Question

a uniform bridge span weighs 50x10^3 and is 40m long. A car weighing 15X10^3 is parked with its center of gravity located 12m from the right pier. what upward support force does the left pier provide?

im lost....this is what i think im supposed to do
40m-12m= 28m to left pier
50.0x10^3- (40m/28m)*15.0x10^3=
28.5x10^3? is this right if not what am i doing wrong please help!

Answer 1

Are the weights in Newtons? It isn't enough to just provide a number. I will assume the bridge weighs 50*10^3 N and the car weighs 15*10^3 N.

Let F1 be the force on the lest pier and F2 be the force on the right pier. One of two equations that you would need to solve for both forces is

F1 + F2 = (50+15)*10^3 = 65*10^3 N

Now write another equation for equilbrium of moments about the right pier. There is a clockwise moment of F1*40 m and two conterclockwise moments of 50*10^3N*20 m due to the weight of the bridge and 15*10^3N*28m. The sum of these moments (with a negative sign for couterclockwise moments)is zero. Use that fact to solve for F1
40 F1 - 50*10^3*20 - 15*10^3*28+0

Solve for F1.

I get 35,500 N, a bit more than half the total weight.