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Tensions in cable

  • Thread starter aftershock
  • Start date
  • #1
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Homework Statement



http://img163.imageshack.us/img163/7491/staticsk.jpg [Broken]

Those weird squiqqles are supposed to represent springs. Not exactly an artist :D

Homework Equations



F = kx

The Attempt at a Solution



After drawing a fbd and breaking the tension into x and y components I got:

ΣFy T(sinθ) =5

ΣFx T(cosθ) = T(cosθ)

T = 5x since the force acting on the spring is T and k is given as 5.

I tried taking the given 4ft into consideration, I tried seeing if there was some substitution I could make to solve any of those equations. Been struggling for like an hour now...
 
Last edited by a moderator:

Answers and Replies

  • #2
1,955
252
If the tension in the spring is kx, then the length of the spring is x+4 ft, because there's no tension if the spring is 4 ft long.

Try to find another expression for [itex] sin(\theta) [/itex] or [itex] cos(\theta) [/itex] just from the goemetry.
 
  • #3
110
0
If the tension in the spring is kx, then the length of the spring is x+4 ft, because there's no tension if the spring is 4 ft long.

Try to find another expression for [itex] sin(\theta) [/itex] or [itex] cos(\theta) [/itex] just from the goemetry.
I should've been more clear. I used x to denote the amount the spring is stretched.. not the total length of the spring.

And I have been trying for a while now. The only thing I can manage to do is sin/cos to get tan but I haven't done that in anyway that helps me out.
 
  • #5
1,955
252
you get a right-angled triangle with corners middle of the block, one end of the block, attachment to the load
 

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