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:
[tex]-Fl*(40m)-Wa*(12m)-Wb(20m)+Fr*(0m)=0[/tex]
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:
[tex]-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[/tex]
The Force which is required for the rotation of the object about an axis is called torque force.
The upward support force provided by the left pier is 29500 N.
Given that the total weight of the bridge is [tex]50\times 10^3 \;\rm N[/tex] and its length is 40 m.
The upward force can be calculated by considering the bridge in equilibrium. In equilibrium, the total force is equal to zero.
[tex]F_1 +F_2 -W_1 -W_2 = 0[/tex]
Where F1 is the force at bridge, F2 is the force at the right pier which is at zero meter, W1 is the weight of the automobile parked at a distance of 12 m and W2 is the weight at the center of the gravity which is at a distance of 20 m.
The upward support force is calculated by the sum of the total torque force at the bridge.
[tex]F_1\times 40 + F_2 \times 0 - 15\times10^3 \times 12 -50\times 10^3\times 20[/tex]
[tex]F_1\times = 15\times 10^3\times 12 + 50\times 10^3\times 20[/tex]
[tex]F_1 = 29500 \;\rm N[/tex]
Hence we can conclude that the upward support force at the left pier is 29500 N.
To know more about the torque force, follow the link given below.
https://brainly.com/question/18992494.
