Ian at 11:11, 17 May 2010

Ian — Mon, 17 May 2010 11:11:18 GMT

←Older revision		Revision as of 11:11, 17 May 2010
Line 19:		Line 19:
	<math>\uparrow :\quad R-mg=0</math>		<math>\uparrow :\quad R-mg=0</math>

-	<math>R=mg=4\times 9.81=39.24\ \text{N}</math>	+	<math>R=mg=4\times 9\textrm{.}81=39\textrm{.}24\ \text{N}</math>
		+
		+	Substituting in the friction equation,
		+
		+	<math>20 = \mu \times 39\textrm{.}24</math>
		+
		+	giving
		+
		+	<math>\mu = 0\textrm{.}51 </math>

Ian: New page: Image:9.4.gif As the block is moving relative to the surface the friction force must satisfy $F = \mu R$ Also as the block has conmstant velocity the forces on it must be...

Ian — Mon, 17 May 2010 11:04:48 GMT

New page: Image:9.4.gif As the block is moving relative to the surface the friction force must satisfy <math>F = \mu R</math> Also as the block has conmstant velocity the forces on it must be...

New page

[[Image:9.4.gif]]

As the block is moving relative to the surface the friction force must satisfy

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

Also as the block has conmstant velocity the forces on it must be in equilibrium.

Resolving horisontally,

<math>\begin{align}
& \to :\quad 20\ \text{N}-F=0 \\
& \\
& F=20\ \text{N} \\
\end{align}</math>

Resolving vertically,

<math>\uparrow :\quad R-mg=0</math>

<math>R=mg=4\times 9.81=39.24\ \text{N}</math>

Solution 9.4 - Revision history

Ian at 11:11, 17 May 2010

Ian: New page: Image:9.4.gif As the block is moving relative to the surface the friction force must satisfy F = \mu R Also as the block has conmstant velocity the forces on it must be...

Ian: New page: Image:9.4.gif As the block is moving relative to the surface the friction force must satisfy $F = \mu R$ Also as the block has conmstant velocity the forces on it must be...