Solution 9.6b
From Mechanics
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| + | From part a) <math>\ T=1380 \text{ N}</math>. | ||
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| + | Resolving perpendicular to the hill, | ||
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| + | <math>\begin{align} | ||
| + | & R+T\sin {{7}^{\circ }}-mg\cos {{5}^{\circ }}=0 \\ | ||
| + | & \\ | ||
| + | & R=mg\cos {{5}^{\circ }}-T\sin {{7}^{\circ }}=1600\times 9\textrm{.}81\times 0\textrm{.}996-1380\times 0\textrm{.}122= \\ | ||
| + | & =15633-168=15465\simeq 15500\ \text{N} \\ | ||
| + | \end{align}</math> | ||
Current revision
From part a) \displaystyle \ T=1380 \text{ N}.
Resolving perpendicular to the hill,
\displaystyle \begin{align} & R+T\sin {{7}^{\circ }}-mg\cos {{5}^{\circ }}=0 \\ & \\ & R=mg\cos {{5}^{\circ }}-T\sin {{7}^{\circ }}=1600\times 9\textrm{.}81\times 0\textrm{.}996-1380\times 0\textrm{.}122= \\ & =15633-168=15465\simeq 15500\ \text{N} \\ \end{align}
