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Vectors II

Applications to mechanics: 3D vectors

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Further kinematics

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Constant acceleration formulae: SUVAT

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Vectors II

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Applications to mechanics: 3D vectors

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Summary

Applications to mechanics: 3D vectors

In a nutshell

3D vectors can be used to model problems in mechanics involving forces and kinematics.

3D vectors and mechanics

3D vectors can be used when modelling mechanics problems in a similar way to problems with 2D vectors. Finding position vectors, parallel and scalar multiples of vectors and the magnitude of a vector can all help solve mechanics problems with 3D vectors. This topic also requires knowledge of forces and kinematics as a prerequisite.

Example 1

A stationary $3kg$ object, is acted upon by three forces as follows:

$F_1 = (5\bold{i}+3\bold{j}-2\bold{k}) N$ , $F_2 = (-\bold{i}-4\bold{j}+\bold{k}) N$ and $F_3 = (6\bold{j}+4\bold{k}) N$

Find:

a) The resultant force.

b) The magnitude of the acceleration.

c) The distance travelled in $10 \ s$ .

d) The time taken to reach a velocity of $10\sqrt{2} \ m.s^{-1}$ .

a) The resultant force is the sum of the individual forces acting:

$\begin{aligned}F_r & = F_1+F_2+F_3 \\& = (5\bold{i}+3\bold{j}-2\bold{k}) + (-\bold{i}-4\bold{j}+\bold{k}) + (6\bold{j}+4\bold{k}) \\& = \underline{(4\bold{i}+5\bold{j}+3\bold{k}) \ N}\end{aligned}$

b) Use $\bold F=m \bold a$ to find the acceleration:

$\begin {aligned}\bold F& =m \bold a \\ \begin{pmatrix} 4 \\ 5 \\ 3 \end{pmatrix} &= 3 \bold a \\ \\\bold a &= \begin{pmatrix} \frac 4 3 \\\\ \frac 5 3 \\\\ 1 \end{pmatrix}\end{aligned}$ 

Find the magnitude:

$| \bold a | = \sqrt{{\frac 4 3}^2 + {\frac 5 3}^2 + 1^2 } = \underline{\dfrac {5\sqrt2}{3}\space ms^{-2}}$ 

c) To find the distance travelled, use the formulae for constant acceleration:

$s = \ ? \quad u = 0 \quad v = \ ? \quad a = \dfrac {5\sqrt2}{3} \quad t=10$

Use $s =ut+ \dfrac 1 2 a t^2$ :

$\begin{aligned}s &=ut+ \dfrac 1 2 a t^2 \\&= 0 + \dfrac 1 2 \times \dfrac {5\sqrt2}{3} \times 10^2 \\& =\underline{ 118 \ m \ (3 \ s.f.)}\end{aligned}$

d) To find the time taken, use $v=u+at$ :

$\begin{aligned} v&=u+at \\10 \sqrt2 &= 0 + \dfrac {5 \sqrt 2 }{3} \times t \\\\t &= \underline{6 \ s}\end{aligned}$

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