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# 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.

### 1 Answer

- WhomeLv 77 years agoFavorite Answer
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