Solution 9.5 - Revision history

Ian at 09:56, 6 March 2011

2011-03-06T09:56:34Z

←Older revision		Revision as of 09:56, 6 March 2011
Line 1:		Line 1:
-	[[Image:9.5.gif]]	+	[[Image:9.5a.gif]]

-	In the figure <math>R1</math> is the air resistance.	+	In the figure <math>F1</math> is the air resistance.

	Resolving up the plane		Resolving up the plane

Ian at 12:42, 18 May 2010

2010-05-18T12:42:00Z

←Older revision		Revision as of 12:42, 18 May 2010
Line 5:		Line 5:
	Resolving up the plane		Resolving up the plane

		+	<math>\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}</math>
		+
		+	Resolving perpendicular to the plane upwards

	<math>\begin{align}		<math>\begin{align}
-	& ~~F1+F~~-mg\~~sin~~ {{30}^{\circ }}=0 \\	+	& R-mg\cos {{20}^{\circ }}=0 \\
	& \\		& \\
-	& F1=mg-F\~~sin~~ {{30}^{\circ }}=60\times 9.81-\~~frac~~{1}{2}F \\	+	& R=mg\cos {{20}^{\circ }}=60\times 9\textrm{.}81\times 0\textrm{.}94=553\ \text{N} \\
	\end{align}</math>		\end{align}</math>
		+
		+	Using the friction equation gives
		+
		+	<math>F=\mu R=0\textrm{.}05\times 553=27\textrm{.}65\ \text{N}</math>
		+
		+	Substituting in the above equation for <math>R1</math> gives
		+
		+	<math>F1=201\textrm{.}3-27\textrm{.}65\approx 173\ \text{N}</math>

Ian: New page: Image:9.5.gif In the figure $R1$ is the air resistance. Resolving up the plane $\begin{align} & F1+F-mg\sin {{30}^{\circ }}=0 \\ & \\ & F1=mg-F\sin {{30}^{\circ ...$

2010-05-18T12:09:41Z

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 ...

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 }}=60\times 9.81-\frac{1}{2}F \\
\end{align}</math>

Solution 9.5 - Revision history

Ian at 09:56, 6 March 2011

Ian at 12:42, 18 May 2010

Ian: New page: Image:9.5.gif In the figure R1 is the air resistance. Resolving up the plane \begin{align} & F1+F-mg\sin {{30}^{\circ }}=0 \\ & \\ & F1=mg-F\sin {{30}^{\circ ...

Ian: New page: Image:9.5.gif In the figure $R1$ is the air resistance. Resolving up the plane $\begin{align} & F1+F-mg\sin {{30}^{\circ }}=0 \\ & \\ & F1=mg-F\sin {{30}^{\circ ...$