How Much Force Do Pillars Exert with a Parked Car on a Bridge?

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The discussion centers on calculating the force exerted by two pillars supporting a bridge with a parked car. The bridge is 20 m long and weighs 4.00x10^5 N, while the car weighs 1.96x10^4 N and is parked 8 m from one end. The initial calculation provided is incorrect due to misidentified distances from the pivot point and the normal force of the pillars. Participants emphasize the importance of accurately determining distances and suggest drawing a diagram for clarity. Correct calculations for each pillar will yield different force values.
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Homework Statement


A uniform bridge 20 m long and weighing 4.00x10^5N is supported by two pillars located 3 m from each end. If a 1.96x10^4 N car is parked 8 m from one end of the brige, how much force does each pillar exert?


Homework Equations


(Fgb*L/2+Fgp*d)/L(sin theta).


The Attempt at a Solution


4x10^5*10+1.96x10^4*8/20(sin 90)=207840
That's what I got as an answer. I don't know if this is correct though. Any help is appreciated. Thank you.
 
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No you don't have all of your distances right. The car is 8 m from one end, but how far is it from the pivot point you've chosen (one of the pillars)? Also, the distance of the normal force of the pillar to the pivot point is not 20 m. Draw a diagram and label everything carefully.

Repeat for the other pillar, the answer will be different.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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