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−1. Calculate the reaction forces at the supports at the two ends of the bridge. (Take g = 10 ms−2)

Answer 1

The mass of the bridge is

$$M=200(100)=20000 kg$$

For the equilibrium:

$$F_1+F_2=Mg+mg=(M+m)g$$

Taking moments about one end of the bridge:

$$mg(25)+Mg\frac{100}{2}=F_2(100)$$

$$1000(10)(25)+(20000)(10)\frac{100}{2}=F_2(100)$$

$$F_2=102500N=102.5 kN$$

$$F_1=(M+m)g-F_2=1000(10)+(20000)(10)-102500$$

$$F_1=107500 N=107.5 kN$$