STEP III 1997 question 9 solution

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STEP III 1997 question 9 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP 1997 Solutions > STEP III 1997 question 9 solution

Let F_1, F_2 be the tensions in the two springs AB, AC, and let their natural lengths be l_1, l_2 and their stretched lengths be

Taking moments about the midpoint of BC, we get

$F_1\sin \angle B = F_2\sin \angle C$

Together with the law of sines, this gives:

F_1 L_2 = F_2 L_1

$\displaystyle \frac{F_1}{L_1} = \frac{F_2}{L_2}$

Let

$\displaystyle \lambda = \frac{F_1}{L_1} = \frac{F_2}{L_2}$ .

Now we have

$\displaystyle \frac{l_1}{l_2} = \frac{L_1 - F_1/\kappa}{L_2 - F_1 / \kappa} = \frac{\kappa L_1 - F_1}{\kappa L_2 - F_2} = \frac{\kappa L_1 - \lambda L_1}{\kappa L_2 - \lambda L_2} = \frac{L_1}{L_2}$

as required.

Solution by ukgea

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