Open main menu

Changes

Resolving Forces

164 bytes removed, 17:32, 6 April 2019
no edit summary
|-
| style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[tension]] as 10N and the [[angle]] from the [[horizontal]] as 37°.
| style="height:20px; width:200px; text-align:center;" |Using the [[horizontal]] as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[tension]] as the hypotenuse to show the two [[perpendicular]] components.
|-
| style="height:20px; width:600px; text-align:center;"; colspan = "2" |
The component needed is the [[Adjacent (Trigonometry)|adjacent]] to the [[angle]] in the [[Right Angle Triangle|right angle triangle]] so [[cosine]] should be used.
<math>F_x = 10\cos(37)</math>
|-
| style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[weight]] as 39N and the [[angle]] of the slope as 22.6°.
| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[weight]] as the hypotenuse to show the two [[perpendicular]] components. The [[angle]] at the bottom of this [[triangle]] is the same as the [[angle]] of the slope.
|-
| style="height:20px; width:600px; text-align:center;"; colspan = "2" |
The component needed is the [[Opposite (Trigonometry)|opposite]] to the [[angle]] in the [[Right Angle Triangle|right angle triangle]] so [[sine]] should be used.
<math>F_x = 39\sin(22.6)</math>