Solution 10.7 - Revision history

Ian at 17:29, 14 June 2010

2010-06-14T17:29:14Z

←Older revision		Revision as of 17:29, 14 June 2010
Line 8:		Line 8:
	& \to :\ 40\cos {{30}^{\circ }}-F=ma \\		& \to :\ 40\cos {{30}^{\circ }}-F=ma \\
	\end{align}</math> ration is represented by the vector <math>a</math>.		\end{align}</math> ration is represented by the vector <math>a</math>.
		+
		+	We need to obtain <math>F</math>.

	Note that <math>R</math> is NOT equal to <math>mg</math> here as the string has a component force in the vertical direction! In fact one has,		Note that <math>R</math> is NOT equal to <math>mg</math> here as the string has a component force in the vertical direction! In fact one has,
Line 22:		Line 24:
	& F=\mu R \\		& F=\mu R \\
	& \\		& \\
-	& F =0\textrm{.}5 \times 9\textrm{.}43 = 4\textrm{.}72 \ \text{N} \\	+	& F =0\textrm{.}5 \times 9\textrm{.}43 = 4\textrm{.}71 \ \text{N} \\
		+	\end{align}</math>
		+
		+	Substituting in the above equation of motion (Newton´s Second Law) gives,
		+
		+	<math>\begin{align}
		+	& 3a=40\times 0\textrm{.}866-4\textrm{.}71=29\textrm{.}93 \\
		+	& \\
		+	& a=9\textrm{.}98\ \text{m}{{\text{s}}^{-2}} \\
	\end{align}</math>		\end{align}</math>

Ian at 17:11, 14 June 2010

2010-06-14T17:11:30Z

←Older revision		Revision as of 17:11, 14 June 2010
Line 8:		Line 8:
	& \to :\ 40\cos {{30}^{\circ }}-F=ma \\		& \to :\ 40\cos {{30}^{\circ }}-F=ma \\
	\end{align}</math> ration is represented by the vector <math>a</math>.		\end{align}</math> ration is represented by the vector <math>a</math>.
		+
		+	Note that <math>R</math> is NOT equal to <math>mg</math> here as the string has a component force in the vertical direction! In fact one has,
		+
		+	<math>\begin{align}
		+	& \uparrow :\ R+40\sin {{30}^{\circ }}-mg=0 \\
		+	& \\
		+	& F=mg-40\sin {{30}^{\circ }}=3\times 9\textrm{.}81-40\times \frac{1}{2}=29\textrm{.}43-20=9\textrm{.}43\ \text{N} \\
		+	\end{align}</math>
		+
		+	Using the friction equation gives
		+
		+	<math>\begin{align}
		+	& F=\mu R \\
		+	& \\
		+	& F =0\textrm{.}5 \times 9\textrm{.}43 = 4\textrm{.}72 \ \text{N} \\
		+	\end{align}</math>

Ian: New page: Image:10.7.gif We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector $a$ . Applying Newton´s Second Law $F=ma<...$

2010-06-14T16:55:02Z

New page: Image:10.7.gif We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector <math>a</math>. Applying Newton´s Second Law <math>F=ma<...

New page

[[Image:10.7.gif]]

We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector <math>a</math>.

Applying Newton´s Second Law <math>F=ma</math> horisontally,

<math>\begin{align}
& \to :\ 40\cos {{30}^{\circ }}-F=ma \\
\end{align}</math> ration is represented by the vector <math>a</math>.

Solution 10.7 - Revision history

Ian at 17:29, 14 June 2010

Ian at 17:11, 14 June 2010

Ian: New page: Image:10.7.gif We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector a. Applying Newton´s Second Law F=ma<...

Ian: New page: Image:10.7.gif We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector $a$ . Applying Newton´s Second Law $F=ma<...$