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Solution 9.5

From Mechanics

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(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 ...)
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}
-
& F1+F-mg\sin {{30}^{\circ }}=0 \\
+
& R-mg\cos {{20}^{\circ }}=0 \\
& \\
& \\
-
& F1=mg-F\sin {{30}^{\circ }}=60\times 9.81-\frac{1}{2}F \\
+
& 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

Image:9.5.gif

In the figure R1 is the air resistance.

Resolving up the plane

F1+Fmgsin20=0F1=mgsin20F=609.810.342F=201.3 NF

Resolving perpendicular to the plane upwards

Rmgcos20=0R=mgcos20=609.810.94=553 N

Using the friction equation gives

F=R=0.05553=27.65 N

Substituting in the above equation for R1 gives

F1=201.327.65173 N