Processing Math: Done
Solution 9.5
From Mechanics
(Difference between revisions)
(New page: Image:9.5.gif In the figure <math>R1</math> is the air resistance. Resolving up the plane <math>\begin{align} & F1+F-mg\sin {{30}^{\circ }}=0 \\ & \\ & F1=mg-F\sin {{30}^{\circ ...) |
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Line 5: | Line 5: | ||
Resolving up the plane | Resolving up the plane | ||
+ | <math>\begin{align} | ||
+ | & F1+F-mg\sin {{20}^{\circ }}=0 \\ | ||
+ | & \\ | ||
+ | & F1=mg\sin {{20}^{\circ }}-F=60\times 9\textrm{.}81\times 0\textrm{.}342-F=201\textrm{.}3\ \text{N}-F \\ | ||
+ | \end{align}</math> | ||
+ | |||
+ | Resolving perpendicular to the plane upwards | ||
<math>\begin{align} | <math>\begin{align} | ||
- | & | + | & R-mg\cos {{20}^{\circ }}=0 \\ |
& \\ | & \\ | ||
- | & | + | & R=mg\cos {{20}^{\circ }}=60\times 9\textrm{.}81\times 0\textrm{.}94=553\ \text{N} \\ |
\end{align}</math> | \end{align}</math> | ||
+ | |||
+ | Using the friction equation gives | ||
+ | |||
+ | <math>F=\mu R=0\textrm{.}05\times 553=27\textrm{.}65\ \text{N}</math> | ||
+ | |||
+ | Substituting in the above equation for <math>R1</math> gives | ||
+ | |||
+ | <math>F1=201\textrm{.}3-27\textrm{.}65\approx 173\ \text{N}</math> |
Revision as of 12:42, 18 May 2010
In the figure
Resolving up the plane
=0F1=mgsin20
−F=60
9.81
0.342−F=201.3 N−F
Resolving perpendicular to the plane upwards
=0R=mgcos20
=60
9.81
0.94=553 N
Using the friction equation gives
R=0.05
553=27.65 N
Substituting in the above equation for
173 N