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The Idiot’s Guide to Mechanics
1. Draw a good clear diagram. (Even if you think it’s not necessary!)
2. Put masses inside the shape.
3. Make sure all units on the diagram are standard units. m, Kg, secs, Watts, ms -1
etc.
4. Label all the forces on your diagram. Indicate directions of velocity and
acceleration. Only attach forces to masses.
5. Label the direction you are using as positive.
6. Resolve forces parallel and perpendicular to any accelerations or motion. If there
is no motion, resolve parallel and perpendicular to the plane.
7. If the forces or accelerations are not constant then you will most likely be using
calculus. There is always one of these on the paper!!
8. Write down all the equations that apply to the diagram. Write them in their
general form first then substitute values . When in doubt use
F = ma.
9. Remember that Distance = speed x time only works for constant speed.
10.Simplify any equations you write as you create them. If you have values for sin
i.e. sin-1 = 4/5 then use those values. Keep the equations as simple as possible.
11.If you have written down all the equations, you are now left with a set of
simultaneous equations to solve.
12.Physicists beware!! Make sure you use the correct value of g as defined on the
front of the paper 9.8 ms-2.
13.In exams, start every question on the top left-hand side of a double page spread.
Simple Harmonic Motion
1. Draw a diagram.
2. Label forces and displacement from centre of motion.
3. Remember to label everything in the x direction as positive. This usually means that the force is negative.
If you are given the motion as S.H.M.
1. The acceleration is proportional to the displacement but in the opposite direction. This is what gives it its negative
value. Therefore acceleration =
 n2 x or   2 x .  and n are interchangeable.
2. The distance from the middle to the end is the amplitude a .
3. The length of the path is 4a.
4. The time of one oscillation is from one end to the other and back again is
2

or
2
n
5. The speed is zero at the ends and maximum at the centre.
6. You may assume all the equations of S.H.M. and use them.
You must learn these!
x   n 2 x or   2 x
7.
v 2  n2 (a 2  x 2 )
v  na only at the centre.
8. If you are being asked about time from one position to another use associated circular motion. This means drawing
a circle lined up with the path so that the diameter and the path are equal. You can then use Time =
that you are working in radians. The value of



Remember
is found by simple trig. The hypotenuse is the radius of the circle
and the amplitude.
If you have to prove S.H.M.
Make sure all your forces are clearly on your diagram.
1. Calculate the position of equilibrium. This usually involves Hooke’s Law T 
. x
a
2. The equilibrium position is the centre of any motion.
3. You must consider a general point that is x from the equilibrium point.
4. Use F = ma in the x direction. This will involve tension(s) and possibly mg. Remember to work away from the
equilibrium point. This will produce the negative required.
5. You will end up with an expression of the form acceleration =  x or  n x This is the condition for S.H.M.
2
2
6. You can now use the equations as before. Remember that you now have a value for n.
Projectiles
1. You must consider the motion in two directions i.e. vertically and
horizontally. The two sets of data should be kept separate on your page.
2. Resolve the velocity into its two component parts ie u.cos
and u sin
3. Set up the values for the equations of motion horizontally and vertically.
4. Keep both sets of values separate.
5. Horizontally:
Acceleration is zero.
6. Vertically:
Displacement s = ucos.t .
Gravity is operating usually -9.8
Velocity changes from
+ on the way up, to  on the way down. Any displacements below the original
starting level are negative.
7. All the usual equations of motion can be used.
8. Remember Initial Velocity is u sin
and gravity is -9.8
9. When a particle hits the level ground vertical displacement is zero. (Not velocity)
10.The vertical velocity at the greatest height is zero ( The horizontal velocity is still
u cos )
11.To calculate the velocity (or direction of travel) at any point combine both vertical
and horizontal velocities.
12.It is vital that you keep horizontal and vertical parts separate in all calculations.
The only common part is Time.
13.Learn how to obtain the equation of the path. It is often asked.
Elastic Impact
3u
2u
Positive
e=½
2m
v
4m
z
Draw a good clear diagram with masses in the particles and directions of motion both
before and after clearly shown.
Conservation of Momentum:
Total momentum before impact = Total momentum after impact
Remember:
directions are important.
State which direction you are going to use as positive. Draw it on your diagram.
Momentum is
mass x velocity .
You must work out each object separately.
Using this and Newton’s Law you can calculate everything.
Impulse: This is the
change in momentum of a particular particle.
An object loses momentum during impact and that momentum is transferred to the
other object. The amount transferred is the impulse.
There are only two basic equations: One involves speeds and the other velocities.
Newton’s experimental law:
e x approach speed = separation speed.
Where e is the coefficient of restitution. NB.
e1
Take care over these. Think about real situations. If two objects are hitting from
opposite directions the approach speed is greater. I.e. added.
Afterwards, the back object cannot be going faster than the one in front. Why?
Energy is not conserved during impacts!
Physicists take note.
Centres of Mass
1. Draw a good clear accurate diagram.
2. Label all relevant measurements
3. Mark the axis or axes you will be working from.
4. Break the shape down into its component parts. I.e. The elements.
5. State the weight per unit area ( possibly length or volume ) usually w
6. Draw up a table showing the component parts
7. Calculate the area (length or volume ) and hence weight of each part.
8. Calculate the total weight of the object.
9. Work out the distances of the centres of mass of each component from one axis.
10.Check distances are being measured from the axis and not part of the shape!
11.Let whole shape have centre of mass as x .
12.Add weight x centre of mass for each element.
13.This equals the whole weight x x
14.Repeat 9. on for the other axis.
15.If the shape is suspended from a point then the vertical goes from that point
through the centre of mass.
16.If an angle with the vertical is required it usually involves a simple right angled
case of trig. Check measurements on your diagram.
Calculus types (M2)
Essentially the same process. Work out the elements. Centre of mass of each element.
Find  of these elements x centres of mass. As x0 it becomes integration. There
are very few of these. Learn them, practice writing them out.
Make sure you know the basic equations for volumes of the solids.
Volumes
Cylinder = r2h
cone = 1/3 r2h
Sphere = 4/3 r3
Curved surface areas
Cylinder = 2rh
cone = rl
Sphere = 4r2
Friction
1. Draw a good clear diagram
2. Make sure the normal reaction is labelled
3. Decide which way the object is about to slip (Is it about to slip??)
4. Put friction in the opposite direction to the motion.
5. Use Fr for friction. When about to slip
Fr = R State this!
6. Resolve forces at right angles to the surface. Look out for forces that are at an
angle to the surface, particularly horizontal forces when on slopes. These are going
to have a component in this equation. They change the value of R
7. Resolve the system parallel to the surface. Use F = ma . in the direction of the
acceleration.
8. If there is no acceleration then the forces resolve to zero.
Therefore forces in one direction equal those in the opposite direction.
(since F = 0 x a) Remember the weight component.
9. If the motion continues then use the acceleration found earlier.
Work Done against friction can be done two ways.
1. Calculate the frictional force and multiply this by the distance travelled against
friction. I.e. Work Done = Force x Distance
2. Calculate the total energy before the motion and compare it with the total energy
after the motion. Since Work done = Change in energy. Therefore the
difference in total energy is accounted for by the work done against friction..
Work Energy and Power
There are three types of energy we are concerned with:
Potential energy`
PE = mgh
KE = 1/2mv2
Kinetic energy
EPE =
Elastic Potential energy
and
. e2
2a
(M2)
where e is the extension and a the natural length.
Work Done is the change in energy
Work done by a force is Force x Distance in the direction of the force.
Power is the rate of doing work. This means it is energy change per second.
The equation involving power is
Power = Driving force x velocity
Important Points
1. Define the point of zero PE
2. Mark the point on your diagram.
3. Energy is conserved by a body therefore the total energy is always the same.
4. An object only loses energy to external forces i.e. friction. The loss is the work done against friction. (During
collisions KE is usually lost.)
5. Always work out the energies of bodies separately. Only combine the masses if they have actually stuck together.
Power
This is not a force. On diagrams indicate the driving (or tractive) force as Df
Use
Power = Df x Velocity to calculate the value of the driving force.
If the velocity is changing (i.e. there is acceleration) then the value of D f is different for every velocity.
If there is acceleration use F = ma You must take into account all the forces (or components) in the direction of the
acceleration.
Power questions often present unusual units. Make sure you are in Watts, kgs, ms -1 . To convert kph to ms-1 multiply by
5
/18.
On slopes you must include the weight component usually mg sin 
Calculus Type Questions
These are on every paper.
Be wary of straight forward equation of motion questions on M2!
The basics are easy:
Velocity is metres per second
This means change in displacement divided by change in time
i.e.
v
ds
dt
To get velocity Differentiate displacement s.
Similarly acceleration is change of velocity compared to time
i.e.
a
dv To get acceleration Differentiate velocity.
dt
To reverse the processes you use the same basic principal.
ds
dt
v. dt  ds
If
then by separating the variables
v
But this cannot be left like this and you must
introduce the integration signs. Therefore:
 v. dt   1. ds
a
dv
dt
Similarly
becomes a .dt = dv
 a. dt   1. dv
Remember to integrate the correct variable.
M2
Special Cases.
dv
a
dt
If
Then
a
dv ds
ds dt
Therefore
a
dv
.v
ds
but
or
 v. dv   a. ds
This gives
ds
v
dt
a V.
dv
ds
Make sure you have any v with dv
Connected Particles
These are usually of the type involving pulleys but include cars and trailers.
Draw a clear diagram labelling all forces that are external to the system.
Label the tension(s) close to the particles. These are internal forces.
If there is a slope involved remember to label the weight of the object.
Decide on the direction of the acceleration and label it for each particle. Check that your diagram is consistent. (The
objects are going the same way)
You can create three equations all using the formula F = ma
A whole system one which will not have any tensions is often the simplest.
Consider all the forces in the direction of the acceleration. Subtract any forces or their components that are in the
opposite direction. This equals the TOTAL mass x acceleration.
Having calculated the acceleration you can now look at individual particles using the tensions.
The weight will always play a part on slopes. (Usually in the form of mgsin )
R
R
mg
m
T
2m
2mg T
The equations from this diagram are:
The whole system:
2mg – mg.sin - R = 3ma. (Remember to use the whole mass)
The 2m mass:
2mg – T = 2m.a
The m mass:
T – mg. sin - R = m.a
Combining these two equations will give the first one.
Motion in a circle
Horizontal Circles.
Always resolve Vertically first and resolve the forces to zero. I.e. Up = Down
The force towards the centre is usually either a tension, friction or a component of the normal reaction.
If the speed is v then the force equates to
the acceleration towards the centre is
hence
mv 2
v2 r
r
If the velocity is angular (often from revs per min) then v = r
Conical pendulum
The tension(s) in the string do two things. They support the mass and they provide the force
towards the centre. You must resolve in both directions.(Up and towards the centre)
Use basic Trig or Pythagoras to find any unknown lengths or
h
angles. Do this first.

T
r
m
m
m
m
mg
Vertically:
There is no acceleration

Tcos = mg
Horizontally: There is an acceleration towards the centre of
v2
r
 Force towards the centre ie T sin  gives
mv 2
Tsin =
r
If there is a second string attached to the mass underneath, its tension must be included in the
vertical and horizontal equations. If the speed slows down then the tension in the bottom string
reduces to a critical value ie zero. At this point the situation is the same as the diagram above. If the
speed reduces further than the angle changes (gets smaller).
The tension in two string questions can only ever be the same if the mass is a smooth ring
threaded on a string.
Motion in a Vertical Circle
1. Mark the zero PE level (Usually the bottom)
2. Calculate the PE and the KE at the starting position.
v1
m
T2
3. Add these energies.
(1)
4. Calculate the new PE and KE at the new position.
You may need trig to calculate the new height.
mg

5.
Remember that the speed has changed to v1
6. Add these new energies. (2)
7. Equate (1) to (2)
r
T1
PE = 0
8. Rearrange to read v1 =
m
(You can rarely get away
with just this)
v
mg
9. Put the forces on the diagram. (The tension may be replaced by a reaction due to contact with
the surface)
10. Work out the force towards the centre of the circle at the point required in (2) It is usually a
combination of a weight component and the tension or reaction.
11. The force mg will almost always have a component towards the centre; usually mg cos 
12. Use F = ma with these forces so that
mv 2
the forces towards the centre =
r
13. The velocity here is the same as v1 in the energies part earlier, so substitute the earlier
mv 2
expression into
r
NOTES:
1. For contact and strings T (or R) must be  0
2. There must also be a speed at the top which agrees with this force under F = ma .
3.
The minimum speed comes from T  mg 
mv12
r
where T = 0 This is at the top.
4. You need a greater initial speed at the bottom to do complete circles with strings than
rods
5. For Rods the tension can be 0 or even negative (If so it is called thrust). This occurs
when the particle is moving slowly so that is gets past “ 3 o’clock “ but would fall if it
were not a rod.
6. These are not difficult but they contain lots of small steps and hence a high “Cock up
Factor”.
Take your time do not miss steps out.
Moments
The moment of a Force about a point is
Force × Distance measured from the point.
The distance must be measured 900 to force.
d
F
can usually be easier.
If a distance which is not at right angles to the force is known resolving
You must consider both resolutions although one often produces no
turning moment.
The moment is F × sin × d
F
d
clockwise.
Note
F × cos
passes through the point and has no turning effect.
Most questions involve real situations which require modelling.
If the system is in equilibrium them the sum of the moments = zero
(They usually are)
This means that the clockwise moments = anticlockwise moments.
You will always need to resolve the forces as well as take moments.
Don’t forget g.
A clear diagram is essential.
General approach:
1.
Resolve
Up = down
2.
Resolve
Left = Right
3.
Take moments about a point.
Choosing points: Pick a point that the most unknowns go through since they will not feature.
If you need to work out a distance use a point that is easy to relate to that distance.
i.e.
A
x
B
C
Use point A or B but not C
Non Uniform Rods:
These have their mass acting “off centre”. You usually have to find out where. Let the distance for the mass be d from
one end and create equations as normal. Take moments about the end you have used.
Modelling:
Masses as particles
Assumptions:
Rods do not bend Masses act in the middle of rods.
Planks beams etc as uniform rods
i and j Notation
i is a vector in the x direction i.e.
j is a vector in the y direction i.e.
Point can have co ordinates i.e. ( 3 , 4 )
In i j form this point would have a position vector of 3i + 4j usually written as r = 3i + 4j
This means r is the position vector. The point is
3
and 4
Position vectors are usually measured from the origin. r may be a position vector it may also be a displacement vector
from a previous position. If r1 is a position vector and r2 a displacement then the new position vector is found by
adding the two vectors.
To add vectors just combine the i s together and the j s together. (Never mix them)
If r1 = 4i – 3j and r2 = 6i + 2j then
r1 + r2 = (4+6)i + ( -3 + 2)j
or 10i - j
Distance is not a vector so the distance is found using Pythagoras.
Watch out for Pythagorean Triples. i.e. Given a distance of 50 in the direction 7i + 24j . You need to realise that the
direction 7i + 24j has a size of 25 (72 + 242 = 252)
2
14i + 48j
2
So the vector with distance 50 must be two times the distance i.e.
2
( 14 + 48 = 50 )
Positions and velocities: ( not involving t)
A velocity of 5i + 3j means the position gains 5i and 3j every second. Therefore the new position after 2 seconds would
be
where it started
plus
2 × (5i + 3j) or 10i + 6j
It the initial position vector was 4i + 7j then the new position vector would be 4i + 7j + 2 × (5i + 3j)
Generally:
New position = Where it started + (Velocity × time)
It is common to find when two different objects meet. This is when they are in the same place at the same time.
Therefore they have the same i values and the same j values at the same time .
These need a good clear diagram!!
Must Learn!!
Positions and velocities involving t :
These are calculus questions. - Very common!!!
e.g. Position vector r = 5t2i – 7tj
Velocity vector would be
A function of t so calculus only
dr written as r
is often
i.e. dr
differentiate r with respect to t
dt
dt
r dot
So r = 10ti – 7j The speed implies Pythagoras i.e. (10 + 7 )
2
2
If acceleration is required you differentiate velocity. Acceleration can be written as r r double dot
Elastic Strings and Springs
These obey Hooke’s Law : The formula is
e l
where
e is the extension (sometimes x
NB.
The extension becomes negative when a spring is compressed.
The tension also is reversed!!
is used)
l
is the natural length ( sometimes a ).
If a particle is tied to the middle of an elastic string treat it as 2 separate strings . Remember to
change the natural length to half the original.
Work done in stretching an elastic sting from one value to another is :
(T1  T2 )  (e2  e1 )
2
Average tension x additional extension
Far easier to remember that “Work done “ is “Change in energy” and use:
Elastic Potential Energy
This is the energy stored in a stretched string, or that needed to stretch a string.
The formula is Lamda extension squared over twice natural length. Ie.
EPE =
 e2
2l
This is very common where any motion is concerned and involves conservation of mechanical
energy. Eg a bungy jump, where initially all the energy is Potential (mgh) ie. at the top.
At the bottom it is all Elastic Potential (EPE) since the object is momentarily stopped.
It follows that:
mgh at the top = EPE at the bottom
Warning! Look out for situations that mean the string is stretched before anything happens.
Catapults are often set up this way. At the point of firing there may still be some EPE.
RTFQ
Elastics and SHM
Usually involving proving SHM. This is always worked around the equilibrium point. (Usually the
first part of the question).
All extensions consist of two parts: the original extension and a displacement from the equilibrium
position. This is
x
and is usually measured in the downwards direction.
The acceleration must be taken in this direction (even though you know it is going the other way!)
This produces the negative you need ie acceleration =  n2 x or   2 x .