… Improvements to simple models (limitations) used in mechanics can include, for example:

- Include air resistance in the model of the motion
- Use a more accurate value for g in the model of the motion
- Include wind effects in the model of the motion
- Include the dimensions of the stone in the model of the motion
- include the mass of any strings and pulleys

Forces and motion

… Become a master diagram-drawer using a pencil and a ruler – it will save a lot of time and make everything much clearer.

… Remember other forces that might not be shown on the diagram given in the question. For example, reaction forces and weights.

… It’s worth noting that:

cos45 = sin45 = 1/√2

cos30 = sin60 = √3/2

cos60 = sin30 = ½

… Keeping ‘g’ as the letter ‘g’ (rather than 9.8) can be useful as you work through the algebra – you will sometimes find that it will cancel out.

… Resolving a force into two perpendicular components:

1) draw a rectangle around the force

2) then put your pencil on the force and rotate to the component you want. If you rotate through the angle, then it is “cos”

… Using “sum of forces” ΣF(→+) technique to analyse bodies in equilibrium. This can be used in two perpendicular directions, e.g. ΣF(→+) = 0 and ΣF(↑+) = 0 , to create two equations.

… Remember that if a Body is not accelerating (at rest or moving at a constant speed) then it is in equilibrium and the ΣF = 0.

… ‘Potato-ing’ a body can help when drawing free body diagrams of forces acting on the object. Try to imagine that YOU are the object – what forces would you experience and in which direction?

… For forces questions involving accelerating bodies, label the positive acceleration direction on each body and then using “ΣF(→+) = ma”. Note that it is useful to define your positive direction as being the same as the acceleration.

… Objects connected by a string or rope can be treated separately (sum of forces), but the tension (pull) in the string will be the same on both bodies, even if it goes round a pulley.

… Note that for an object accelerating up or down a plane: the object is in equilibrium perpendicular to the plane.

… Free body diagrams example involving a lift of 500kg accelerating upwards due to an upwards cable force of 6.06kN. The lift carries a passenger of 80kg. What is the magnitude of the reaction force experienced by the passenger?

There are three ways of analysing this system using free body diagrams (aka potatoes!). Try drawing out the free body diagrams in each case:

1) Drawing a free body diagram of the Lift (potato the lift only)… But imagine that the passenger is NOT in the lift, but hanging from a rope under the lift!…

Upwards cable tension = 6060

Downwards weight of lift = 500g

Downwards tension in rope = T

(Note that acceleration, a, is upwards)

Using ΣF(↑+) = ma

6060 – 500g – T = 500a

T = 1160 – 500a … Equation 1

Note that we have NOT included the mass of the passenger here as this is taken into account by the rope tension.

2) Drawing a free body diagram of the (now dangling!) passenger (potato the passenger, not the lift)

Upwards rope tension = T

Downwards weight of passenger = 80g

Using ΣF(↑+) = ma

T – 80g = 80a … Equation 2

3) Drawing a free body diagram for the whole system. In this case, the mass of the lift and the passenger are treated a one body of mass 580kg (potato the whole system):

Upwards cable tension = 6060N

Downwards (combined) weight = 580g

Using ΣF(↑+) = ma

6060 – 580g = 580a

a = 0.648 m/s²

Note that modelling the passenger as a dangling mass helps us avoid the (intuitive) mistake of including his mass in free body diagram 1).

Moments

… Using sum of moments about a pivot = 0, for bodies in equilibrium. For example:

ΣM(P)clockwise+ = 0

Starting a calculation by stating this tells us that we are summing up all the moments about the pivot point P, with clockwise moments considered to be positive (and anticlockwise moments negative).

This is especially useful for questions involving beams and is often used with ΣF= 0, often making simultaneous equations.

… It is often useful when using ΣM(P) = 0, to place the pivot P at a location where you can ‘get rid’ of an unknown force which might be acting at the pivot.

… If the system is not in rotational equilibrium, then the sum of all the moments about a pivot will add up to a resultant moment:

ΣM(pivot) = resultant moment

… Try the ‘does it feel right’ test at the end of the question. If it doesn’t feel right, check your working out :).

… If the beam is ‘about to tilt’, then one of the reaction forces (from a support) will equal 0N.

… If the question asks you to ‘show that the beam will tilt…’, then take ΣM(P) about the pivot it is rotating about. You should find that ΣM is not equal to zero, i.e. the beam is not in equilibrium.

… The maximum friction force between an object and a surface is given by:

F = μR, where μ is the …?… of …?… , which is a measure of the ‘stickiness’ between the two surfaces. μ typically has a numerical value between 0 and 1 (sometimes more!). This is sometimes called ‘limiting friction’ and is used in the context of ‘limiting equilibrium’ – i.e. the object is just about to slide, but not quite.

… Using “sum of moments = 0” and “sum of forces = 0” for statics problems. Keep an eye on how many variables you have. Generally speaking, most problems will involve a maximum of two unknowns (forces and/or distances).

This means that the two equations you generate using ΣF and ΣM will be sufficient to solve for the variables.

Note that the third equation we were using involving the ΣM about ANOTHER point is effectively the SAME equation as the first ΣM equation, just with different variables. That is why you it is sometimes possible to show that 1 = 1 !

In other words, you should be able to solve a problem using only ONE equation for ΣF and ONE equation for ΣM.

… The magnitude and direction of a Reaction force of a pulley can be calculated by first resolving tensions into horizontal and vertical components and then using ΣF = 0 for vertical and horizontal directions.

Momentum

… Momentum = mass x velocity,

or

p = mv

… The units of momentum are kgm/s

… Momentum is a vector as direction is important. For example, an object might have a momentum of 5 kgm/s moving East. If it reverses direction and moves at the same speed, it will have a momentum of …?… kgm/s.

… The conservation of momentum states that:

“The sum of momentums before a collision or explosion = the sum of momentums afterwards (as long as no external force acts on the system)”

So

sum of (p) before = sum of (p) afterwards

Σp (before) = Σp (after)

… Always draw a before and after diagram, making sure that you define which direction is positive.

… Momentum is a vector. This means that you may be required to resolve velocities of moving masses into perpendicular components.

Then you can use the conservation of momentum by considering, for example, the vertical momentum components only.

… Impulse = the change in momentum of a body = Δp, units Ns (or kgm/s)

… Δp = mv – mu

Where v is the final velocity and u is the initial velocity.

… Also,

Impulse = average force x time the force is applied

… Force is therefore the rate of change of impulse

… The area under a force-time graph tells us the impulse applied to the body.

… A perfectly elastic collision means that no kinetic energy was lost. For example, an atom bouncing off another atom would be considered to be a perfectly elastic collision. However, in normal ‘everyday size’ collisions, kinetic energy is usually lost.

… The coefficient of restitution, e = speed of separation / speed of approach of a collision between two objects.

It represents how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects.

If e = 1 the collision is perfectly elastic (no kinetic energy lost).

If e = 0 the collision is perfectly inelastic (the objects coalesce)

Kinematics and Projectile Motion

‘Suvat’ equations for working out constant acceleration problems:

… Often a labelled diagram helps to ‘digest’ the question at the start (neat and large please!)

… You may need to split up the initial velocity vector into its horizontal and vertical components using the ‘through the angle = cos’ pencil trick.

… First start by defining the direction you’re looking at (horizontal or vertical – usually you’ll be looking vertically to start with) and also note down which is the positive direction by using an arrow and a plus sign.

… Then make a suvat LIST of data with the units.

… Select the appropriate suvat equation (from your formulae sheet) which fits your list.

v = u + at

s = ut + 0.5at²

s = vt – 0.5at²

v² = u² + 2as

s = (v+u)t/2

… Sub in the numbers and then Solve.

… Remember to State the units of the answer at the end (they should match the units in your data List).

Typical questions on projectile motion often ask for:

Maximum height reached… Consider vertical components and use v = 0 m/s (final velocity)

Time of flight… Consider vertical components and use s = 0 m (displacement)

Range (for projectile motion in 2D)… Use s = 0 (displacement) to find the time of flight, t for the vertical component of the motion only. Then consider the horizontal motion (which will be at a constant velocity) to find the distance travelled using s = ut

Remember that you may need to consider a negative displacement if the projectile falls below the initial point of projection.

… Sometimes you may need to create two equations and solve them simultaneously:

1) A vertical motion equation involving two unknowns (usually one of them will be time, t)

2) a horizontal motion equation (e.g. s = ut) involving the same two unknowns.

This is often the case when a projectile needs to pass over an obstacle of a certain height and distance away.

… When two particles are moving together at different speeds and accelerations, then we can ‘link’ the two particles together with a stopwatch time that is started when one of them passes a certain point.

For example:

Particle Q passes a starting line 1 second after particle P.

We start the global clock (Tp seconds) when P passes the starting line.

This means that if P passes the starting line at Tp = 0 seconds, then Q passes the line at Tp = 1 seconds.

But… Q’s ‘personal clock’ only starts from when it passes the starting line.

So we can write two suvats:

For P: For Q:

s s

u u

v v

a a

t = Tp t = Tp – 1

Motion graphs.

… The gradient of a displacement – time graph at a certain time = velocity

… The gradient of a velocity – time graph at a certain time = acceleration.

… For questions involving the motion of TWO separate objects, draw a velocity time graph (use a ruler!).

… The displacement travelled in a velocity – time graph is found by calculating the area under the graph. Either use the area of a trapezium formula or split the area up into squares and triangles.

… If the objects pass each other (starting from the same position) then draw a speed-time graph of both objects on the same axes. Then form and solve an equation in terms of the time t (when one passes the other) by equating the distance travelled by each objects (area under the graphs).

Position and velocity vector questions

… Unit vectors i and j have magnitude 1 and can be used to describe the motion of an object in two dimensions or the forces acting on it.

i is in the x-axis direction, j is in the y-axis direction and k is in the z-direction (usually vertical)

… How to find a unit vector in the direction of, for example, the vector:

a = i – 2j + 3k

unit vector (“a hat”) = (i – 2j + 3k) / |a|

where |a| is the magnitude of vector a, i.e. its length, which can be found using Pythagoras.

… Position, velocity and acceleration of a body can be written using i and j unit vector notation. This splits up an expression into two perpendicular directions which are completely independent of each other.

It (usually!) makes the expressions for displacement, velocity and acceleration easier to work with :).

So an acceleration vector

a = (3)i

Means that the acceleration of the body is constant in the i direction (getting faster at a steady rate). There is no acceleration in the j direction (so could be moving at a constant speed in the j direction)

… Diagrams are really useful (again) here – use a large pair of axes and sketch the approximate position of bodies with coordinates, you can also add velocity vectors as arrows to show the initial directions of the objects.

… To find a resultant vector, add the ‘i’ components together, the ‘j’ components together, and the the ‘k’ components together.

You can also work out a resultant vector graphically with a scale diagram by placing individual vector components (the horizontal i and vertical j components) head to tail and then finding the vector for the start of the first vector to the end of the final vector.

… Whenever you read the word ‘speed’ or ‘distance’ in a question, think ‘magnitude’, i.e. you will have to use Pythagoras.

… Differentiation can be used on the position vector at time t, to find the velocity vector at a time t. For example, if:

r = i(4t -1) + j(3 + 2t²)

then

v = dr/dt = i(4) + j(4t) = 4i + 4tj

You can also differentiate again to find the acceleration

a = dv/dt = 4j

… Integration can be used on an acceleration vector to find the velocity vector, but you will need to use some given velocity information to find the constant of integration. For example, if:

a = 4j

then v = ∫(4j)dt = 4tj + c

If we know that v = 4i when t = 0, then:

4i = 0 + c

So

v = 4tj + 4i

In the same way, we can integrate again to find the position vector at time t, but will also need to use some position information to find the new constant of integration.

… The velocity vector at time t:

v = u + at

Here vector ‘u’ is the initial velocity vector and ‘a’ is the acceleration vector at time t = 0.

… A neat way to approach constant acceleration questions involving vectors is using “rsuvat” and the constant acceleration equations:

ro = initial position vector

s = displacement vector

u = initial velocity vector

v = final velocity vector

a = acceleration vector (if constant speed, this will be zero)

t = time

The two most useful constant acceleration equations for vectors are:

v = u + at

r = ro + s = ro + ut + ½(at²)

For example, r = ro + (4i – 3j)t + (1/2)(5i + 2j)t²

… When you have formed an expression for the position or velocity vector of an object at time t, it is usually a good idea to factorise for i and j components, for example:

r = i(3 + 2t) + j(6t – 1)

… You can show that two bodies collide by equating the i components to find the time, t. Then equate the j components and you should find that the time, t, is the same value.

… If one body, A, is due North of another body, B, then their i components must be equal. You can use this to find the time at which this due North position occurs.

If A is due East of another body, they their j components must be equal.

… To find the time when a velocity is parallel to a given vector, factorise the velocity equations (e.g. from v = u + at) for the i and j terms. Then examine the coefficients of the i and j terms and how they relate to the given vector.

For example, if our expression for velocity at time t is given by:

v = 2i + t(2i + 5j)

then

v = i(2 + 2t) + j(5t)

and if this velocity is parallel to the vector i + 2j at time t, then the j component must be twice the i component:

5t = 2(2 + 2t)

5t = 4 + 4t

t = 4

… The position vector of body B relative to body A:

rBA = rB – rA = vector AB

This means you are standing at A and looking at B.

If you need to find the distance between two bodies, then you will first need to find rBA (or rAB) and then use pythagoras on the i and j components.