Question

A uniform bridge span weighs 50.0 x 10^3 N and is 40.0 m long. An automobile weighing 15.0 x 10^3 N is parked with its center of gravity located 12.0 m from the right pier. What upward support force does the left pier provide?

Answer 1

Answer:

(Left Force absolute value) Fl=29500N

Explanation:

when you consider the bridge static, you must also consider it is in static equilibrium, then you can use the torques to solve it as shown below:

$-Fl*(40m)-Wa*(12m)-Wb(20m)+Fr*(0m)=0$

where if you consider the torque from the right pier, then the distance of Fl is 40 m, the Wa(weight of the car) is 12 m from it, the Center of mass of the bridge Wb is 20 m from it, and eh Right Force pier is at 0m, so it just turns 0.

the possitive and negative signs, comes from the direction of your positive or negative torque respectively your torque point (in this case upward with the torque point placed at the right pier).

then you just have to replace the values and get Left Force:

$-Wa*(12m)-Wp*(20m)-Fl*(40m)+Fr*(0m)=0-Wa*(12m)-Wp*(20m)=Fl*(40m)\frac{-Wa*(12m)-Wp*(20m)}{40m}=Fl \frac{-15000N*12m-50000N*20m}{40m}=Fl-29500N=Fl$