Vector problem

**astartleddeer** · July 8th 2012, 05:31 PM

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 $18^\tiny o$ and $30^ \tiny o$ 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 $\sin (90) = 1$ . 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:

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

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

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

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

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

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

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

$z = 0.309(y)$

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

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

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

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

$1.4 - 0.309y = 1.4 - 0.309y$

$0 = 0.618y\ ?$

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

Any help would brilliant.

Thank you for your attention.

**skeeter** · July 8th 2012, 05:41 PM

vertical components of the two tensions are in equilibrium with the load's weight ...

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

horizontal components of tension are equal and opposite in direction

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

solve the system of equations ...

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

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

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

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

$T_1 \approx 2.91 N$

**astartleddeer** · July 9th 2012, 02:24 AM

@ Skeeter

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

Thanks!

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