From Mechanics

Revision as of 11:04, 17 May 2010 (edit) Ian (Talk | contribs) (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...) ← Previous diff		Current revision (11:11, 17 May 2010) (edit) (undo) Ian (Talk | contribs)
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>

---

Current revision

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

\displaystyle F = \mu R

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

Resolving horisontally,

\displaystyle \begin{align} & \to :\quad 20\ \text{N}-F=0 \\ & \\ & F=20\ \text{N} \\ \end{align}

Resolving vertically,

\displaystyle \uparrow :\quad R-mg=0

\displaystyle R=mg=4\times 9\textrm{.}81=39\textrm{.}24\ \text{N}

Substituting in the friction equation,

\displaystyle 20 = \mu \times 39\textrm{.}24

giving

\displaystyle \mu = 0\textrm{.}51