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202207081631 Statics Figures (Elementary) Q1

In a lift, the panel screen displays signs on an array of seven bars of light-emitting diodes (LEDs):

such that the numeric digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 when lit up are as follow:

For each configuration of lit-up cells, assuming that the mass distribution is uniform and the plate thickness negligible, find the centre of mass (CM), and hence the centre of gravity (CG).

Intuition.

Try arguing \textrm{\scriptsize{NOT}} by symmetry in mathematics, \textrm{\scriptsize{BUT}} by moment in physics.

We state without proof the centre of mass (CM) in each cell is as shown below:

For N particles in 2-D:

\begin{aligned} x_{\textrm{CM}} & = \frac{m_1x_1+m_2x_2+\cdots +m_Nx_N}{m_1+m_2+\cdots +m_N} = \frac{\sum_{i=1}^{N}m_ix_i}{M} \\ y_{\textrm{CM}} & = \frac{m_1y_1+m_2y_2+\cdots +m_Ny_N}{m_1+m_2+\cdots +m_N} = \frac{\sum_{i=1}^{N}m_iy_i}{M} \\ \end{aligned}

Define the center at the origin O(0,0), and let \mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3, \mathbf{r}_4, \mathbf{r}_5, and \mathbf{r}_6 be the vectors pointing to the CM of each cell:

such that

\begin{aligned} \mathbf{r}_1 & = 2\,\hat{\mathbf{j}} \\ \mathbf{r}_2 & = -1\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \mathbf{r}_3 & = 1\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \mathbf{r}_4 & = -1\,\hat{\mathbf{i}} -1\,\hat{\mathbf{j}} \\ \mathbf{r}_5 & = -2\,\hat{\mathbf{j}} \\ \mathbf{r}_6 & = 1\,\hat{\mathbf{i}} -1\,\hat{\mathbf{j}} \\ \end{aligned}

To illustrate how to get the CM for the number signs, we first begin with 1:

\mathbf{1}_{\textrm{CM}} = (x_{\textrm{CM,1}},y_{\textrm{CM,1}}).

\begin{aligned} \mathbf{1}_{\textrm{CM}} & = \frac{m\mathbf{r}_3+m\mathbf{r}_6}{m+m} \\ & = \frac{1}{2}\mathbf{r_3} + \frac{1}{2}\mathbf{r_6} \\ & = \frac{1}{2}(1,1) + \frac{1}{2}(1,-1) \\ & = (1,0) \\ \end{aligned}

The remaining are left the reader as an exercise.