Introduction to mechanics

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.

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|

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 $(s)$

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.

Newton's equations of motion

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

• $v=u+at$
• $s=ut+0.5at^{2}$
• $s=0.5t(u+v)$
• $v^{2}=u^{2}+2as$

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 $s=ut+0.5at^{2}$ 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 $s=ut+0.5at^{2}$.
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 $v^{2}=u^{2}+2as$.
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.

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)

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.

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:
$0=0-0+0+C\Rightarrow \ C=0$
$v={\frac {t^{4}}{3}}-{\frac {t^{3}}{2}}+2t$

Questions on introduction to mechanics

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 $t^{2}-16t+64$. Find the velocity function and therefore the acceleration applied to the object.

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 $a=2ms^{-2}$.

Credits

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