91Ó°ÊÓ

Leonardo da Vinci and the resolution of forces. Leonardo (1452–1519) asked himself how the weight of a body, supported by two strings of different length, is apportioned between the two strings.

Three forces are acting at the point D: the tensions ${\mathit{F}}_{\mathbf{1}}$and ${\mathit{F}}_{\mathbf{2}}$in the strings and the weight $\stackrel{\mathbf{→}}{\mathbf{W}}$. Leonardo believed that

$\frac{\mathbf{\left|}\mathbf{\right|}{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{1}}\mathbf{\left|}\mathbf{\right|}}{\mathbf{\left|}\mathbf{\right|}{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{2}}\mathbf{\left|}\mathbf{\right|}}\mathbf{=}\frac{\stackrel{\mathbf{¯}}{\mathbf{E}}\stackrel{\mathbf{¯}}{\mathbf{A}}}{\stackrel{\mathbf{¯}}{\mathbf{E}}\stackrel{\mathbf{¯}}{\mathbf{B}}}$

Was he right? (Source: Les Manuscripts de Léonard de Vinci, published by Ravaisson-Mollien, Paris, 1890.)

Hint: Resolve${\mathit{F}}_{\mathbf{1}}$into a horizontal and a vertical component; do the same for ${\mathit{F}}_{\mathbf{2}}$. Since the system is at rest, the equation${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{1}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{2}}\mathbf{+}\stackrel{\mathbf{→}}{\mathbf{W}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{0}}$holds. Express the ratios

$\frac{\mathbf{\left|}\mathbf{\right|}{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{1}}\mathbf{\left|}\mathbf{\right|}}{\mathbf{\left|}\mathbf{\right|}{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{2}}\mathbf{\left|}\mathbf{\right|}}$and $\frac{\stackrel{\mathbf{¯}}{\mathbf{E}}\stackrel{\mathbf{¯}}{\mathbf{A}}}{\stackrel{\mathbf{¯}}{\mathbf{E}}\stackrel{\mathbf{¯}}{\mathbf{B}}}$. In terms of$\mathit{Î\pm }$and $\mathit{β}$, using trigonometric functions, and compare the results.

Step by step solution

01

Step by step solution Step 1: Horizontal components

To find out the horizontal components consider the terms below.

The Horizontal components of ${\stackrel{→}{F}}_1→\left|\right|{\stackrel{→}{F}}_1\left|\right|sin\mathrm{β}$

The Horizontal components of ${\stackrel{→}{F}}_2→\left|\right|{\stackrel{→}{F}}_2\left|\right|sin\mathrm{Î\pm }$

The Horizontal components of ${\stackrel{→}{F}}_3$is zero.

$\begin{array}{rcl}-\left|\right|{\stackrel{→}{F}}_1\left|\right|\sin \mathrm{β}+\left|\right|{\stackrel{→}{F}}_2\left|\right|\sin \mathrm{Î\pm }& =& 0\\ \left|\right|{\stackrel{→}{F}}_1\left|\right|\sin \mathrm{β}& =& \left|\right|{\stackrel{→}{F}}_2\left|\right|\sin \mathrm{Î\pm }\\ \frac{\left|\right|{\stackrel{→}{F}}_1\left|\right|}{\left|\right|{\stackrel{→}{F}}_2\left|\right|}& =& \frac{\sin \mathrm{Î\pm }}{\sin \mathrm{β}}\end{array}$

Find $\frac{\stackrel{¯}{E}\stackrel{¯}{A}}{\stackrel{¯}{E}\stackrel{¯}{B}}$as:

$\begin{array}{rcl}\frac{\stackrel{¯}{E}\stackrel{¯}{A}}{\stackrel{¯}{E}\stackrel{¯}{B}}& =& \frac{tan\mathrm{Î\pm }}{tan\mathrm{β}}\\ & =& \frac{sin\mathrm{Î\pm }}{sin\mathrm{β}}.\frac{cos\mathrm{β}}{cos\mathrm{Î\pm }}\\ & =& \frac{\left|\right|{\stackrel{→}{F}}_1\left|\right|cos\mathrm{β}}{\left|\right|{\stackrel{→}{F}}_2\left|\right|cos\mathrm{Î\pm }}\end{array}$

Thus,$\frac{\left|\right|{\stackrel{→}{F}}_1\left|\right|}{\left|\right|{\stackrel{→}{F}}_2\left|\right|}=\frac{sin\mathrm{Î\pm }}{sin\mathrm{β}}$and$\frac{\stackrel{¯}{E}\stackrel{¯}{A}}{\stackrel{¯}{E}\stackrel{¯}{B}}=\frac{\left|\right|{\stackrel{→}{F}}_1\left|\right|cos\mathrm{β}}{\left|\right|{\stackrel{→}{F}}_2\left|\right|cos\mathrm{Î\pm }}$

02

Compare

Since, $\mathrm{Î\pm }$and $\mathrm{β}$are two distinct acute angles, it follows that:

$\frac{\left|\right|{\stackrel{→}{F}}_1\left|\right|}{\left|\right|{\stackrel{→}{F}}_2\left|\right|}≠\frac{\stackrel{¯}{E}\stackrel{¯}{A}}{\stackrel{¯}{E}\stackrel{¯}{B}}$

Thus, Leonardo was mistaken and$\frac{\left|\right|{\stackrel{→}{F}}_1\left|\right|}{\left|\right|{\stackrel{→}{F}}_2\left|\right|}≠\frac{\stackrel{¯}{E}\stackrel{¯}{A}}{\stackrel{¯}{E}\stackrel{¯}{B}}$

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