Solution 10.1a
From Mechanics
(Difference between revisions)
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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= | ||
- | + | 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
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}