Resolving Forces - FAST

Here is an object of 2kg lying on a slop raised from the horizontal of an angle of 'a'.
From this we know that there is a force of 2g downwards as a result of (M)x(G).
We also know that the reaction force will be perpendicular to the slope as shown.
Therefore this triangle can be drawn with the bottom line parallel to that of the slope itself.

 We can now look at this triangle in more detail. The angle joining the vertical force of gravity and the reaction force will be the same as the angle of which the slope is raised from the horizontal.

From this we can draw the triangle to the right to show the different forces acting upon the object on the slope.

 As we know the values of the downward force (gravity) and the angle a; we can derive the values of the force parallel to the slope and the value of the reaction force.

The reaction force is worked out using SOHCAHTOA.
Cos = Adjacent/Hypotenuse.
Cos a = Hypotenuse / 2g
Hypotenuse = 2g Cos a

The force parallel to the slope is worked out the same way:
Sin = Opposite / Hypotenuse
Sin a = Opposite / 2g
Opposite = 2g Sin a

The trick to resolving forces is to learn that:

• The force next to the angle will always use 'Cos'.
• The force that is not next to the angle will use 'Sin'.

Therefore we can assume the general formulae:

• 'mg cos a'

for the reaction force of an object.

• 'mg sin a'

for the force parallel to the slope.