Question

A gymnast with mass m₁ = 48 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m₂ = 124 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.

*How much extra mass could the gymnast hold before the beam begins to tip?

Now the gymnast (not holding any additional mass) walks directly above the right support.

*What is the force the left support exerts on the beam?

*What is the force the right support exerts on the beam?

At what location does the gymnast need to stand to maximize the force on the right support?

_at the center of the beam

_at the right support

_at the right edge of the beam

Answer 1

Gurcen data:- m = 4869 m₂ = 124 kg (a) The beam starts to tip when FR =0 0-1 (in, m') > m = mah om - m - m - m - m. .: sm, + lay - 48 (om, = 14 kg (6) when the system is at rest the systemetre o=fret - ft FR - m,g-mag. force of the left {=?. Fa (4m, +mg. F = + (4648) + 124) (9.8). TF₂ = 1548.4 N) ( Force of the right: FR = 1 (my -2m, 29 fra (1a4-a(48))C9-8) FR = 134.2 N (6) At the right edge of the beam.