Question

A pan balance is made up of a rigid, massless rod with a hanging pan attached at each end. The rod is supported at and free to rotate about a point not at its center. It is balanced by unequal masses placed in the two pans. When an unknown mass mis placed in the left pan, it is balanced by a mass ${\mathit{m}}_{\mathbf{1}}$ placed in the right pan; when the mass mis placed in the right pan, it is balanced by a mass ${\mathit{m}}_{\mathbf{2}}$in the left pan. Show that$\mathit{m}\mathbf{}\mathbf{=}\sqrt{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}$

Short answer

$\mathrm{m}=\sqrt{{\mathrm{m}}_1{\mathrm{m}}_2}$ is proved.

Step-by-step solution

Understanding the given information

The unknown mass$\mathrm{m}$in the left pan is balanced with the mass${{\text{m}}_{\text{1}}}_{}$in the right pan.

The unknown mass $\text{m}$ in the right pan is balanced with the mass ${\text{m}}_{\text{2}}$ in the left pan.

Concept and formula used in the given question

You draw the free body diagram. The system is at equilibrium. For such a system, the vector sum of the forces acting on it is zero. The vector sum of the external torques acting on the pan at one point is zero. The formulas used are given below.

$\begin{array}{l}\mathbf{\sum }{\mathbf{F}}_{\mathbf{x}}\mathbf{=}\mathbf{0}\\ \mathbf{\sum }{\mathbf{F}}_{\mathbf{y}}\mathbf{=}\mathbf{0}\\ \mathbf{\sum }\mathbf{\tau }\mathbf{=}\mathbf{0}\end{array}$

Calculation for the  m =m1m2

This is the free body diagram when the unknown mass $\mathrm{m}$ is placed to the left and ${\text{m}}_{\text{1}}$ is placed to the right. Consider ${\mathrm{l}}_1$and ${\mathrm{l}}_2$ as lengths of the two arms of the balance as shown in the figure. The weight of the masses acts in the downward direction as shown in the figure. This is the static equilibrium condition for the system, hence the vector sum of the external torques acting on the pan at one point is zero.

$\begin{array}{c}\mathrm{\Sigma }{\overrightarrow{\mathrm{\tau }}}_{\mathrm{net}}=0\\ \mathrm{mg}\times {\mathrm{l}}_1-{\mathrm{m}}_1\mathrm{g}\times {\mathrm{l}}_2=0\\ \mathrm{mg}\times {\mathrm{l}}_1={\mathrm{m}}_1\mathrm{g}\times {\mathrm{l}}_2\\ {\mathrm{ml}}_1={\mathrm{m}}_1{\mathrm{l}}_2\left(1\right)\end{array}$

A similar condition can be applied when an unknown mass $\mathrm{m}$ is placed in the right pan and a mass ${\text{m}}_{\text{2}}$is placed in the left pan. Hence,

$\begin{array}{c}{\mathrm{m}}_2\mathrm{g}\times {\mathrm{l}}_1-\mathrm{mg}\times {\mathrm{l}}_2=0\\ {\mathrm{m}}_2\mathrm{g}\times {\mathrm{l}}_1=\mathrm{mg}\times {\mathrm{l}}_2\\ {\mathrm{m}}_2{\mathrm{l}}_1={\mathrm{ml}}_2\left(2\right)\end{array}$

Divide equation (2) by equation (1) as

$\begin{array}{c}\frac{{\mathrm{m}}_2{\mathrm{l}}_1}{{\mathrm{ml}}_1}=\frac{{\mathrm{ml}}_2}{{\mathrm{m}}_1{\mathrm{l}}_2}\\ {\mathrm{m}}^2={\mathrm{m}}_1{\mathrm{m}}_2\\ \mathrm{m}=\sqrt{{\mathrm{m}}_1{\mathrm{m}}_2}\end{array}$