Forces_Moments AQA Summary

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.


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


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.


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.


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.


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.


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).


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.


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.


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


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.


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


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.


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!)


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.


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.


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


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


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.


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.


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 ]


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.)


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


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


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.


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.)


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.


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


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


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.


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.


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.


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.


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.


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)


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.


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.


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

Forces_Moments AQA Summary

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