Answer:
76440 N.
Explanation:
Given that the length of uniform bridge is 8 m, and the mass of this bridge is 15000 kg.
And the car of mass 1200 kg is position at a distance of 6 m from the left side of the bridge.
Now according the figure W(c) force by the car, W is the force by the bridge on the center of mass, and N is the normal force on the left side of bridge.
Now the normal torque should be balance with respect to the torques on the bridge.
[tex]W(c)\times 2 + W\times 4=N\times 8[/tex]
Now put all the given values.
[tex]8N=1200\times 9.8(2)+15000\times 9.8(4)\\8N=(2400+60000)\times 9.8\\N=\frac{62400}{8}\times 9.8\\ N=7800\times 9.8 N[/tex]
Therefore,
[tex]N=7800\times 9.8N\\N=76440 N[/tex]
Therefore, the normal force on the left side of the bridge is 76440 N.
