In the Cavendish balance apparatus shown in Figure, suppose that m₁ = 1.10 kg, m₂ = 25.0 kg, and the rod connecting the m₁pairs is 30.0 cm long. If, in each pair, m₁ and m₂ are 12.0 cm apart center to center, find (a) the net force and (b) the net torque (about the rotation axis) on the rotating part of the apparatus. (c) Does it seem that the torque in part (b) would be enough to easily rotate the rod? Suggest some ways to improve the sensitivity of this experiment.

• The mass of$m_1=1.10\mathrm{kg}$
• The mass of$m_2=25\mathrm{kg}$
• The distance between the masses,$r=12\mathrm{cm}$
• The length of the connecting rods, $l=30\mathrm{cm}$

The net force refers to the gravitational force given by the Newton’s law of Gravitation.

The Net force,

$\begin{array}{r}{\mathrm{F}}_{\mathrm{g}}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}\\ =\frac{\left(6.67\ast {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\ast \left(1.10\mathrm{kg}\right)\ast \left(25\mathrm{kg}\right)}{\left(0.120{\mathrm{m}}^2\right)}\\ =\frac{183.425\ast {10}^{-11}\mathrm{N}}{0.0144{\mathrm{m}}^2}\\ =12,737.84\ast {10}^{-11}\mathrm{N}\\ =1.273\ast {10}^{-7}\mathrm{N}\end{array}$

But the net force here is Zero, because the force directed in the opposite directions.

The net torque isgiven by the formula,

$\begin{array}{r}\tau =2F_{g}l\\ =2\ast 1.27\ast {10}^{-7}\mathrm{N}\ast 0.30\mathrm{m}\\ =0.762\ast {10}^{-7}\mathrm{Nm}\\ =7.62\ast {10}^{-8}\mathrm{Nm}\end{array}$

The net torque is ,$\tau =7.62*{10}^{-8}\mathrm{Nm}$

From Part b), it can be clearly seen that the torque is very small, which implies that the equipment is exceedingly sensitive.

To improve the sensitivity of the experiment, we can either increase the mass of the objects or we can reduce the separation between the masses m₁ and m₂.