Year 11 Specialist (Unit 1 & 2) Vectors in the Plane

Applications of vectors - forces and equilibrium

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Questions

Question 1

61414

A mass of 10kg is suspended from two strings of length 8cm and 6cm. The ends of the strings being attached to two points in a horizontal line 12cm apart. Find the tension in each string.

Answer

$T_1=9.06kg\,wt \quad T_2=10.08kg\,wt$

Worked Solution

$$\begin{aligned}
\cos A&=\frac{12^{2}+8^{2}-6^{2}}{2 \times 12 \times 8}\\
\angle A&=26^{\circ} 23^{\prime}\\
\angle M C A&=90^{\circ}-26^{\circ} 23^{\prime}\\
&=63^{\circ} 37^{\prime}\\
\therefore\ \angle D C E&=63^{\circ} 37^{\prime}\ (vert opp \angle s)\\
\cos B&=\frac{12^{2}+6^{2}-8^{2}}{2 \times 12 \times 6}\\
\angle B&=36^{\circ} 20^{\prime}\\
\angle M C B&=90^{\circ}-36^{\circ} 20^{\prime}\\
&=53^{\circ} 40^{\prime}\\
\therefore\ \angle F C E&=53^{\circ} 40^{\prime}\ (vert opp \angle s)\\
\therefore\ \angle C E D&=53^{\circ} 40^{\prime}\ (a / t\ \angle s)\\
\angle C D E &=180^{\circ}-\left(63^{\circ} 37^{\prime}+53^{\circ} 40^{\prime}\right) \\ &=62^{\circ} 93^{\prime} \\ \frac{T_{1}}{\sin 53^{\circ} 40^{\prime}} &=\frac{10}{\sin 62^{\circ} 43^{\prime}} \\ T_{1} &=9.06 \mathrm{~kgwt} \\ \frac{T_{2}}{\sin 63^{\circ} 37^{\prime}} &=\frac{10}{\sin 62^{\circ} 43^{\prime}} \\ T_{2} &=10.08 \mathrm{~kgwt}
\end{aligned}
$$

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