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Applications to mechanics: 3D vectors

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 3kg3kg object, is acted upon by three forces as follows:

F1=(5i+3j2k)NF_1 = (5\bold{i}+3\bold{j}-2\bold{k}) N, F2=(i4j+k)NF_2 = (-\bold{i}-4\bold{j}+\bold{k}) N and F3=(6j+4k)NF_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 s10 \ s.

d) The time taken to reach a velocity of 102 m.s110\sqrt{2} \ m.s^{-1}.


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

Fr=F1+F2+F3=(5i+3j2k)+(i4j+k)+(6j+4k)=(4i+5j+3k) N\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 F=ma\bold F=m \bold a to find the acceleration:

F=ma(453)=3aa=(43531)\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:

a=432+532+12=523 ms2| \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= ?u=0v= ?a=523t=10 s = \ ? \quad u = 0 \quad v = \ ? \quad a = \dfrac {5\sqrt2}{3} \quad t=10


Use s=ut+12at2s =ut+ \dfrac 1 2 a t^2:

s=ut+12at2=0+12×523×102=118 m (3 s.f.)\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+atv=u+at:

v=u+at102=0+523×tt=6 s\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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