A horizontal truss, symmetric around its center, of mass 15,000 kg and total length 22m, is placed on two posts as shown in the figure .
Find the downward force that it exerts on each post.

http://s1168.photobucket.com/albums/r497/amy_parker1/?action=view&current=Untkkitled.jpg

Suggest you start by taking the sum of moments about one of the supports. Consider the left support: the weight of the truss can be modelled as a 15000 Kg mass with a center of gravity that is 9 meters from the left support. The moment produced by that weight is counteracted by an upward force from the right support, which is 16m from the left. So:


\Sigma M = 0 = 15000 kg \ \times g \ \times 9 m - F_R \times 16 m


This gives you the reaction force at the right support in newtons. Then use


\Sigma F = 0


to find the reaction force at the left support.

So the they would be the same?

so the they would be the same?

No. Since the truss is shifted toward the right support you should find that the reaction force from the right support is greater than the left. To clarify: since


\Sigma F = 0


means that


F_R + F_L = mg

Ohh I see, thank you so much for you help

do these answers make sense?

so F Right post= 82771.875 N

and F Left post = 64378.125 N

do these answers make sense??

so F Right post= 82771.875 N

and F Left post = 64378.125 N

Yes, but...

Do not take your calculation to 8 significant digits - the data you were given is good to only 2 digits. So you should properly show the answers as 8.3 x 10^4 N and 6.4 x 10^4 N.

Thank you such a life savor! This question just confused me so much!