Calculating Bridge Pillar Forces: Solving a Seesaw Problem

[jjjsarnis]

the problem reads:

a uniform bridge 20 m long and weighing 400,000 N is supported by two pillars located 3 m from each end. if a 19,600 N car is parked 8 m from one end of the bridge, hoe much force does each pilliar exert?

i was thinking about using something like (W1)(d1) = (W2)(d2) that we used early for a seesaw problem. also if the sum of forces could be added up to equal zero and place it in equilibrium.

thanks for any help

 

[Parth Dave]

Well you're sort of on the right track. First of all, the net force would be zero. This is because the system is in static equilibrium. I'm not sure which seesaw problem you're referring to but (W1)(d1) = (W2)(d2) would not help. If a system is in static equilibrium, there are two things that must be true. First of all, the net force is zero. What is the other one?

 

[Curious3141]

Consider the forces acting on the bridge. You should have the weight of the car and two normal reaction forces (one each from each of the pillars). Make a simple equatio relating those three forces. Now consider moments about one of the pillars (it doesn't matter which, just pick one). Make another equation for the balance of the moments. Now you have two simple simultaneous equations in two variables (the two reaction forces). Solve for them and you have your answer.

 

[jjjsarnis]

thanks guys, the other thing in static equilibrium would be net torque would equal zero but i didnt think there was any torque in this sytem, thanks again for your help

 

[Parth Dave]

Great job!