Solution 10.1b
From Mechanics
(Difference between revisions)
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<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{.} | + | ma = 300\times 0 \textrm{.}2 \\ |
& \\ | & \\ | ||
- | & T=2943- | + | & T=2943-60\approx 2880\ \text{N} \\ |
\end{align}</math> | \end{align}</math> |
Current revision
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} & \downarrow :\ \text{Resultant}\ \text{Force}=300\times 9 \textrm{.}81-T= ma = 300\times 0 \textrm{.}2 \\ & \\ & T=2943-60\approx 2880\ \text{N} \\ \end{align}