Agymnast with mass m1 = 44 kg is on a balance beam that sits on (but is not attached to) two supports. the beam has a mass m2 = 117 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.
1)what is the force the left support exerts on the beam?
2)what is the force the right support exerts on the beam?
3)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.

4)what is the force the left support exerts on the beam?
5)what is the force the right support exerts on the beam?

Please refer to the figure showing the free body diagrams. 
We apply the law of conservation of momentum wherein the summation of momentum equals zero. Momentum is force times distance. Force equals mass times gravity, if mass is known. g=9.8 m/s2

Note: I will not indicate the units. Since all units are uniform and in SI, the answer would also be in SI units. 

1. What is the force the left support exerts on the beam?

Take the right support to be the fixed point (F3=F4=0). The equation would become



F2 = 945.7 N

2. What is the force the right support exerts on the beam? 

Take the point where the gymnast is standing (F1=0) to be the fixed point. The equation would become



F4 = 387.1 N

3. 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.

Take the the left beam to be the fixed point (F2=0). The equation would become



F4 = 1004.5 N,   this should be equal to the force exerted by the man so the beam won't tip.

The force exerted by man is,



The excess force the gymnast could carry would be 1004.5 N - 431.2 N = 573.3 N. Converting this to mass,

Extra mass = 573.3 N ÷ 9.8
Extra mass = 58.5 kg

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.