Solution 9.5

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(New page: Image:9.5.gif In the figure <math>R1</math> is the air resistance. Resolving up the plane <math>\begin{align} & F1+F-mg\sin {{30}^{\circ }}=0 \\ & \\ & F1=mg-F\sin {{30}^{\circ ...)
Current revision (09:56, 6 March 2011) (edit) (undo)
 
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[[Image:9.5.gif]]
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[[Image:9.5a.gif]]
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In the figure <math>R1</math> is the air resistance.
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In the figure <math>F1</math> is the air resistance.
Resolving up the plane
Resolving up the plane
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<math>\begin{align}
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& F1+F-mg\sin {{20}^{\circ }}=0 \\
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& \\
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& F1=mg\sin {{20}^{\circ }}-F=60\times 9\textrm{.}81\times 0\textrm{.}342-F=201\textrm{.}3\ \text{N}-F \\
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\end{align}</math>
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Resolving perpendicular to the plane upwards
<math>\begin{align}
<math>\begin{align}
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& F1+F-mg\sin {{30}^{\circ }}=0 \\
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& R-mg\cos {{20}^{\circ }}=0 \\
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& \\
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& F1=mg-F\sin {{30}^{\circ }}=60\times 9.81-\frac{1}{2}F \\
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& R=mg\cos {{20}^{\circ }}=60\times 9\textrm{.}81\times 0\textrm{.}94=553\ \text{N} \\
\end{align}</math>
\end{align}</math>
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Using the friction equation gives
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<math>F=\mu R=0\textrm{.}05\times 553=27\textrm{.}65\ \text{N}</math>
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Substituting in the above equation for <math>R1</math> gives
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<math>F1=201\textrm{.}3-27\textrm{.}65\approx 173\ \text{N}</math>

Current revision

Image:9.5a.gif

In the figure \displaystyle F1 is the air resistance.

Resolving up the plane

\displaystyle \begin{align} & F1+F-mg\sin {{20}^{\circ }}=0 \\ & \\ & F1=mg\sin {{20}^{\circ }}-F=60\times 9\textrm{.}81\times 0\textrm{.}342-F=201\textrm{.}3\ \text{N}-F \\ \end{align}

Resolving perpendicular to the plane upwards

\displaystyle \begin{align} & R-mg\cos {{20}^{\circ }}=0 \\ & \\ & R=mg\cos {{20}^{\circ }}=60\times 9\textrm{.}81\times 0\textrm{.}94=553\ \text{N} \\ \end{align}

Using the friction equation gives

\displaystyle F=\mu R=0\textrm{.}05\times 553=27\textrm{.}65\ \text{N}

Substituting in the above equation for \displaystyle R1 gives

\displaystyle F1=201\textrm{.}3-27\textrm{.}65\approx 173\ \text{N}