Introduction to mechanics

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[edit] Introduction

Mechanics is the study of the motion of objects using equations. This can be achieved by using standard formulae or mathematical techniques such as differentiation or integration.

[edit] Definitions

A few definitions need to be discussed.

  • Displacement - How far an object is from its starting position. This is given the symbol s
  • Distance - How far an object has travelled. This is not usually used in calculations.
  • Initial Velocity - The starting rate of change of displacement. This is given the symbol u
  • Final Velocity - The ending rate of change of displacement. This is given the symbol v
  • Speed - The rate of change of distance. This is not usually used in calculations.*
  • Acceleration - The rate of change of velocity. This is given the symbol a
  • Time - How long has elapsed from the start of the event. This is given the symbol t

Speed can be calculated as the absolute value of velocity, ie spd = |u| or spd = |v|

[edit] Standard Units

The SI units (standard units) of the quantities used in this section are given below:

  • Displacement - metres (m)
  • Velocity - metres per second (ms − 1)
  • Acceleration - metres per second squared(ms − 2)
  • Time - seconds (s)

[edit] Motion in a straight line with constant velocity

To calculate the displacement moved by an object with constant velocity we use the following formula:

s = ut

Example: A car is travelling at 10ms − 1. How far will it travel in 12s?
s = 10 x 12 = 120.
Thus the car will travel 120m in 12s.

Note. This formula is based on a formula shown below.

[edit] Newton's equations of motion

In mechanics, the following formulae are VERY important. These are:

  • v = u + at
  • s = ut + 0.5at2
  • s = 0.5t(u + v)
  • v2 = u2 + 2as


These can be used to solve any motion question where there is a constant (or assumed to be constant) acceleration. Even if the acceleration is 0.
The equation from the above section (s = ut) is derived from s = ut + 0.5at2 with a = 0 and therefore s = ut + 0 etc.

To decide which equation to use, look at the data you are given. Write down a list of the quantities given (the s,u,v,a and t values). Then cross out the quantity that is not mentioned in the question. All you have to do now is see which equation has the 4 quantities that you have left in and then solve.

Example 1: A car is travelling at 10ms − 1 when it starts accelerating at 1.5ms − 2. How far will it travel in 15s?
s = ?, u = 10, v =, a = 1.5, t = 15
Thus use s = ut + 0.5at2.
s = (10 x 15) + (0.5 x 1.5 x 15²)
s = 150 + 168.75 = 318.75
Thus the car will travel 318.75m

Example 2: A cyclist is initially at rest and then accelerates to 12ms − 1 in 6s. Calculate the acceleration.
s =, u = 0, v = 12, a = ?, t = 6
Thus use v = u + at.
12 = 0 + (a x 6)
12 = 6a
Thus a = 2
The cyclist is accelerating at 2ms − 2

Example 3: A van accelerates at 1.5ms − 2 over a distance of 10m to 20ms − 1. How fast was the van travelling before it started accelerating?
s = 10, u = ?, v = 20, a = 1.5, t =
Thus use v2 = u2 + 2as.
20² = u² + (2 x 1.5 x 10)
400 = u² + 30
u² = 370
u = \sqrt{370} = 19.2 (3sf)
The van was travelling at 19.2ms − 1 (3sf)

Example 4: A lorry can come to a stop from 10ms − 1 in 3 seconds. How far will it travel before it stops?
s = ?, u = 10, v = 0, a =, t = 3
Thus use s = 0.5t(u + v).
s = (0.5 x 3) x (10 + 0)
s = 1.5 x 10
s = 15
The lorry will cover 15m before stopping.

[edit] Applied force

Whenever a force is applied to an object it accelerates proportionally to the size of the force applied.

ie F \alpha\ a.
and therefore F = ka where k is a constant.

Via calculations the constant k, is actually the mass (m) of the object being accelerated therefore:
F = ma.

Force is measured in Newtons (N) and is given the symbol F
Mass is measured in Kilograms (kg) and is given the symbol m


Example: A freighter weighing 100000kg is accelerating at 10ms − 2. What force is required?
Use F = ma
F = 100000 x 10
F = 1000000N
A force of 1000kN is required. (Where 1kN = 1000N)

[edit] Vector motion

So far we have only really looked at motion in 1 dimension. Some times you need to look at motion in 2 and maybe 3 dimensions.
To achieve this, all you have to do is convert the equations of motion into vector equations. Therefore the equations of motion become:

  • \underline{v}=\underline{u}+\underline{a}t
  • \underline{s}=\underline{u}t+0.5\underline{a}t^2
  • \underline{s}=0.5t(\underline{u}+\underline{v})

Where the underlined quantities are vector quantities none-underlined quantities are scalar.

Note: the last equation of motion is missing as this requires a more complex method of solving as it involves vector products.

Example: A ball is moving with an initial velocity of (\underline{i}+3\underline{j}-6\underline{k})ms − 1 for 2 seconds. Assume the only acceleration is that of gravity, which can be taken to be 9.8ms − 2. Find the position vector of the ball at time t = 2s. (i, j and k represent the x,y and z axis of the cartesean system)

\underline{s} = ?
\underline{u}=(\underline{i}+3\underline{j}-6\underline{k})
v =
\underline{a}=-9.8\underline{j}
t = 2
Thus use \underline{s}=\underline{u}t+0.5\underline{a}t^2
\underline{s}=2\underline{i}+6\underline{j}-12\underline{k})-39.2\underline{j}
\underline{s}=2\underline{i}-33.2\underline{j}-12\underline{k}
Thus the ball is at (2\underline{i}-33.2\underline{j}-12\underline{k})

To find the distance of the position vector, take the modulus (|a|) of the vector using Pythagoras' theorem.

[edit] Mechanics using calculus

More often than not you will be presented with a problem that gives you the displacement/velocity/acceleration as a function rather than just values. To solve these you must use differentiation and integration techniques.

Acceleration, velocity and Displacement are related as follows:

  • Displacement = \int(Velocity)dt
  • Velocity = \int(Acceleration)dt

And therefore..

  • Acceleration = \frac{d}{dt}(Velocity)
  • Velocity = \frac{d}{dt}(Displacement)

Example:
A car's acceleration describes the function a=\frac{4t^3}{3}-\frac{3t^2}{2}+2. State the function of the velocity given that the car was at rest before accelerating.
a=\frac{dv}{dt}=\frac{4t^3}{3}-\frac{3t^2}{2}+2
Therefore v=\int(\frac{4t^3}{3}-\frac{3t^2}{2}+2)dt
v=\frac{t^4}{3}-\frac{t^3}{2}+2t+C where C is the constant of integration</math>
At time t = 0, v = 0. Therefore:
0=0-0+0+C \Rightarrow \ C = 0
v=\frac{t^4}{3}-\frac{t^3}{2}+2t

[edit] Questions on introduction to mechanics

[edit] Questions

  1. A ball is rolled 15m in 10s. How fast was it travelling?
  2. A Car accelerates from rest to 25ms − 1 while travelling 15m. How long did this take?
  3. An athlete accelerates from rest to 10ms − 1 in 1s. State the acceleration.
  4. Calculate the acceleration created when a 10kg block has a force of 200N exerted on it.
  5. Using the acceleration found above, how fast will a object travelling at 10ms − 1 be travelling after 10 seconds?
  6. A rock is thrown with velocity \underline{i}-5\underline{j}-\underline{k}. Assuming that the only acceleration acting upon the rock is gravity, (9.8ms − 2) state the position vector after 13s and therefore the distance after 13s.
  7. An object's path is described by the function t2 − 16t + 64. Find the velocity function and therefore the acceleration applied to the object.

[edit] Answers

  1. 1.5ms − 1
  2. 1.2s
  3. 10ms − 2
  4. 20ms − 2
  5. 210ms − 1
  6. \underline{s}=13\underline{i}-403.1\underline{j}-13\underline{k}. And therefore the distance is 403.52m (2dp)
  7. v = (2t − 16)ms − 1 and therefore a = 2ms − 2.

[edit] Credits

Written by Richardtebbs 00:21, 23 January 2007 (UTC)
Please find any mistakes and update them to keep content accurate.