Content

Mechanics

*Disclaimer: Let me be frank - I don't really understand this subject. If you see something that doesn't make sense, let me know, because I've probably made a mistake!*

Before starting on this topic, you should familiarise yourself with the Key Words.


Kinematics

The study of a particle's motion is called kinematics. There are five initials that you need to be familiar with before we go into this topic:

s - displacement/distance (meters)
t - time (secs)
a - acceleration (ms-2)
u - initial velocity (ms-1)
v - final velocity (ms-1)

When a particle is moving with constant acceleration, a, initial velocity, u, to a final velocity, v, we can show this on a velocity/time graph:

For this type of graph, the gradient is the acc'n.
The displacement/distance is the area of the graph.

We can also have a distance/time graph:

We all know the formula: speed = distance / time, so therefore, the gradient here is the speed.

Assumptions:

  1. The particle is travelling in a straight line.
  2. The particle travels at a constant speed (no acc'n).
  3. If we were considering vectors, we would assume there is no vertical movement.
  4. There is no friction.

Constant Acc'n equations

When we have a constant speed, there are several equations that apply. You MUST learn these:

  1. v = u + at

  2. v2 = u2 + 2as
  3. s = ut + at2
  4. s = vt - at2

Example:

A car is travelling along a straight road with constant speed 36 kmh-1 and is brought to rest with uniform deceleration in 60 m. How long did it take to stop?

Whenever we have a constant speed question, the first step it to write down the variables that we have and need:

u = 36 kmh-1 (convert into ms-1 = 10 ms-1)
v = 0 ms-1 (it comes to rest)
s = 60 m
t = ?

So, we can see from the letters that we need the formula:

Finally, stick the numbers into the formula and rearrange to find t:

60 = (10 + 0) x t

60 = 5t

t = 12 seconds

In the question, we needed to convert from kmh-1 into ms-1:

kmh-1 to ms-1 = 3.6

ms-1 to kmh-1 = x 3.6

Vertical motion under gravity

If a particle moves freely under its own weight, it experiences a constant downwards acc'n due to gravity.

We call this "g" and it is approximately 9.8 ms-2.

We use constant acc'n equations for these types of questions again.

Example:

A ball is thrown into the air with a velocity of 10ms-1 from the top of a ladder of height 10 m. Find the distance to the greatest height and the time taken to reach the floor.

First step is to draw a simple diagram:

Because the ball is thrown into the air, gravity is acting against it. Therefore "g" is negative = -9.8.

The arrow shows which way the ball is travelling.

The greatest height occurs when v=0.

We write down the variables that we have:

u = 10 ms-1
v = 0 ms-1
a = - 9.8 ms-2
s = ?

From this, we know we need the formula: v2 = u2 + 2as.

Stick in the figures and rearrange to find s:

0 = 100 - 19.6s

s = 100 19.6

s = 5.10 (3 s.f.)

For the next part of the question, we have a number of options. We could find the time to the greatest height and add on the time taken to hit the floor.

However, because we're dealing with displacement, it's easier to find the time to reach -10m. (10 metres less than the point it was thrown from).

Therefore, we have the variables:

u = 10 ms-1
s = - 10 m
a = - 9.8 ms-2
t = ?

From this, we see we need the formula: s = ut + at2.

Stick in the numbers and rearrange for t:

- 10 = 10t - 4.9t2

4.9t2 - 10t - 10 = 0

We have a quadratic - solve by the quadratic formula...

t = 2.78 sec (3 s.f.)

And that's pretty much all there is to Kinematics. Learn those formula! They're used throughout Mechanics...

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Vectors

A vector is any quantity that has a magnitude (size) and a direction. Take a look at the following, showing vectors and their scalar equivalents.

Vector Scalar
Velocity Speed
Displacement Distance
Weight Mass
Acceleration "magnitude of acceleration"

Vectors are represented by lines with arrows:

The vector from A and B is called: . If you give it a letter name, such as c, make sure to underline it.

Vectors can be added or subtracted.

If you want to find the vector , you need to add the two vectors given. Therefore, = b + a.

If you want vector , though, you need to go in the opposite direction to a and b. Therefore = - a - b.

Vectors are sometimes given in the form i + j. If you draw them on a graph, i means how for across the x-axis it is from the start point, and j is how far up the y-axis it is.

For example: a = 5i + 6j looks like this:

The unit vector = i + j. (just going up one and going across one).

It can be found from the following:

= unit vector

= magnitude of vector a.

To find magnitude of a vector, (i2 + j2).

For example:

Find the vector with magnitude 5 in the direction of -8i + 6j.

Magnitude of vector: (-82 + 62). = 10

Divide to get unit vector: =

Times by magnitude 5: -4i + 3j.

Parallel Vectors

To be parallel, the is and js must be multiples of each other. For example, 2i + 3j and 4i + 6j are parallel, because they have a common multiple of 2.

If a = and b = λ, find λ when a + λb is parallel to i.

(note in this question that λ is just a number and i is the vector .)

+ λ = k

+ =

Considering the is and js, we get two equations:

2 + λ = k
1 + 3λ = 0

Therefore, λ = -1/3 and k = 5/3

Components of a vector

From this, we can see:

OA = OPcosθ
OB = OPsinθ
= +

= (OPcosθ)i + (OPsinθ)j

This is a way of expressing vector in its horizontal and vertical components.

For example:

The displacement, r, has magnitude 10km and acts at the bearing 060o. Express r in the form ai + bj.

The first step is to draw a diagram of the vector, labelling what you have:

Next, it's just a case of using trig to work out the i and the j.

10cos30 = 5i

10sin30 = 5j

Put them together and you get: 5i + 5j.

Easy, no?

From the question above, we see a vector coming from the origin (O). is the position vector for P relative to the origin. We call the position vector r.

If we are given another vector, we can find the position of A relative to B.

On a diagram:

rA =

rB =

To find the position of A relative to B means that we are going from B to A.

Therefore:

= rA - rB

The distance from B to A is simply the magnitude: || = |rA - rB|

Constant Velocity

Constant velocity occurs when the acceleration is 0.

The velocity of a vector, A:

VA= xi + yj ms-1
(for every second, it travels x metres up and y metres across).

From this, we can see the direction and the speed of the vector, but NOT its position.

To find the direction of the vector, if we draw it out:

direction:

tanΘ =

To find the speed, we need the magnitude:

(x2 + y2)

Scalar connection:

Speed = distance / time.

Vector form:

Velocity = displacement / time

From this, we can rearrange the vector form to:

r = Vt

Example: If a particle is moving with velocity 2i + 3j ms-1, what displacement will occur after three seconds?

r = (2i + 3j) x 3

r = 6i + 9j

For a position vector, this formula is slightly different:

rF = rI + Vt

rI is the starting point of the vector (initial position), V is the velocity, t is the time and rF is the final position of the vector.

Example

At noon, a ship, s, is at position vector 600j and velocity 7i + 8j.
At the same time a speedboat, b, has position vector 120j and velocity 7i + 24j.
Show the two will collide and find the time when this occurs.

First, we need to work out the final position equations of the two, using the formula: rF = rI + Vt.

For the ship:

rF = 600j + (7i + 8j)t

For the speedboat:

rF = 120j + (7i + 24j)t

For them to collide, the displacement at some point of time must be equal. Therefore, we can say:

600j + (7i + 8j)t = 120j + (7i + 24j)t

Expand and rearrange to find t:

600j + 7ti+ 8tj = 120j + 7ti + 24tj

480j = 16tj

30 = t

Both i and j must be equal for them to collide.

Using i: 7t = 7t

Using j: 600 + 8t = 120 + 24t

Rearranging j will also find t=30.

Because both i and j are equal when t=30, we know they will collide.

After travelling for 15 secs, the boat changes its velocity to 7i + 30j. Find its new position vector after 30 seconds, and find the distance between s and b.

First we need to use the formula: rF = rI + Vt.

rF = 600j + (7i + 8j)15

rF = 105i + 720j

This is its position vector after 15 secs. Then use this vector with the new velocity:

rF = 105i + 720j + (7i + 30j)15

rF = 210i + 1170j

To find the displacement of s relative to b, we use the earlier formula = rS - rB.

After 30secs, rB = 120j + (7i + 24j)30 = 210i + 840j

210i + 1170j - (210i + 840j) = 330j

Magnitude=distance, so: |330j| = 330m.

S and b are 330m apart.

Minimum distance

At time t, the following have position equations:

rB = (-1 + 11t)i + (-4 + 3t)j

rA = (2 + 3t)i + (1 + t)j

Finding rB - rA will give us: (-3 + 8t)i + (-5 + 2t)j

From this, we can't work out t or know how far apart they are. However, we can find the time and minimum distance from each other.

Distance is the magnitude, so: |(-3 + 8t) + (-5 + 2t)|

d = [(-3 + 8t)2 + (-5 + 2t)2]

d = (64t2 - 48t + 9 + 4t2 - 20t + 25)

d = (68t2 - 68t + 34)

This is messy, because of the square root. Therefore, we square:

d2 = (68t2 - 68t + 34)

To solve this, we can differentiate.

The minimum distance occurs when the differential equals 0. So, differentiating, we get:

136t - 68 = 0

t = 0.5

So, a half an hour, if it's in minutes. (If they set off at noon, it might now be: 12:30.)

Now we have t, we can sub in to find d2:

d2 = 17

Therefore the distance = 17 m

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Equilibrium

The study of forces is called statics. "Force" is measured in Newtons and is a vector quantity. To find a resultant force, we simply gave to add the vectors:

Example:

3 forces, (2i - j)N, 3iN and (-i + 4j)N act on a particle. Find the magnitude and direction of the resultant force, R.

(2i - j) + 3i + (-i + 4j) = 4i + 3j N

You'll recall from the vectors section that magnitude is (42 + 32) = 5 N

and direction is the angle, Θ , when the vector is drawn:

tan Θ =

Θ = 36.9o (3 s.f.)

 

If we are given a force that's not in vector form, such as 10 N, we will be given the its direction. From this we can work out the horizontal and vertical componants.

|F1| = |R|cosΘ ... F1 = |R|cosΘi

|F2| = |R|sinΘ ... F1 = |R|sinΘj

|R| = |R|cosΘi + |R|sinΘj

We resolve horizontally and vertically to find the resultant force, once we have split the forces into their components.

Example:

Find the resultant force of the following system:

Resolving --->

5 N + 10cos60oN - 8cos30oN = 4.86 N (3 s.f.)

Resolving

- 30 + 10sin60oN + 8sincos30oN = - 15.2 N (3 s.f.)

Turning this into a vector, we get:

4.86i - 15.2j

Finally, it's just a case of finding the magnitude, 15.9 N, and direction, 72.3o.

When the forces do not cause a change of motion, they are said to be in equilibrium.

If we resolve them, x components and y components must equal 0.

Example: Given that the following system is in equalibrium, find p and Θ:

Resolving --->

- 8cos30 + 5 + pcosΘ = 0

pcosΘ = 1.9282

Resolving

8sin30 + psinΘ = 0

psinΘ = 4

If we divide the second equation by the first, we get:

tanΘ = 2.07447

Θ = 64.3o (3 s.f.)

Now, if we sub Θ into psinΘ = 4, we can find p:

0.9008p = 4

p = 4.44 N (3 s.f.)

It's as simple as that!

Types of forces

There are various different forces we need to consider in equilibrium situations.

  1. Weight, W
    This is the force exerted on a particle due to gravity.
    Remember: W = mg = 9.8
  2. Tension, T
    This is the force found in a cable or string.
  3. Thrust, P
    This is the pulling or pushing force on a particle.

  4. Normal reaction, R
    This is the resisting force a plane exerts on a particle, to oppose its weight.

    Normal means at right angles with the plane, so if it's on a slope:

  5. Friction, F
    Force caused by resistance when surfaces contact, which always resists a motion.

As P increases, so does F. This continues until the point where there particle is about to move... F reaches its maximum value, which is called FMAX.

The maximum force depends upon two things:

  1. The weight of the particle, and its R reaction.
  2. The coefficient of friction, μ, which depends on the the "stickiness" of the surface.
    F = μR

When a particle is on the point of moving, it is still in equailibrium. Therefore, we can solve by resolving.

Example:

A block of weight 20N rests in equilibrium on a rough plane inclined at 30o to the horizontal. A force of p N is acting on the particle up the plane. The coefficient of friction between the plane and the particle is 0.3.

Given that the particle is on the point of slipping down the plane, find the force p.

First off, we want to turn all that waffle into a nice diagram:

Gravity is pulling the weight downwards, but we're resolving parallel and perpendicular to the plane, so it's no use. Instead, we split W into the components, 20cos30 and 20sin30.

Because its on the point of slipping down the plane, F acts upwards.

Once we have the diagram, we resolve and :

Resolve

F + p - 20sin30 = 0

p = 10 - F

Resolve

R - 20cos30 = 0

R = 17.32 N

We still can't find p, but now we have R and μ, we can use the formula F = μR:

F = 17.32x0.3

F = 5.196 N

Now, sub F into the p = 10 - F:

p = 4.80 N (3 s.f.)

Remember, if a question gives you a mass, 4kg, convert it to weight: 4g or 39.2 N.

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Dynamics

Newton was a pretty smart guy. Besides his discovery of gravity, he also had a few things to say about motion...

  1. A particle remains in a state of rest (or at constant velocity) unless acted upon by a force.
  2. A (resultant) force produces an acceleration (change of velocity) in the direction of the force, proportional to the magnitude of the force.
    Using this, we can produce the formula:
  3. netF a
    netF = ma

The units are SI.
F = force, Newtons.
m = mass, kg
a = acc'n, ms-2

   3.  Every force has an oppostite or equal reaction.  For example:
The particle exerts a weight on the floor and the floor exerts an oppostite force on the particle.

By using Newton's second law (NSL) we can work out lots of problems.

Example:

A carof mass 600 kg is travelling along a road with acc'n 2 ms-2. The engine exerts a force of 1500 N. Find the magnitude of resistance, R, given that acc'n is constant.

First, we want a little diagram of what we have:

Now, if we think of NSL --> and recall that netF = ma, we can put in the numbers to the equation:

1500 - R = 2x600
R = 1500 - 1200
R = 300 N

Remember that we're only using mass here, not weight like in equilibrium questions.

If a particle is moving freely under its own weight, its acceleration is g (9.8) ms-2.

A bucket of mass 2 kg is suspended by a light inextensible rope. Find the tension in the rope if the bucket is ascending with acc'n of 3 ms-1.

Use NSL

T- 2g = 2 x 3
T = 6 + 2g
T = 25.6 N

Remember to say in which direction you are using NSL. This defines positive. If you were using NSL , but a particle were descending with 3ms-1, you would say it had acc'n of -3ms-1.

Connected Particles

When string connects two particles, we need to consider tension...

An engine of mass 200,000kg pulls a carriage of 50,000kg with an acc'n of a ms-2. There is no resistance to motion. Find a and the tension in the couplings, if the driving force is 50,000N.

So, a little diagram to start:

Use NSL --> on the engine:

50,000 - T = 200,000a

Use NSL --> on the carriage:

T = 50,000a

Now, we merely sub in, to eliminate T for the moment:

50,000 - 50,000a = 200,000a
50,000 = 250,000a
a = 0.2 ms-2

Now, if we sub a into the second equation, we can find T:

T = 50,000 x 0.2
T= 10,000 N

T is the same, but in opposite directions, as seen in the diagram, due to Newton's third law.

Pulleys

When a question involves pulleys, there are a number of assumptions:

  • Pulley:
    1. Light
    2. Smooth
    3. Fixed on a vertical plane
  • String:
    1. Light
    2. Inextensible

Quite a few assumptions!

A pulley:

If the weight (mg) of particle B is greater than A, the forces and accelation are as shown in the diagram. The same method is used for solving these as in connected particles.

Greatest height: If the heavier pulley hits the ground, the lighter will continue travelling up to a greatest height. In the example above we'd find this using constant acc'n equations from:

u = (velocity that B hits the ground)
v = 0
a = -9.8
s = ?

Therefore, the constant acc'n equation: v2 = u2 + 2as.

The forces acting on the pulley shown above look like this:

Therefore, the total forces of the pulley is: 2T.

Other pulleys:

Another type of pulley... the only difference is the imput of F, R and μ from the previous chapter.

In this type, the total forces are R = 2T2 as seen to the right:

If we use NSL on all three ms, we get the following equations:

m1g - T1 = m1a
T1 - T2 = m2a
T2 - m3g = m3a

These added produce:

m1g - m3g = a(m1 + m2 + m3)

The previous pulley system could also throw friction into the question, of course...

From this one, we use NSL to get the equations:

m1g - T = m1a
T - m2gsinΘ = m2a

And solve from there.

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Moments

The moment of a force is its turning effect.

Moment = force x distance (distance being perpendicular from the pivot).

For example, opening a door:

0.8 x 10 = 8 Nm... It requires 8Nm to open a door.

If we decrease the distance, the force will naturally require increasing to make the same Nms... that's why handles are as far from the hinges as possible.

If the handle were right on the hinges, the distance would be 0... you'd need an infinate force to turn, which is impossibe...

We can add moments together. If we have the following, and we want to rotate around the point A:

We state what direction we're taking moments about: anticlockwise or clockwise.

=
- (0.8x13) + (1.2x4) + (1.6x10)
= 10.4 Ns

Notice how any forces rotating it in the oppostite direction become negative?

Moments can either apply to a light, unbending, thin rod (such as the previous example) or a lamina.

Lamina are rigid areas with negliable width... like a piece of card. These are used for vector moments questions...

Example: Find the moment of the force, F, acting on the point (5,7) about the origin.

The first step is to draw it out, and split F into its components:

=

Multiply the distance along i from the origin by the j component (5x3). It turns anticlockwise = -15.

Multiply distance along j from O by i component (7x2) and it's clockwise so = 14.

- 15 + 14 = - 1 Ns.

The pivot doesn't have to be the origin, though, so remember to work out the distance.

Equilibrium

...occurs when the total moments = 0, naturally. With equalibrium situations you might be asked to find a distance or a force...

A uniform beam of 3m and weight 50N has weights of 20N and 30N attached to its ends. Find the point where a support should be placed to make the beam rest horizontally.

The key word here is "uniform". This means the weight is found at the midpoint (a large assumption). Also, due to NTL, a support produces a R force upwards.

First step, draw a diagram and call the support something... A.

(If the question wanted us to find R, we'd just resolve ).

 

Because it's in equalibrium, we can say:

=
30(1.5 - x) - 50x - 20(1.5 + x) = 0

Now, expand brackets and solve to find x:

45 - 30x - 50x - 30 - 20x = 0
15 = 100x
x = 0.15m

So, 1.65m away from the 20N end of the beam.

A question might also say "on the point of turning about support B"... this means that for support A, the reaction, R, is now 0.

And that's all there is to Moments!

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Momentum and Impulse

Imagine an elephant and a mouse running. Which would you rather be hit by? Unless your sucidal... probably the mouse. This is because, though they're both travelling at the same speed, the elephant has a much greater momentum due to its mass.

The equation for momentum is quite simple:

I = mv - mu

Impulse is the final velocity times mass, take away the initial velocity times mass.

Thinking about constant acc'n equations, I also equals Ft (net force times time).

Example:

A ball of mass 0.05 kg travels towards a wall at 12 ms-1 and rebounds with 10m and rebounds with 10ms-1. Find the impulse on the ball by the wall.

Be careful here... little diagram first, and label I to define positive:

Because I is going -->, the 12ms-1 needs to be negative. So we get:

I = 0.05x10 - 0.05x-12
0.5 - - 0.6
1.1 Nm

Conservation of momentum means that the momentum is the same before and after (either a collision or a load being added).

There is a new formula for this: m1u1 + m2u2 = m1v1 + m2v2

Example: Find v in the diagram.

Use the formula:

2x5 + -2x1 = 3x2 + vx1
10 - 2 = 6 + v
v = 2 ms-1

 

If particles are connected by a string, we use exactly the same formula. Before the jerk, one will have a velocity, while the other lies at rest. After the jerk, they will both travel at the same velocity.

Remember... if a particle's direction is changed, you reverse the sign!

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That's mostly Mechanics done...