Introduction to mechanics

From Wikiversity

[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.

$D i s p l a c e m e n t$ - How far an object is from its starting position. This is given the symbol $s$
$D i s t a n c e$ - How far an object has travelled. This is not usually used in calculations.
$I n i t i a l$ $V e l o c i t y$ - The starting rate of change of displacement. This is given the symbol $u$
$F i n a l$ $V e l o c i t y$ - The ending rate of change of displacement. This is given the symbol $v$
$S p e e d$ - The rate of change of distance. This is not usually used in calculations.*
$A c c e l e r a t i o n$ - The rate of change of velocity. This is given the symbol $a$
$T i m e$ - 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:

$D i s p l a c e m e n t$ - metres $(m)$
$V e l o c i t y$ - metres per second $(m s - 1)$
$A c c e l e r a t i o n$ - metres per second squared $(m s - 2)$
$T i m e$ - 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:

$s = u t$

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:

$v = u + a t$
$s = u t + 0.5 a t 2$
$s = 0.5 t (u + v)$
$v 2 = u 2 + 2 a s$

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 $(s = u t)$ is derived from $s = u t + 0.5 a t 2$ with $a = 0$ and therefore $s = u t + 0$ 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 $s = u t + 0.5 a t 2$ .
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 $v = u + a t$ .
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 $v 2 = u 2 + 2 a s$ .
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 $s = 0.5 t (u + v)$ .
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 $F = k a$ where k is a constant.

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

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 $F = m a$
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

A ball is rolled 15m in 10s. How fast was it travelling?
A Car accelerates from rest to 25ms $- 1$ while travelling 15m. How long did this take?
An athlete accelerates from rest to 10ms $- 1$ in 1s. State the acceleration.
Calculate the acceleration created when a 10kg block has a force of 200N exerted on it.
Using the acceleration found above, how fast will a object travelling at 10ms $- 1$ be travelling after 10 seconds?
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.
An object's path is described by the function $t 2 - 16 t + 64$ . Find the velocity function and therefore the acceleration applied to the object.

[edit] Answers

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

[edit] Credits

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

Introduction to mechanics

From Wikiversity

Contents

[edit] Introduction

[edit] Definitions

[edit] Standard Units

[edit] Motion in a straight line with constant velocity

[edit] Newton's equations of motion

[edit] Applied force

[edit] Vector motion

[edit] Mechanics using calculus

[edit] Questions on introduction to mechanics

[edit] Questions

[edit] Answers

[edit] Credits

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