Ian: New page: The force on the particle down the plane is $mg\sin {{30}^{\circ }}=m\times 9\textrm{.}81\times \frac{1}{2}=m\times 4\textrm{.}9\ \text{N}$ Using Newton´s Second Law $F...$

2010-05-21T15:06:28Z

New page: The force on the particle down the plane is <math>mg\sin {{30}^{\circ }}=m\times 9\textrm{.}81\times \frac{1}{2}=m\times 4\textrm{.}9\ \text{N}</math> Using Newton´s Second Law <math>F...

New page

The force on the particle down the plane is

<math>mg\sin {{30}^{\circ }}=m\times 9\textrm{.}81\times \frac{1}{2}=m\times 4\textrm{.}9\ \text{N}</math>

Using Newton´s Second Law <math>F=ma\ </math> gives

<math>\begin{align}
& m\times 4\textrm{.}9=ma \\
& \text{ or} \\
& a=4\textrm{.}9\ \text{m}{{\text{s}}^{-2}} \\
\end{align}</math>

Note this result is true independent of the mass of the particle. The fact that all bodies slide down a smooth plane with the same acceleration was an important discovery in the development of modern mechanics.

Solution 10.3 - Revision history

Ian: New page: The force on the particle down the plane is mg\sin {{30}^{\circ }}=m\times 9\textrm{.}81\times \frac{1}{2}=m\times 4\textrm{.}9\ \text{N} Using Newton´s Second Law F...

Ian: New page: The force on the particle down the plane is $mg\sin {{30}^{\circ }}=m\times 9\textrm{.}81\times \frac{1}{2}=m\times 4\textrm{.}9\ \text{N}$ Using Newton´s Second Law $F...$