For a summary of the conditions for equilibrium of three forces click here. 



Example 1 

A uniform beam is placed on a pivot directly under
its centre of gravity, as shown below. 

The two masses on the left are m_{1} =
2kg and m_{2} = 1.2kg 



There are two slightly different approaches which, of course, come down to
the same thing in the end. 



You can either find the perpendicular distance of the line of action of the
force F from the pivot and use that to calculate the moment
of the force or find the magnitude of the component of F
which acts at 90° to the beam. 



It is clear that 



and the component of F which is perpendicular to the beam
is F_{p} given by 



so, either way you get sin50 into the
calculation. 



Taking moments about the centre of the beam (and using the first method) we
have 

clockwise moment = anticlockwise moment 



which gives the magnitude of F to be 73.2N 



Example 2 

A mass m = 1.5kg hangs on two strings
as shown below. Find the magnitudes of the tensions (forces) T_{1}
and T_{2 }in the strings. 





Here we have (at least) two different methods to find the forces. 



First method: Resolving the forces into their horizontal and vertical
components. 



The horizontal forces acting on the mass are the horizontal components of
T_{1} and T_{2} 

These must have equal magnitudes, therefore 



The vertical forces are also in equilibrium so the sum of the vertical
components of the two forces must have the same magnitude as the force of
gravity acting on the mass. 



This gives us 



From these two equations, the magnitudes of the two forces can be
calculated. 



Alternative method: Drawing a scale (vector) diagram. 

See rule 4 (equilibrium) and/or adding vectors
if necessary 








The three forces acting on the mass are shown here. 
















First draw to scale an arrow representing the force for which we
know both the direction and the magnitude, in this case, mg. 



Then draw lines in the directions of the other two forces, at their
correct angles to the known force. 










Complete the vector diagram. 



The magnitudes of the unknown forces can
now be found by measuring the lengths of the arrows and using the scale. 



Having said that, you will have noticed that, in this case, the triangle
is a right angled triangle so we could easily work out the magnitudes using
trigonometry. 





Whatever method you use, you should find the magnitude of T_{1}
to be about 12.73N
and that of T_{2} to be about 7.35N.




Looking back on all this... it would appear that this last method is the
easiest, in this case. 
