Step 1

We are given a wire which is bent in a shape as shown in the given diagram. The dimensions of the bend wire are shown in the diagram.

We are required to locate the centroid $\left( {\overline x ,\overline y ,\overline z } \right)$ of the given bent wire.

 
Step 2

The diagram of the wire having 3 different side lengths represented by $1,2{\rm{ and 3}}$ is given below,

The side of the wire parallel with $x$ axis is represented by side 1, similarly the side of the wire parallel with $y$ axis is represented by side 2 and the side of the wire parallel with $z$ axis is represented by side 3.

 
Step 3

For side 1:

The length of the side 1 is given by,

\[{L_1} = 300{\rm{ mm}}\]

The expression for $\overline x $ component of the centroid for the side 1is given by,

\begin{array}{l} {\overline x _1} = \frac{{300{\rm{ mm}}}}{2}\\ {\overline x _1} = 150{\rm{ mm}} \end{array}

Similarly, the expression for $\overline y $ component of the centroid for the side 1 is given by,

\[{\overline y _1} = 0\]

The expression for $\overline z $ component of the centroid for the side 1 is given by,

\[{\overline z _1} = 0\]
 
Step 4

For side 2:

The length of the side 2 is given by,

\[{L_2} = 600{\rm{ mm}}\]

The expression for $\overline x $ component of the centroid for the side 2 is given by,

\[{\overline x _2} = 300{\rm{ mm}}\]

Similarly, the expression for $\overline y $ component of the centroid for the side 2 is given by,

\begin{array}{l} {\overline y _2} = \frac{{600{\rm{ mm}}}}{2}\\ {\overline y _2} = 300{\rm{ mm}} \end{array}

The expression for $\overline z $ component of the centroid for the side 2 is given by,

\[{\overline z _2} = 0\]
 
Step 5

For side 3:

The length of the side 3 is given by,

\[{L_3}{\rm{ = 40}}0{\rm{ mm}}\]

The expression for $\overline x $ component of the centroid for the side 3 is given by,

\[{\overline x _3} = 300{\rm{ mm}}\]

Similarly, the expression for $\overline y $ component of the centroid for the side 3 is given by,

\[{\overline y _3} = 600{\rm{ mm}}\]

The expression for $\overline z $ component of the centroid for the side 3 is given by,

\begin{array}{l} {\overline z _3} = \frac{{ - 400{\rm{ mm}}}}{2}\\ {\overline z _3} = - 200{\rm{ mm}} \end{array}
 
Step 6

Using the centroid formula, the expression for $\overline x $ component of the centroid of the bent wire is given by,

\begin{array}{l} \overline x = \frac{{\sum {\overline x L} }}{{\sum L }}\\ \overline x = \frac{{{{\overline x }_1}{L_1} + {{\overline x }_2}{L_2} + {{\overline x }_3}{L_3}}}{{{L_1} + {L_2} + {L_3}}}\\ \overline x = \frac{{\left( {{\rm{150}} \times 300{\rm{ m}}{{\rm{m}}^2}} \right) + \left( {300 \times 600{\rm{ m}}{{\rm{m}}^2}} \right) + \left( {300 \times 400{\rm{ m}}{{\rm{m}}^2}} \right)}}{{\left( {{\rm{300 mm}}} \right) + \left( {600{\rm{ mm}}} \right) + \left( {400{\rm{ mm}}} \right)}}\\ \overline x = 265.38{\rm{ mm}} \end{array}

Similarly, the expression for $\overline y $ component of the centroid of the bent wire is given by,

\begin{array}{l} \overline y = \frac{{\sum {\overline y L} }}{{\sum L }}\\ \overline y = \frac{{{{\overline y }_1}{L_1} + {{\overline y }_2}{L_2} + {{\overline y }_3}{L_3}}}{{{L_1} + {L_2} + {L_3}}}\\ \overline y = \frac{{\left( {0 \times 300{\rm{ m}}{{\rm{m}}^2}} \right) + \left( {300 \times 600{\rm{ m}}{{\rm{m}}^2}} \right) + \left( {600 \times 400{\rm{ m}}{{\rm{m}}^2}} \right)}}{{\left( {{\rm{300 mm}}} \right) + \left( {600{\rm{ mm}}} \right) + \left( {400{\rm{ mm}}} \right)}}\\ \overline y = 323.08{\rm{ mm}} \end{array}

The expression for $\overline z $ component of the centroid of the bent wire is given by,

\begin{array}{l} \overline z = \frac{{\sum {\overline z L} }}{{\sum L }}\\ \overline z = \frac{{{{\overline z }_1}{L_1} + {{\overline z }_2}{L_2} + {{\overline z }_3}{L_3}}}{{{L_1} + {L_2} + {L_3}}}\\ \overline z = \frac{{\left( {0 \times 300{\rm{ m}}{{\rm{m}}^2}} \right) + \left( {0 \times 600{\rm{ m}}{{\rm{m}}^2}} \right) + \left\{ {\left( { - 200} \right) \times 400{\rm{ m}}{{\rm{m}}^2}} \right\}}}{{\left( {{\rm{300 mm}}} \right) + \left( {600{\rm{ mm}}} \right) + \left( {400{\rm{ mm}}} \right)}}\\ \overline z = - 61.54{\rm{ mm}} \end{array}