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Revision as of 16:38, 4 April 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (16:03, 4 March 2011) (edit) (undo) Ian (Talk | contribs)
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	<math>\begin{align}		<math>\begin{align}
-	& N=mg-5000\cos {{65}^{\circ }}=500\times 9.8-5000\times 0.423 \\	+	& N=mg-5000\cos {{65}^{\circ }}=500\times 9\textrm{.}8-5000\times 0\textrm{.}423 \\
	& =4900-2113=2787\ \text{N} \\		& =4900-2113=2787\ \text{N} \\
	\end{align}</math>		\end{align}</math>

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Current revision

In the vertical direction the forces on the tank must be in equilibrium.

\displaystyle \uparrow \ -mg+N+5000\cos {{65}^{\circ }}=0

or

\displaystyle \begin{align} & N=mg-5000\cos {{65}^{\circ }}=500\times 9\textrm{.}8-5000\times 0\textrm{.}423 \\ & =4900-2113=2787\ \text{N} \\ \end{align}