Solution 10.1a

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Current revision (10:22, 20 May 2010) (edit) (undo)
 
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<math>\begin{align}
<math>\begin{align}
& \uparrow :\ \text{Resultant}\ \text{Force}=T-300\times 9 \textrm{.}81=
& \uparrow :\ \text{Resultant}\ \text{Force}=T-300\times 9 \textrm{.}81=
-
mg = 300\times 0 \textrm{.}1 \\
+
ma = 300\times 0 \textrm{.}1 \\
& \\
& \\
& T=30+2943\approx 2970\ \text{N} \\
& T=30+2943\approx 2970\ \text{N} \\
\end{align}</math>
\end{align}</math>

Current revision

Image:10.1.gif

Here \displaystyle T is the tension in the cable and \displaystyle a is the acceleration of the packet.

Applying Newton's second law

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