physics question

Question

A seesaw of mass 47.9 kg and total length 4.55 is allowed to pivot about an axis through its center. Sadly, the age and neglect of this seesaw is such that we can't ignore friction at the pivot point. Two children, with masses 25.9 kg and 29.9 kg, seat themselves at extreme opposite ends of the seesaw. They notice that when the seesaw makes an angle of 13.9 degrees with respect to the horizontal that they are motionless even when neither of them are in contact with the ground. What is the magnitude of the torque due to friction at the pivot point? Use units of N*m.

Answer 1

Hi Zachariah W.,

So, assume the see-saw is a simple bar of uniform mass-distribution along its length. It therefore balances, in and of itself (at any angle). The net effect of the kids is the difference of their masses, times g, exerted at 4.55/2 m distance, times a correction factor for the cos of that angle (since gravity isn't acting exactly perpendicular to the see-saw beam). That's balanced by the torque of the friction, right?

Reminds me of a method I saw in Lambertville, NJ at the town park playground to prevent kids from climbing onto the see-saw and jolting one another senseless from acceleration due to grossly unequal masses. The see-saw was simply lifted completely free of the pivot and laid onto the ground. Only an adult had the strength to lift it back into place (and would presumably have the sense to monitor, and restore to storage after!).

-- Cheers, -- Mr. d.