Solution 10.1b

From Mechanics

Revision as of 10:03, 20 May 2010 by Ian (Talk | contribs)
(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)
Jump to: navigation, search

Image:10.1b.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= mg = 300\times 0 \textrm{.}1 \\ & \\ & T=30+2943\approx 2970\ \text{N} \\ \end{align}