Solution 10.6 - Revision history

Ian at 10:10, 6 March 2011

2011-03-06T10:10:53Z

←Older revision		Revision as of 10:10, 6 March 2011
Line 44:		Line 44:

	<math>\begin{align}		<math>\begin{align}
-	& a=9\textrm{.}81\times 0\textrm{.}643-0\textrm{.}2\times 9\textrm{.}81\times 0~~.76~~\textrm{.}= \\	+	& a=9\textrm{.}81\times 0\textrm{.}643-0\textrm{.}2\times 9\textrm{.}81\times 0\textrm{.}77= \\
-	& =9\textrm{.}81\times (0\textrm{.}643-0\textrm{.}154)=9\textrm{.}81\times 0\textrm{.}489=0\textrm{.}80\ \text{m}{{\text{s}}^{-2}} \\	+	& =9\textrm{.}81\times (0\textrm{.}643-0\textrm{.}154)=9\textrm{.}81\times 0\textrm{.}489=4\textrm{.}80\ \text{m}{{\text{s}}^{-2}} \\
	\end{align}</math>		\end{align}</math>

	Note by waiting to substitute the values of the various quantities we minimized the amount of calculation needed. Also the result is independent of the mass of the particle as we could cancel it out.		Note by waiting to substitute the values of the various quantities we minimized the amount of calculation needed. Also the result is independent of the mass of the particle as we could cancel it out.

Ian at 16:07, 14 June 2010

2010-06-14T16:07:11Z

←Older revision		Revision as of 16:07, 14 June 2010
Line 45:		Line 45:
	<math>\begin{align}		<math>\begin{align}
	& a=9\textrm{.}81\times 0\textrm{.}643-0\textrm{.}2\times 9\textrm{.}81\times 0.76\textrm{.}= \\		& a=9\textrm{.}81\times 0\textrm{.}643-0\textrm{.}2\times 9\textrm{.}81\times 0.76\textrm{.}= \\
-	& =9\textrm{.}81\times (0\textrm{.}643-0\textrm{.}154)=9\textrm{.}81\times 0\textrm{.}489=\textrm{.}80\ \text{m}{{\text{s}}^{-2}} \\	+	& =9\textrm{.}81\times (0\textrm{.}643-0\textrm{.}154)=9\textrm{.}81\times 0\textrm{.}489=0\textrm{.}80\ \text{m}{{\text{s}}^{-2}} \\
	\end{align}</math>		\end{align}</math>

	Note by waiting to substitute the values of the various quantities we minimized the amount of calculation needed. Also the result is independent of the mass of the particle as we could cancel it out.		Note by waiting to substitute the values of the various quantities we minimized the amount of calculation needed. Also the result is independent of the mass of the particle as we could cancel it out.

Ian: New page: Image:10.6.gif The figure shows the forces acting on the particle. Assuming the particle has an acceleration $a$ down the plane Newton´s Second Law applied down the plan...

2010-06-14T12:20:17Z

New page: Image:10.6.gif The figure shows the forces acting on the particle. Assuming the particle has an acceleration <math>a</math> down the plane Newton´s Second Law applied down the plan...

New page

[[Image:10.6.gif]]

The figure shows the forces acting on the particle.

Assuming the particle has an acceleration <math>a</math> down the plane Newton´s Second Law applied down the plane gives,

<math>mg\sin {{40}^{\circ }}-F=ma</math>
(equation of motion)

We need to obtain
<math>F</math>.

Resolving perpendicular to the plane the sum of the forces on the particle must be zero as the particle has no motion in that direction.

<math>\begin{align}
& R-mg\cos {{40}^{\circ }}=0 \\
& \\
& R=mg\cos {{40}^{\circ }} \\
\end{align}</math>

The friction condition gives

<math>F=\mu R</math>
which gives

<math>F=\mu mg\cos {{40}^{\circ }}</math>

Substituting in the equation of motion above

<math>mg\sin {{40}^{\circ }}-\mu mg\cos {{40}^{\circ }}=ma</math>

We can cancel the mass
<math>m</math>
in this equation giving

<math>g\sin {{40}^{\circ }}-\mu g\cos {{40}^{\circ }}=a</math>
or

<math>\begin{align}
& a=9\textrm{.}81\times 0\textrm{.}643-0\textrm{.}2\times 9\textrm{.}81\times 0.76\textrm{.}= \\
& =9\textrm{.}81\times (0\textrm{.}643-0\textrm{.}154)=9\textrm{.}81\times 0\textrm{.}489=\textrm{.}80\ \text{m}{{\text{s}}^{-2}} \\
\end{align}</math>

Note by waiting to substitute the values of the various quantities we minimized the amount of calculation needed. Also the result is independent of the mass of the particle as we could cancel it out.

Solution 10.6 - Revision history

Ian at 10:10, 6 March 2011

Ian at 16:07, 14 June 2010

Ian: New page: Image:10.6.gif The figure shows the forces acting on the particle. Assuming the particle has an acceleration a down the plane Newton´s Second Law applied down the plan...

Ian: New page: Image:10.6.gif The figure shows the forces acting on the particle. Assuming the particle has an acceleration $a$ down the plane Newton´s Second Law applied down the plan...