Figure 8.14 shows a square table of side 2 m which is supported by four legs, each 1 m high, situated at the corners of a 1m square. The mass of each leg is 1 kg and that of the table top 2 kg. The legs and the top are uniform. A small heavy object of mass 6 kg is placed on one corner of the table.

Question

Figure 8.14 shows a square table of side 2 m which is supported by four legs, each 1 m high, situated at the corners of a 1m square. The mass of each leg is 1 kg and that of the table top 2 kg. The legs and the top are uniform. A small heavy object of mass 6 kg is placed on one corner of the table.
i) Find the position of the centre of mass relative to the point O shown in the diagram using the axes indicated with unit length 1 m.
ii) What is the significance of your result?

Accepted Answer

i) Let the co-ordinates of the centre of mass be [latex](\bar{x} , \bar{y} , \bar{z})[/latex].

You can put the information in a table and then write the equations in vectors as follows, or deal with each co-ordinate separately.

	leg A	leg B	leg C	leg D	table top	object	whole
mass	1	1	1	1	2	6	12
x	0	1	1	0	[latex]\frac{1}{2}[/latex]	1[latex]\frac{1}{2}[/latex]	[latex]\bar{x}[/latex]
y	0	0	1	1	[latex]\frac{1}{2}[/latex]	1[latex]\frac{1}{2}[/latex]	[latex]\bar{y}[/latex]
z	[latex]\frac{1}{2}[/latex]	[latex]\frac{1}{2}[/latex]	[latex]\frac{1}{2}[/latex]	[latex]\frac{1}{2}[/latex]	1	1	[latex]\bar{z}[/latex]

Taking moments about O:

[latex]12 × \left(\begin{matrix} \bar{x} \ \bar{y} \ \bar{z} \end{matrix} ight)=\overset{A}{\left(\begin{matrix} 0 \ 0 \ \frac{1}{2} \end{matrix} ight) } + 1 \overset{B}{\left(\begin{matrix} 1 \ 0 \ \frac{1}{2} \end{matrix} ight) } + 1 \overset{C}{\left(\begin{matrix} 1 \ 1 \ \frac{1}{2} \end{matrix} ight) } + 1 \overset{D}{\left(\begin{matrix} 0 \ 1 \ \frac{1}{2} \end{matrix} ight) } + 2\left(\begin{matrix} \frac{1}{2} \ \frac{1}{2} \ 1 \end{matrix} ight) + 6 \left(\begin{matrix} 1 \frac{1}{2} \ 1 \frac{1}{2} \ 1 \end{matrix} ight) [/latex]

⇒ 12 [latex]\left(\begin{matrix} \bar{x} \ \bar{y} \ \bar{z} \end{matrix} ight)= \left(\begin{matrix} 12 \ 12 \ 10 \end{matrix} ight)[/latex]

⇒The centre of mass is at (1, 1, 0.83).

ii) This means that the centre of mass is in the leg nearest the object and so the table is on the point of toppling over. All the weight is being taken by that one leg.

Question 8.7: Figure 8.14 shows a square table of side 2 m which is suppor...

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