# Any help on this statics problem?

This is the exact problem I'm having trouble on for my Building Tech II Class. Any help on where to begin would be great.

http://www.chegg.com/homework-help/questions-and-a...

Relevance

Let tension in each section be labeled with a T followed by the end points involved.

The horizontal forces about D must equal zero, assume right is positive direction

Tdesin75 - Tcd(4/5) = 0

Tde = 4Tcd/5sin75

Summing vertical forces about D to zero, up is assumed positive direction.

Tcd(3/5) + Tdecos75 - 200 = 0

substitute the result from the first equation

Tcd(3/5) + (4Tcd/5sin75)cos75 - 200 = 0

Tcd[(3/5) + (4/5sin75)cos75] = 200

Tcd = 245.6 lb < === ANS a

Tde = 4(245.6)/5sin75

Tde = 203.4 lb < === ANS a

Summing horizontal forces about C to zero

Tcd(4/5) - Taccos60 = 0

Tac = 245.6(4/5) / cos60

Tac = 393 lb < === ANS a

Summing vertical forces about C to zero, assume BC is in compression

Tbc - Tacsin60 - Tcd(3/5) = 0

Tbc = 393sin60 + 245.6(3/5)

Tbc = 488 lb < === ANS a

as this result is positive, our assumption about the pole being in compression is valid.

The highest tension exists in section AC. If we reduce its load to a maximum of 320 lb. the other forces will decrease in the same proportion.

320 / 393 = 0.814

Tac = 393(0.814) = 320 lb T

Tbc = 488(0.814) = 397 lb C

Tcd = 245.6(0.814) = 200 lb T

Tde = 203.4(0.814) = 165.5 lb T

W = 200(0.814) = 162.8 lb

As the pole stress is below its maximum compression stress of 450 lb, it holds as well