Ian: New page: Resolving vertically upwards, $T\cos {{30}^{\circ }}+S-mg=0$ where $S$ is the spring force '''which we assume is upwards'''. This gives $S=mg-\ T\cos {{30}...$

2010-04-06T11:33:14Z

New page: Resolving vertically upwards, <math>T\cos {{30}^{\circ }}+S-mg=0</math> where <math>S</math> is the spring force '''which we assume is upwards'''. This gives <math>S=mg-\ T\cos {{30}...

New page

Resolving vertically upwards,

<math>T\cos {{30}^{\circ }}+S-mg=0</math>

where <math>S</math> is the spring force '''which we assume is upwards'''.

This gives

<math>S=mg-\ T\cos {{30}^{\circ }}=40\times 9\textrm{.}8-90\times 0\textrm{.}866=314\ \text{N}</math>

As <math>S</math> is positive this means the spring is pushing upwards on the mass and thus the spring is in compression. (If <math>S</math> had been negative the spring would have been pulling downwards on the mass and thus would be in tension).

Solution 5.7b - Revision history

Ian: New page: Resolving vertically upwards, T\cos {{30}^{\circ }}+S-mg=0 where S is the spring force '''which we assume is upwards'''. This gives S=mg-\ T\cos {{30}...

Ian: New page: Resolving vertically upwards, $T\cos {{30}^{\circ }}+S-mg=0$ where $S$ is the spring force '''which we assume is upwards'''. This gives $S=mg-\ T\cos {{30}...$