Question

A uniform bridge 20.0m long and weighing 4.00 x 10^5 N is supported by two pillars located 3.00m from each end. If a 1.96 x 10 ^4N car is parked 8.00m from one end of the bridge. How much force does each pillar exert?

Answer 1

We have 2 conditions of balance. First of all, the total forces have to have a net sum of 0 since the bridge is balancing. Hence, if we denote by F1 the force of the first pole, F2 the power of the second pole (the one closer to the car), W the weight of the bridge and w the weight of the car, we have that W+w=F1+F2=4.196*10^5 N.

We also have that it does not rotate. Hence, taking as origin of our frame of reference the car, we have that 5*F2+4*W=9*F2 by calculating the distances from our point of reference. Thus yields 5F2+8*10^5=9F2. When we solve the system of equations that is created above (best way here is by substitution), we get that F1=2.07*10^5 N while F2=2.126*10^5 N . Each pole takes up around half the weight but due to the car the pole closer to it has more weight to bear; nevertheless the car does not weigh a lot so the difference is small.