Solution 10.1b - Revision history

Ian at 10:35, 20 May 2010

Ian — Thu, 20 May 2010 10:35:21 GMT

←Older revision		Revision as of 10:35, 20 May 2010
Line 7:		Line 7:
	<math>\begin{align}		<math>\begin{align}
	& \downarrow :\ \text{Resultant}\ \text{Force}=300\times 9 \textrm{.}81-T=		& \downarrow :\ \text{Resultant}\ \text{Force}=300\times 9 \textrm{.}81-T=
-	ma = 300\times 0 \textrm{.}1 \\	+	ma = 300\times 0 \textrm{.}2 \\
	& \\		& \\
-	& T=2943-30\approx ~~2910~~\ \text{N} \\	+	& T=2943-60\approx 2880\ \text{N} \\
	\end{align}</math>		\end{align}</math>

Ian at 10:25, 20 May 2010

Ian — Thu, 20 May 2010 10:25:21 GMT

←Older revision		Revision as of 10:25, 20 May 2010
Line 6:		Line 6:

	<math>\begin{align}		<math>\begin{align}
-	& \~~uparrow~~ :\ \text{Resultant}\ \text{Force}=T-300\times 9 \textrm{.}81=	+	& \downarrow :\ \text{Resultant}\ \text{Force}=300\times 9 \textrm{.}81-T=
-	mg = 300\times 0 \textrm{.}1 \\	+	ma = 300\times 0 \textrm{.}1 \\
	& \\		& \\
-	& T=~~30+~~2943\approx ~~2970~~\ \text{N} \\	+	& T=2943-30\approx 2910\ \text{N} \\
	\end{align}</math>		\end{align}</math>

Ian: New page: Image:10.1b.gif Here $T$ is the tension in the cable and $a$ is the acceleration of the packet. Applying Newton's second law $\begin{align} & \uparrow :\ ...$

Ian — Thu, 20 May 2010 10:03:56 GMT

New page: Image:10.1b.gif Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet. Applying Newton's second law <math>\begin{align} & \uparrow :\ ...

New page

[[Image:10.1b.gif]]

Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet.

Applying Newton's second law

<math>\begin{align}
& \uparrow :\ \text{Resultant}\ \text{Force}=T-300\times 9 \textrm{.}81=
mg = 300\times 0 \textrm{.}1 \\
& \\
& T=30+2943\approx 2970\ \text{N} \\
\end{align}</math>

Solution 10.1b - Revision history

Ian at 10:35, 20 May 2010

Ian at 10:25, 20 May 2010

Ian: New page: Image:10.1b.gif Here T is the tension in the cable and a is the acceleration of the packet. Applying Newton's second law \begin{align} & \uparrow :\ ...

Ian: New page: Image:10.1b.gif Here $T$ is the tension in the cable and $a$ is the acceleration of the packet. Applying Newton's second law $\begin{align} & \uparrow :\ ...$