Ian: New page: First the maximum friction must be calculated. To do this we need the normal reaction force $R$ . Resolving in the normal direction $\begin{align} & R-mg\cos {{20}^{\circ...$

Ian — Mon, 05 Apr 2010 17:47:09 GMT

New page: First the maximum friction must be calculated. To do this we need the normal reaction force <math>R</math>. Resolving in the normal direction <math>\begin{align} & R-mg\cos {{20}^{\circ...

New page

First the maximum friction must be calculated.

To do this we need the normal reaction force <math>R</math>.

Resolving in the normal direction

<math>\begin{align}
& R-mg\cos {{20}^{\circ }}=0 \\
& R=72\times 9\textrm{.}8\times 0\textrm{.}94=663\ \text{N} \\
\end{align}</math>

Then the maximim friction <math>F</math> is

<math>\mu R=0\textrm{.}2\times 663=132\ \text{N}</math>

The gravitational force which we denote as <math>G \text 1</math>, down the slope is

<math>G1=mg\sin {{20}^{\circ }}=72\times 9\textrm{.}8\times 0\textrm{.}342=241\ \text{N}</math>

The forces are in equilibrium down the slope. If we denote the air resistance force as <math>F \text 1</math>, we get,

<math>G1-F1-F=0</math>

giving

<math>F1=G1 -F=241-132=109\ \text{N}</math>

Solution 5.6c - Revision history

Ian: New page: First the maximum friction must be calculated. To do this we need the normal reaction force R. Resolving in the normal direction \begin{align} & R-mg\cos {{20}^{\circ...

Ian: New page: First the maximum friction must be calculated. To do this we need the normal reaction force $R$ . Resolving in the normal direction $\begin{align} & R-mg\cos {{20}^{\circ...$