m2] ?A. pressure
is constant throughout (ignoring differences in depth), so
p=12/0.01=1 200 [Pa], so the output force is 1200 x 0.06=72
[N]
(Or simply take the ratios: F1A1
=F2A2
)
Demo: hydraulic lift
Page 25 of 38 G.F.Farrelly
-
Buoyancy/UpthrustLess dense objects will float in/on fluids that
are denser, e.g. cork on water, oil on water, hot air in cold air,
helium in air (balloons). A ship floats because its average density
is less than that of the seawater, although the density of the
steel is greater than that of the water.
Archimedes PrincipleWhen a body is completely or partially
immersed in a fluid, the fluid exerts an upward force [upthrust] on
the body equal to the weight of the fluid displaced by the body.The
reason for this is due to the difference in pressure in a fluid
between the bottom and top of an object submerged in the fluid:
Pressure at top= p1=gh1 Pressure at bottom= p2=gh2So there is
net upward pressure:
p2p1=g (h2h1) , or p=g h
So the upthrust (force) is F= p A=g h A
But W=mg and m=V, thus the weight of fluid displaced is: W=Vg= A
h g , so the weight of fluid displaced is equal to the upthrust
(Q.E.D.);
NB V is the immersed volume.The 'effective' weight of an
immersed object of density in a liquid of density ' is thus:
W=( ' )Vg
Story of Archimedes and the
crown:http://en.wikipedia.org/wiki/Archimedes#The_Golden_Crown
Expt: upthrust using perspex block with 1 cm markings, attached
to newton-meter, gradually immersed in water beaker.
Demo: as the newton-meter reduces, the reading of the whole
apparatus on a top-pan balances increases (N1: upthrust=force down
on water)
For an object to float, it must displace its own weight of
fluid.
Page 26 of 38 G.F.Farrelly
-
HydrometerA hydrometer is an instrument used to measure the
relative density (or specific gravity) of liquids; that is, the
ratio of the density of the liquid to the density of water.
A hydrometer is usually made of glass and consists of a
cylindrical stem and a bulb weighted with mercury or lead to make
it float upright. The liquid to be tested is poured into a tall
container, often a graduated cylinder, and the hydrometer is gently
lowered into the liquid until it floats freely. The point at which
the surface of the liquid touches the stem of the hydrometer is
noted. Hydrometers usually contain a scale inside the stem, so that
the specific gravity can be read directly.
Q. Why is there air in the tube?A. So that it will float
Q. Why is there a weight at the bottom?A. So that it will be
vertical when floating (very low c. of g.)
Page 27 of 38 G.F.Farrelly
-
ViscosityViscosity is internal friction in a fluid. A 'viscous
fluid' tends to cling to a solid and flow less readily, e.g. oil,
honey. There is a thin boundary layer of near-stationary fluid in
contact with the surface (i.e. moving with the surface). By
considering laminar motion between parallel plates it is clear that
there is a continuously increasing shearstrain as the motion
continues.
StokessLawThe viscous (drag) force, F, exerted on sphere, radius
r, speed v, with laminar flow (i.e. fairly slow), moving through a
fluid with coefficient of viscosity, , is given by Stokess Law
(after George Gabriel Stokes, Irish): F=6 rv (very difficult to
derive - requires Navier-Stokes equations). Note that this is of
the form F=kv. Also, F r and F .
Expt. using glycerol and magnetic ball bearings with light gates
or simply time with stopwatch for equal intervals (assume it
reaches terminal velocity a fairly short distance/time after being
dropped);try this for different radii - use Vernier callipers to
measure diameter (see my Practical worksheet)
NB. Viscosity decreases with temperature for liquids over a
reasonable range
Expt: time taken for small glass bead to sink to bottom of
measuring cylinder filled with Radox (bubble bath)/ washing-up
liquid heated to different temperatures (by immersion in beaker of
hot water). Test whether the viscosity follows an exponential decay
law: =0exp (bT ) , where b is a
constant. Do this by taking logs: ln =ln 0bT , and (cf. supra),
v1 kv
1=kv01bT , so a
plot of time t (=d/v) vs. T should be linear with a negative
gradient, if viscosity follows a negative exponential temperature
dependence.
Page 28 of 38 G.F.Farrelly
-
Fluid resistance: Drag
An object moving through a fluid experiences a
resistive/frictional force, often called drag, which increases with
the speed of the object and is always in the opposite direction
(i.e. opposes motion).Drag increases with speed because the faster
the relative motion, the more atoms/molecules of the fluid have to
be moved out of the way in a given time (greater rate of momentum
change means greater force: (N2). The resistance also depends on
the viscosity (see infra), the 'stickiness' of the liquid. Fluid
resistance is due to contact forces; here the forces between
surface and fluid and between the 'layers' of the fluid itself.
Drag increases with: speed (for relatively low speeds, the drag
is proportional to the speed: D=kv) 'shadow' area (the area
presented by the object in profile) viscosity
Terminalvelocity
At equilibrium: mg=kv so Terminal speed/velocity: v t=mgk
N.B. With air resistance, heavy objects fall more rapidly than
light ones !
Air (etc) resistance on a moving body is reduced by streamlining
the shape of the body.N.B. drag reduces both range and max. height
for projectile motion
Page 29 of 38 G.F.Farrelly
-
In fact, v=v t [1e k /m t ] (using N2 with calculus), a ge k m t
( / )
Acceleration graph
k is greater for a parachute (greater surface area - moves more
air out of the way, so more resistance) than for the human body,
thus giving a lower terminal velocity:
The terminal velocity of a skydiver in a belly to earth
free-fall position is about 55 [m/s] (195 km/h or 120 mph). This is
reached after about 10 [s].With a normal parachute, the terminal
velocity of a human is about 10 [m/s] (35 km/h or 20 mph).
Animation:
http://waowen.screaming.net/revision/force&motion/skydiver.htm
Page 30 of 38 G.F.Farrelly
-
MomentsThe turning effect of a force about an axis through O is
the moment of a force. If rotation results, itis often described as
torque.
Units: Nm; N.B. [Nm, NOT Joules because moment is a vector
9cross product) not a scalar like work].Convention used:
anticlockwise rotation = positive sense of moment.Perpendicular
distance, l, of a line of action of force F from axis is the
lever/moment arm.Moment = force x perpendicular distance of line of
action from force.M=Fl=Fr sin , where r is distance from axis and
is the angle between F and r. can be used, rather than M, depending
on whether the term torque or moment is being used.
=rF =r F sin (vector cross product)right-hand (screw) rule: curl
fingers from r to F, thumbpoints along torque vector.
Work is done by a turning force only when the object rotates. It
can be shown that the work done is:W=.
Levers
Examples: spanners, scissors, wheelbarrows.
These can reduce the force (effort) required by increasing
thedistance from the pivot/fulcrum to obtain the same moment.
Page 31 of 38 G.F.Farrelly
-
Theprincipleofmoments(UK; the lever principle): In equilibrium,
the sum of the anticlockwise moments is equal to the sum of
clockwise moments. 0.
If these moments are not equal, there is a resultant moment,
thus the system will rotate (if free to do so).
If the lines of action of the forces pass through a single
point, there is no resultant moment, thus the forces are in
equilibrium, e.g. a ladder leaning against a wall of negligible
friction (so the resultant force here is the normal force) and
standing on a rough surface (so here there is a resultant force
dueto the normal vertical force and the frictional force parallel
to the ground). The angle of the resultantforce at the ground can
be found from this diagram:
It is easier mathematically to take moments about an axis
through which several forces act as the moments of those forces
must be zero. Similarly, if there are unknown forces, take moments
about axes through which they act.
Page 32 of 38 G.F.Farrelly
-
Q. Find d.A. 6d =2x0.25+5x0.50, so d = 0.50 [m]Q. Find the force
at the fulcrum, NA. N=6+2+5=13 [N]
Q. If d1=0.1 [m] and d2=0.4 [m], find the forces at the
frictionless supports, given that the uniform beam weighs 500
[N].
A. Taking moments about the left support:
Since the forces are in equilibrium, N 1+N 2=mg N 1=500100=400[N
]
CoupleTwo collinear forces acting in opposite directions may
give tension or thrust.A pair of non-collinear forces equal in
size, opposite in direction acting on a body form a couple:
This gives pure rotation (no translational force), e.g. hands on
steering wheel. The resultant torque is Fd where d is the
perpendicular distance between the two forces (show diagram),
independent of the (perpendicular) axis of rotation.
Page 33 of 38 G.F.Farrelly
F
F
O
d
mg
N1 N2
Gd1 d2
mg d1=N 2(d1+d 2)5000.1=N 20.5 N 2=100 [N ]
-
Centre of mass/gravityThis is the mass-weighted average position
of particles: r CM=
mi r i mi
.
This must lie on any axis of symmetry.
For solid bodies we need integrals: M rcm= r dm where M=dm
If the bodies are homogeneous, then the centre of mass must lie
at the geometric centre, or along an axis of symmetry. It may lie
outside the body (e.g. the centre of a ring).
All real materials are elastic, i.e. they deform and return to
their original shape, cf. rigid body model.
Motion of CM: M vCM=i mi v i where M=imi
Similarly, for external forces: M aCM=i mi ai= F ext=d pdt ;
F 0int (by N3)When a body or a collection of particles is acted
on by external forces, the centre of mass moves as though all the
mass were concentrated at that point and as if it were acted on by
a net force equal to the sum of the external forces on the
system.It is the point through which any external force will
produce translation but no rotation.Although we often regard the
moon as orbiting the earth, with the earth stationary (i.e. using
the earth's frame of reference), in fact the earth and moon move in
orbits around their centre of mass (the barycentre).
Centroid areas/volumes are equally distributed about this point
(geometric centre): r= r in
The centroid of a triangle is at the intersection ofthe medians.
This occurs 2/3 (NOT 1/2) of thedistance from a vertex along the
median to themidpoint of the opposite side.
Page 34 of 38 G.F.Farrelly
-
Centre of mass - the point about which mass is equally
distributed: r CM= mi r i mi
Q. What is the CM if two masses of 2 [kg] and 3 [kg] are placed
4 [cm] and 6 [cm] respectively from the origin along a line?A. M
x=mi x i5 x=2(4)+3(6) x=5.2[cm ]
Centre of gravity the point about which weight is equally
distributed: r CG=W i r iW i
,
where W is the magnitude of the weight of each point. It is the
point about which the weight of the body is held to act.
If g has the same value throughout the body (i.e. a uniform
gravitational field), then the centre of gravity is identical to
the centre of mass. Since this is true for bodies on the surface of
the earth, the terms centre of gravity and centre of mass are
usually used interchangeably.The centre of gravity of a homogeneous
body is at its centroid.
At equilibrium, the centre of gravity (CG) is always directly
above/below the point of support/suspension. If supported at
several points then it must lie within the area defined by these
points.
To find the CG experimentally: suspend the body and draw a
vertical line from point of suspension, then suspend it from
another point. The intersection of the lines is the CG. More
realistically, suspension from several points determines a region
in which the CG lies.
Page 35 of 38 G.F.Farrelly
-
Conditions for EquilibriumFirst condition for equilibrium: F 0
(for external forces). This implies a closed polygon of forces
[translational equilibrium: acceleration =0]. For the common case
of three forces this results in a vector triangle (shown on the
right):
For forces not in equilibrium, the vector diagram is not
closed:
Second condition for equilibrium
0 (external moments wrt any point) [rotational equilibrium:
angular acceleration = 0]This implies that the forces lines of
action meet: concurrent forcesStatic equilibrium: rigid body at
rest, cf. rigid body in uniform translational motion (without
rotation)
Page 36 of 38 G.F.Farrelly
-
Q. A point object is acted on by a 6[N] force at 30 to the
x-axis, and an 8 [N] force along the x-axis. What third force is
required to maintain equilibrium?
A. Method a) draw a diagram and use sine/cosine ruleMethod b)
use components:x: 6 cos30 + 8 =13.19....y: 6 sin30 = 3
So resultant is (13.19 ....2+ 32)=13.5 [N] (3 s.f.) at tan1(
313.19 ...)=12.80 3 s.f.So the required third force is 13.5 [N] at
12.8 below the negative x-direction.(This is why it is so important
to draw a diagram)
Ladder Problems
Here we have to consider the weight of theladder (usually at the
midpoint - i.e. uniformladder), the normal forces at the wall and
theground, the weight of the person on the ladderand the frictional
force at the ground (andpossible at the wall):
For equilibrium, the forces are in equilibriumand the moments
are in equilibrium.
Page 37 of 38 G.F.Farrelly
-
StabilityToppling occurs when the centre of gravity falls
outside an objects base; a low centre of gravity and/or a wide base
ensure greater stability. The angle at which the object is about to
topple is called the critical angle.Neutral equilibrium: the body
can be moved to another position: the centre of gravity is at the
same height: there is no resultant moment.Stable equilibrium: the
body returns to its initial position due to the resultant moment as
the CG. moves.Unstable equilibrium: the body moves away from its
initial position rapidly, due to the resultant moment as the CG.
moves
A plumb-line will show true vertical because otherwise there
would be a moment which would move the weight until it is beneath
the point of suspension.
Page 38 of 38 G.F.Farrelly
Forces (AQA) SummaryTerms used for force systemsForces and
MotionThe fundamental forces
Newtons laws of motionNewtons first lawSuperposition of
forcesFree body diagrams
Newton's second lawMassInertial massMass and WeightFree fall
Gravitation Field Strength
Statics
Newton's third lawConnected Bodies
FrictionFriction on an inclined planeDensity
PressurePascal's law or Pascal's principle
Buoyancy/UpthrustArchimedes PrincipleHydrometer
ViscosityStokess Law
Fluid resistance: DragTerminal velocity
MomentsLeversThe principle of momentsCoupleCentre of
mass/gravityConditions for EquilibriumLadder ProblemsStability