Question

A lorry of weight 15000N is parked on a metal bridge of length 10m at a distance of 2m from one end. If the weight of the bridge is 5000, what are the reactions at the two ends of the bridge?

Answer 1

Let lorry is parked a distance of 2m from 1 end.

For the equilibrium:

$$R_1+R_2=W+F_L$$

$$R_2L=W(0.5L)+F_L(d)$$

$$10R_2=5000(5)+15000(2)$$

$$R_2=5500\ N$$

Thus,

$$R_1+5500=5000+15000$$

$$R_1=14500\ N$$