Solution 5.8

From Mechanics

(Difference between revisions)
Jump to: navigation, search
(New page: Image:5.8.gif)
Current revision (15:15, 6 April 2010) (edit) (undo)
 
Line 1: Line 1:
[[Image:5.8.gif]]
[[Image:5.8.gif]]
 +
 +
Resolving horisontally to the right
 +
 +
<math>\begin{align}
 +
& T2\cos {{40}^{\circ }}-T1\cos {{50}^{\circ }}=0 \\
 +
& \\
 +
\end{align}</math>
 +
 +
giving
 +
 +
<math>T2=\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1</math>
 +
 +
Resolving vertically upwards.
 +
 +
<math>T1\cos {{40}^{\circ }}+T2\cos {{50}^{\circ }}-mg=0</math>
 +
 +
Substituting for
 +
<math>T2</math>
 +
in this equation gives an equation only containing
 +
<math>T1</math>.
 +
 +
<math>\begin{align}
 +
& T1\cos {{40}^{\circ }}+\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1\cos {{50}^{\circ }}-mg=0 \\
 +
& \\
 +
& T1\left( \cos {{40}^{\circ }}+\frac{{{\cos }^{2}}{{50}^{\circ }}}{\cos {{40}^{\circ }}} \right)=mg \\
 +
& \\
 +
& T1\left( 0\textrm{.}766+0\textrm{.}539 \right)=50\times 9\textrm{.}8 \\
 +
& \\
 +
& T1=\frac{490}{1\textrm{.}3}=375\ \text{N} \\
 +
\end{align}</math>
 +
 +
 +
Also using
 +
<math>T2=\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1</math>
 +
 +
<math>T2=\frac{0\textrm{.}642}{0\textrm{.}766}\times 375=315\ \text{N}</math>

Current revision

Image:5.8.gif

Resolving horisontally to the right

\displaystyle \begin{align} & T2\cos {{40}^{\circ }}-T1\cos {{50}^{\circ }}=0 \\ & \\ \end{align}

giving

\displaystyle T2=\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1

Resolving vertically upwards.

\displaystyle T1\cos {{40}^{\circ }}+T2\cos {{50}^{\circ }}-mg=0

Substituting for \displaystyle T2 in this equation gives an equation only containing \displaystyle T1.

\displaystyle \begin{align} & T1\cos {{40}^{\circ }}+\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1\cos {{50}^{\circ }}-mg=0 \\ & \\ & T1\left( \cos {{40}^{\circ }}+\frac{{{\cos }^{2}}{{50}^{\circ }}}{\cos {{40}^{\circ }}} \right)=mg \\ & \\ & T1\left( 0\textrm{.}766+0\textrm{.}539 \right)=50\times 9\textrm{.}8 \\ & \\ & T1=\frac{490}{1\textrm{.}3}=375\ \text{N} \\ \end{align}


Also using \displaystyle T2=\frac{\cos {{50}^{\circ }}}{\cos {{40}^{\circ }}}T1

\displaystyle T2=\frac{0\textrm{.}642}{0\textrm{.}766}\times 375=315\ \text{N}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Solution_5.8"