absurdly easy tension problem

[StephenPrivitera]

The book and I aren't getting along tonight. Maybe you can help.
A rope of length L has a tension T. Someone pushes on the rope with a force F at its midpoint and deflects the rope by a distance d. What is T is terms of L,d and F.
This is so simple I won't even explain my work.
2Tsinθ=F
sinθ=d/(L/2)=2d/L (approx)
so
T=\frac{FL}{4d}
Right?
The prob. in the book had numbers, but in the end I was off by a factor of 2.

[ShawnD]

Ok well this is what I'm getting. instead of using some greek letter for the angle, i'll use H.

forces balancing:
F = 2Tsin(H)
T = F/2sin(H)

length of rope used by the angle is L/2.
since L is the hypotinuse, sin(H) = d/(L/2)
sin(H) = 2d/L

sub that into the first equation:
T = F/2(2d/L)
T = F/(4d/L)
T = FL/4d

I get the same thing.

[himanshu121]

I too got the same value i.e. t=FL/4d [g)]

[StephenPrivitera]

BOOK's WRONG AHAHAHAHA!

[Doc Al]

Originally posted by StephenPrivitera
BOOK's WRONG AHAHAHAHA!
What book are you using? Seems like it has quite a few mistakes.

[StephenPrivitera]

I know. I've found at least four in the last two chapters, and I've verified these with my professor, so it's not just stupid Stephen being less smart than the author. The book is PHYSICS by Ohanian 2ed.