If we say that an object is under the influence of forces which are in
equilibrium, we mean that the object is not accelerating. 



The following rules help to solve problems in which a body is acted on by
three forces. 

1. 
The sum of all the forces acting vertically upwards must have the same
magnitude as the sum of all the forces acting vertically downwards 

The sum of all the forces acting horizontally to the right must have the
same magnitude as the sum of all the forces acting horizontally to the left. 


2. 
The principle of moments: the sum of all the clockwise moments
about any point must have the same magnitude as the sum of all the
anticlockwise moments about the same point. 


3. 
The lines of action of the three forces must intersect at a point. 


4. 
If the three forces are represented by arrows, drawn to scale and at the
correct angles to each other (with the arrows drawn "head to tail"), they
will form a closed triangle. 




In practice, when using condition 1, we will often be considering the
vertical and horizontal components of forces. 



Condition 3 is related to condition 2. 

If you can find a point about which all the turning
effects are zero, then there can be no net moment and the body will
not have any tendency to rotate. 



In condition 4 we are using the triangle of forces
idea: any one of the forces must be equal in magnitude but opposite in sense
to the vector sum of the other two forces. 



Example to illustrate the use of these rules 


A uniform ladder is leaning against a smooth* wall, as shown. 

The top point of the ladder is 3m above the
ground and the other end is 2.5m away from
the base of the wall. The mass of the ladder is 20kg. 



Find the magnitude and direction of the force exerted by the ground on
the ladder and the magnitude of the force exerted by the wall on
the ladder. 



* In this context, when we say smooth, we mean there is no force of
friction between the ladder and the wall. 








We start by adding to the diagram the two force whose directions we know. 



The wall is described as smooth so it exerts no force of friction.
Therefore, the force it exerts on the ladder must be at 90° to the wall. 



The ladder is described as uniform so the force of gravity can be considered
to act at its mid point and, of course, acts vertically downwards. 










If we extend the lines of action of these two forces they meet at a point.
Rule 3 above says that the line of action of the third force must also pass
through this point. 



The geometry of the diagram allows us to easily find the angle
θ. 



so, the angle θ
= 67.4° 










To find the magnitude of the force F. 



We will first use the fact that the forces acting on the ladder are in
equilibrium vertically. 



The two forces with vertical effects are, mg and the
vertical component of F. These two forces must therefore
have equal magnitudes, rule 1, above. 









So, we can write (considering magnitudes only) 





which gives the magnitude of F to be 212.4N 



We can now use rule 2, above, to find the magnitude of R. 



Imagine the ladder to have a pivot at point, p
(where it touches the ground). 



The force R has a clockwise turning effect about
this point and the weight, mg, has an anticlockwise turning effect
about the same point. 

These two moments must have equal
magnitudes. 



The perpendicular distance of R from the "pivot" is
3m and the perpendicular distance of mg
from the "pivot" is 1.25m. 

Therefore we have 



which means that the magnitude of R is 81.7N 



Note that, in principle, we could have chosen any point
as our imagined "pivot". 

However, the advantage of choosing point p as a
"pivot" point is that the perpendicular distance between point p and the force
F (and therefore its
vertical and horizontal components) is zero. 

This means that F has no
turning effect about p and so the "moments equation" will not have F
in it. 



For a couple more examples of equilibrium situations, click here 
