Answer to Question #122785 in Math for Felix Coleman

Question

A number of particles with masses m1, m2, m3, · · · , mn are situated at the points with position vectors
r1, r2, r3, · · · , rn relative to an origin O. The center of mass G of the particles is defined to be the
point of space with position vector
R =
m1r1 + m2r2 + m3r3 + · · · + mnrn
m1 + m2 + m3 + · · · + mn
.
Show that if a different origin O0 were used, this definition would still place G at the same point of
space.
(b) An object of mass 40kg is supported in equilibrium by four cables. The forces, in Newtons, exerted
by three of the cables, F1, F2 and F3, are given in terms of the unit vectors, i, j and k as F1 =
80i + 20j + 100k, F2 = 60i − 40j + 80k and F3 = −50i − 100j + 80k. The unit vectors i and j are
perpendicular and horizontal and the unit vector k is vertically upwards.
i. Find F4, the force exerted by the fourth cable, in terms of i, j and k. Also find its magnitude to
the nearest Newton.
ii. Find the angle between F1 and F4.

Answer 1

(a) Given in the system with origin $O$

$\overline{r}_G=\dfrac{\displaystyle\sum_{i=1}^nm_i\overline{r}_{i}}{\displaystyle\sum_{i=1}^nm_i}=\dfrac{m_1\overline{r}_1+m_2\overline{r}_2+...+m_n\overline{r}_n}{m_1+m_2+...+m_n}$

In new system with origin $O'$

$\overline{r}_G'=\dfrac{\displaystyle\sum_{i=1}^nm_i\overline{r}_{i}'}{\displaystyle\sum_{i=1}^nm_i}=\dfrac{m_1\overline{r}_1'+m_2\overline{r}_2'+...+m_n\overline{r}_n'}{m_1+m_2+...+m_n}$

We have that for new system

$\overline{r}_G'=\overline{r}_G+\overline{OO'}$ $\overline{r}_1'=\overline{r}_1+\overline{OO'}$ $\overline{r}_2'=\overline{r}_2+\overline{OO'}$ $...$ $\overline{r}_n'=\overline{r}_n+\overline{OO'}$

Then

$\overline{r}_G'=\dfrac{\displaystyle\sum_{i=1}^nm_i\overline{r}_{i}'}{\displaystyle\sum_{i=1}^nm_i}=\dfrac{m_1\overline{r}_1'+m_2\overline{r}_2'+...+m_n\overline{r}_n'}{m_1+m_2+...+m_n}=$

$=\dfrac{m_1(\overline{r}_1+\overline{OO'})+m_2(\overline{r}_2+\overline{OO'})+...+m_n(\overline{r}_n+\overline{OO'})}{m_1+m_2+...+m_n}=$

$=\dfrac{m_1\overline{r}_1+m_2\overline{r}_2+...+m_n\overline{r}_n}{m_1+m_2+...+m_n}+$

$+\dfrac{m_1+m_2+...+m_n}{m_1+m_2+...+m_n}\overline{OO'}=$

$=\dfrac{m_1\overline{r}_1+m_2\overline{r}_2+...+m_n\overline{r}_n}{m_1+m_2+...+m_n}+\overline{OO'}=$

$=\overline{r}_G+\overline{OO'}$

The center of mass G of the particles is at the same point of space.

(b) Given

$\overline{F}_1=80\overline{i}+20\overline{j}+100\overline{k},$

$\overline{F}_2=60\overline{i}-40\overline{j}+80\overline{k},$

$\overline{F}_3=-50\overline{i}-100\overline{j}+80\overline{k}.$

The force of gravity is $\overline{F}_G=m\overline{g}=-40\cdot9.8\overline{k}=-392\overline{k}$

Let $\overline{F}_4=F_{4x}\overline{i}+F_{4y}\overline{j}+F_{4z}\overline{k}.$

i.

An object of mass 40kg is supported in equilibrium

$\overline{F}_1+\overline{F}_2+\overline{F}_3+\overline{F}_4+\overline{F}_G=0$

Substitute

$80\overline{i}+20\overline{j}+100\overline{k}+60\overline{i}-40\overline{j}+80\overline{k}+$

$+(-50\overline{i}-100\overline{j}+80\overline{k})+F_{4x}\overline{i}+F_{4y}\overline{j}+F_{4z}\overline{k}+$

$+(-392\overline{k})=0$

$80+60-50+F_{4x}=0=>F_{4x}=-90$

$20-40-100+F_{4y}=0=>F_{4y}=120$

$100+80+80+F_{4z}-392=0=>F_{4z}=132$

$\overline{F}_4=-90\overline{i}+120\overline{j}+132\overline{k}.$

$|\overline{F}_4|=\sqrt{(-90)^2+(120)^2+(132)^2}=\sqrt{39924}=$

$=6\sqrt{1109}\approx200(N)$

ii.

$\overline{F}_1=80\overline{i}+20\overline{j}+100\overline{k},$

$\overline{F}_4=-90\overline{i}+120\overline{j}+132\overline{k}.$

$\overline{F}_1\cdot\overline{F}_4=|\overline{F}_1||\overline{F}_4|\cos \theta$

$\theta=\arccos{\dfrac{\overline{F}_1\cdot\overline{F}_4}{|\overline{F}_1||\overline{F}_4|}}$

$\overline{F}_1\cdot\overline{F}_4=80(-90)+20(120)+100(132)=8400$

$|\overline{F}_1|=\sqrt{(80)^2+(20)^2+(100)^2}=\sqrt{16800}=$

$=20\sqrt{42}(N)$

$\theta=\arccos{\dfrac{8400}{20\sqrt{42}(6\sqrt{1109})}}=\arccos{70\over \sqrt{46578}}\approx71\degree$

The angle between F1 and F4 is approximately $71\degree .$