Answer to Question #86807 in Mechanics | Relativity for Eric

Question

a 1000 kg truck is parked on a bridge 100 m long at a distance of 25 m from one end of the bridge. the mass of the bridge is 200 kg m. calculate the reaction forces at the supports at the two ends of the bridge. (take g 10 ms )

Answer 1

The equilibrium conditions are as follows: R₂L = mga + 0.5MgL, R₁L = mg (L-a) + 0.5MgL and R₁ + R₂ = mg + Mg, from which follows R₂ = mga/L + 0.5MgL, R₁ = mg (1-a/L) + 0.5Mg, where M = ρL, therefore, R₁ = 10 (250 + 100*100) = 102500 (N), R₂ = 10 (750 + 100*100) = 107500 (N).

Finally, R₁ = 102500 N, R₂ = 107500 N.