Truss/equilibrium homework help

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The discussion revolves around calculating the downward force exerted by a horizontal truss on two support posts. The truss has a mass of 15,000 kg and a total length of 22 meters. Participants suggest using torque calculations about each post to determine the force reactions accurately. The initial calculated forces were approximately 82,772 N for the right post and 64,378 N for the left post, which were later adjusted to 83,000 N and 64,000 N, respectively, for significant figures. The conversation emphasizes the importance of proper support placement in truss design, despite the calculations being valid.
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Homework Statement



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.

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The Attempt at a Solution



do i just take the force (weight and multiply it by distances?)

I have no clue

can someone give me a hint?
 

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please!>!>!>
 


do I calculate torque?
 


amyparker30 said:
do I calculate torque?
Yes, sum torques about one of the support posts = 0 to calculate the force reaction of the other post. Do the same thing in summing torques = 0 about the other post to calculate the force reaction of the first post. Check your work using Newton 1 (the equilibrium condition) in the y direction.

Note: In actuality, a truss should never be supported this way, the supports should be located at a joint. Doesn't change your answer. though.
 


do these answers make sense??

so F Right post= 82771.875 N

and F Left post = 64378.125 N
 


amyparker30 said:
do these answers make sense??

so F Right post= 82771.875 N

and F Left post = 64378.125 N
Almost spot on! :cool: I say almost, because there should be just 2 significant figures in your answer. Implying Force right post = 83000 N, and F Left post = 64000 N. Or something like that. At least, get rid of the decimal points! :smile:
 


thanks so much for your help!
 
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