Solution 10.1a
From Mechanics
(Difference between revisions)
| Line 6: | Line 6: | ||
<math>\begin{align} | <math>\begin{align} | ||
| - | & \uparrow :\ \text{Resultant}\ \text{Force}=T-300\times 9 \textrm{.}81=300\times 0 \textrm{.}1 \\ | + | & \uparrow :\ \text{Resultant}\ \text{Force}=T-300\times 9 \textrm{.}81= |
| + | mg = 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> | ||
Revision as of 09:57, 20 May 2010
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= mg = 300\times 0 \textrm{.}1 \\ & \\ & T=30+2943\approx 2970\ \text{N} \\ \end{align}
