1. ## statics

a particle of weight w is attached to two light inextensible strings each of length 40cm. the other ends of the strings are attached to two points on the same horizontal ceiling 64 cm apart. find the tension of the strings in terms of w

so i am trying to solve this resolving forces vertical and horizontal and then equaling them to each other. but its not working out the answer is 5/6 w. i can post my method is people need it

2. ## Re: statics

You have an isosceles triangle with sides of length 40, 40, and 64. Dropping a perpendicular divides that into two right triangles with hypotenuse of length 20 and horizontal leg of length 32 so the vertical leg has length $\sqrt{32^2- 20^2}= \sqrt{624}= 4\sqrt{39}$. Taking the tension in each string to be 'T', the vertical and horizontal components are $\frac[4\sqrt{39}}{40}T= \frac{\sqrt{39}}{10}T$ and $\frac{32}{40}T= \frac{4}{5}T$. The total of the two vertical components is $\frac{\sqrt{39}}{5}T$ and that must be equal to the weight, w. That is, $T= \frac{5}{\sqrt{39}}w$, not 5/6 w, though it is pretty close, 0.801w as compared to 5/6w= 0.833w.

3. ## Re: statics

Originally Posted by HallsofIvy
You have an isosceles triangle with sides of length 40, 40, and 64. Dropping a perpendicular divides that into two right triangles with hypotenuse of length 20 and horizontal leg of length 32 so the vertical leg has length $\sqrt{32^2- 20^2}= \sqrt{624}= 4\sqrt{39}$. Taking the tension in each string to be 'T', the vertical and horizontal components are $\frac[4\sqrt{39}}{40}T= \frac{\sqrt{39}}{10}T$ and $\frac{32}{40}T= \frac{4}{5}T$. The total of the two vertical components is $\frac{\sqrt{39}}{5}T$ and that must be equal to the weight, w. That is, $T= \frac{5}{\sqrt{39}}w$, not 5/6 w, though it is pretty close, 0.801w as compared to 5/6w= 0.833w.
I'm don't think this is correct.

You do have a 40,40,64 isosceles triangle but splitting that in two gets you a 32,24,40 triangle

4. ## Re: statics

I get the vertical leg has length $\sqrt{40^2-32^2} = \sqrt{576}=24$. Thus the vertical components are $\dfrac{24}{40}T = \dfrac{3}{5}T$. Thus, $\dfrac{6}{5}T = w \Longrightarrow T = \dfrac{5}{6}w$. I think the book's solution is correct.

5. ## Re: statics

$2T\sin{\theta} = W$

$2T \cdot \dfrac{3}{5} = W$

$T = \dfrac{5W}{6}$

6. ## Re: statics

Thanks i was doing it wrong