A 14,000 kg bridge of length 16 m is supported at both ends. If a 1500 kg car is parked on the bridge 3.5 m?

Question

A 14,000 kg bridge of length 16 m is supported at both ends. If a 1500 kg car is parked on the bridge 3.5 m from the left support, what are the supporting forces at the left and right ends?

-----N (left end)

-----N (right end)

Answer 1

FL + FR = 14,000+1500 = 15,500

FL*16 = 1500*(16-3.5) + (14,000/2)*16

FL = 130,750/16

FL = 8,171.875 kg

FR = 7,328.125 kg

Answer 2

The weight of the bridge exerts equal pressure on each support.  14,000/2 x9.8 = 7,000 x9.8 = 68,600 Newtons.



the car exerts a total force of 1500 kg x 9.8 = 14,700 Newtons

the question realy comes down to how the cars weight is distributed between the ends. I believe it to be inversly proportionally to the distance. (the closer the more weight)

from the left side it is 3.5 meters, from the right 12.5 a ratio of  7:25     7+ 25 = 32    14,700/32 =    459.375 N

7 x 459.375 = 3,215.6 N ; 25 x 459.375 = 11,484.4 N

left side =  68,600 +11,484.4 N = 80,084.4N

right side = 68,600 +3,215.6 N =  71,815.6 N