# Vector problem

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• Jul 8th 2012, 05:31 PM
astartleddeer
Vector problem
Hello,

I'm having trouble answering the following question. The answer (s) is a given, but I still can't derive the end result. The chapter this question's at in my book, I think the author may want you to return to it with more knowledge after the next few chapters, but I am too stubborn to move on until I have solved the question.

Anyway:

A load of 4.32 N is lifted by two strings making angles of $\displaystyle 18^\tiny o$ and $\displaystyle 30^ \tiny o$ with the vertical. If for this system the vectors representing the forces form a closed triangle when in equilibrium, calculate the tensions in the strings.

Answer [1.8 N, 2.91 N ]

Now, I've drawn this to scale and it is a scalene triangle inside two right angled triangles. I've made the assumption in that the vertical line will have a tension of 4.32 N because $\displaystyle \sin (90) = 1$. However, I can't see how to set up the equations in order to use the cosine or conversely the sine rule. My best attempt is the following (using trignometric ratios in order to solve for one of the tensions in the ropes which corresponds to 18 degrees):

Let x, y and z equal a decrease in the adjacent, hypotonuse and opposite side respectively.

Then:

$\displaystyle \cos (\theta) = \frac {A}{H}$

$\displaystyle \cos (18) = \frac {4.32 - x}{\left (\frac {4.32}{\cos (18)}\right) - y}$

$\displaystyle 4.32 - 0.951(y) = 4.32 - x$

$\displaystyle x = 0.951(y)\ or\ y= \frac{x}{0.951}$

$\displaystyle \sin (\theta) = \frac{O}{H}$

$\displaystyle \sin (18) = \frac {1.4 - z}{\left (\frac{4.32}{\cos(18)} - y \right)}$

$\displaystyle 4.32\tan(18) - \sin (18)y = 1.4 - z$

$\displaystyle z = 0.309(y)$

$\displaystyle \tan (\theta) = \frac{O}{A}$

$\displaystyle \tan(18) = \frac {1.4 - z}{4.32 - x}$

$\displaystyle Substituting\ for\ x\ and\ z\ in\ terms\ of\ one\ variable\ only$

$\displaystyle \tan(18) = \frac{1.4 - 0.309y}{4.32 - 0.951y}$

$\displaystyle 1.4 - 0.309y = 1.4 - 0.309y$

$\displaystyle 0 = 0.618y\ ?$

Sidenote: I do feel how this end result looks, it should have cancelled.

Any help would brilliant.

Thank you for your attention.
• Jul 8th 2012, 05:41 PM
skeeter
Re: Vector problem
vertical components of the two tensions are in equilibrium with the load's weight ...

$\displaystyle T_1 \cos(18) + T_2 \cos(30) = 4.32$

horizontal components of tension are equal and opposite in direction

$\displaystyle T_1 \sin(18) - T_2 \sin(30) = 0$

solve the system of equations ...

$\displaystyle T_1 = \frac{T_2 \sin(30)}{\sin(18)}$

$\displaystyle \frac{T_2 \sin(30)}{\sin(18)} \cdot \cos(18) + T_2 \cos(30) = 4.32$

$\displaystyle T_2 \left[\sin(30) \cot(18) + \cos(30) \right] = 4.32$

$\displaystyle T_2 = \frac{4.32}{\sin(30) \cot(18) + \cos(30)} \approx 1.8 N$

$\displaystyle T_1 \approx 2.91 N$
• Jul 9th 2012, 02:24 AM
astartleddeer
Re: Vector problem
@ Skeeter

Aha, I see. I think my mechanics needs a service.

Thanks!