Two forces act on point A along the sides AC | vedclass

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(B) Let $AC = b$ and $AB = c$. The forces are $F_1 = 1/b$ acting along $AC$ and $F_2 = 1/c$ acting along $AB$.Since $\angle A = 90^{\circ}$,the result

Vedclass · Accepted Answer

$\frac{1}{AD}$

Vedclass · Answer

(B) Let $AC = b$ and $AB = c$. The forces are $F_1 = 1/b$ acting along $AC$ and $F_2 = 1/c$ acting along $AB$.Since $\angle A = 90^{\circ}$,the resultant force $F_{	ext{net}}$ is given by:$F_{	ext{net}} = \sqrt{F_1^2 + F_2^2} = \sqrt{\frac{1}{b^2} + \frac{1}{c^2}} = \sqrt{\frac{b^2 + c^2}{b^2 c^2}}$In $\Delta ABC$,$BC^2 = b^2 + c^2$,so $F_{	ext{net}} = \frac{\sqrt{BC^2}}{bc} = \frac{BC}{bc}$.The area of $\Delta ABC$ can be expressed as $	ext{Area} = \frac{1}{2} 	imes b 	imes c = \frac{1}{2} 	imes BC 	imes AD$,where $AD$ is the altitude to the hypotenuse $BC$.Therefore,$bc = BC 	imes AD$,which implies $\frac{BC}{bc} = \frac{1}{AD}$.Thus,$F_{	ext{net}} = \frac{1}{AD}$.

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