Agymnast with mass 46.0 kg stands on the end of a uniform balance beam as shown. the beam is 5.00 m long and has a mass of 250 kg (excluding the mass of the two supports). each support is 0.540 m from its end of the beam. what are the forces on the beam due to support 1 (f1) and support 2 (f2)?

Sum up the moments of the right and the left support:
∑ M2 = 0
F1 · 3.92 m = 46 kg · 9,81 m/s² · 4.46 m + 250 kg · 9.81 m/s² · 1.96 m
F 1 · 3.92 = 2012.6 + 4806.9
F 1 = 6819.5 : 3.92
F 1 = 1739.67 N
∑ M 1 = 0
F 2 · 3.92 m + 46 kg · 9.81 m/s² · 0.54 m = 250 kg · 9.81 m/s² · 1.96 m
F 2 · 3.92 + 243.68 = 4806.9
F 2 · 3.92 = 4563.22
F 2 = 4563.22 : 3.92
F 2 = 1164 N
The forces on the beam are: F 1 = 1739.67 N and F 2 = 1164 N.