Question

In the Cavendish balance apparatus shown in Figure 13.4 in the textbook, suppose that m1 = 1.00 kg, m2 = 27.0 kg, and the rod connecting each of the pairs of masses is 28.0 cm long. Once the system reaches equilibrium, each pair of masses, m1 and m2, are separated by a distance of 12.0 cm center-to-center.

A) Find the magnitude (in N) of the net force on one of the small masses, m1

B) Find the gravitational torque (about the rotation axis) on the rotating part of the apparatus. Ignore any forces on the connecting rods in your calculation (answer in Nm)

C) Suggest some ways to improve the sensitivity of this experiment.

Answer 1

G = 6.67* 10^-11 N m /kg

The moment arm for the torque due to each force is 0.140 m

Fb = 6.67* 10^-11 N m /kg *1*27*/0.120^2m = 1.25*10^-7

For each pair of spheres, we see that the forces for each pair are in opposite directions, so net F net = 0

(b) The net torque is  rnet = 2(1.25*10^-7N)(0.140 m) = 3.5 *10^-8 N m

(c) The torque is very small and the apparatus must be very sensitive. The torque could be increased by increasing the mass of the spheres or by decreasing their separation