Question

A gymnast with mass m1 = 49 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 120 kg and length L = 5 m. Eachsupport is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.

1)What is the force the left support exerts on the beam?
2)How much extra mass could the gymnast hold before the beam begins to tip?
3)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?
4)What is the force the right support exerts on the beam?

Answer 1

Let N1 be the force exerted by the left support and N2 be the force exerted by the right support.

The gymnast exerts a force of 49*g = 490N and the beam exerts a force of 120g = 1200N.

1) Balancing forces, 490 + 1200 = N1 + N2
=>1690 = N1 + N2 .........................................(1)

Balancing moments about the point where N2 is acting...

490*(10/3) - N1*(5/3) + 1200*(5/2 - 5/3) = 0

=> N1 = 1580N

2) for the beam to tip, N2 = 0
=> N1 = 1690N

let m be the extra mass

(490 + mg) (10/3) - 1690*(5/3) + 1200 (5/2 - 5/3) = 0

=> mg = 550
=> m = 55kg

3) Now the gymnast is standing above the right support.

Balancing moments about that point again

-N1*(5/3) + 1200*(5/2 - 5/3) = 0
=> N1 = 600N

4) N2 = 1690 - N1 = 1090N ....................................from (1)

Answer 2