# Tensions in cable

## Homework Statement

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

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

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...

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## Answers and Replies

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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 $sin(\theta)$ or $cos(\theta)$ just from the goemetry.

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 $sin(\theta)$ or $cos(\theta)$ 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.

Anybody?

you get a right-angled triangle with corners middle of the block, one end of the block, attachment to the load