The Open Door Web Site
HOME PAGE BIOLOGY CHEMISTRY PHYSICS ELECTRONICS HISTORY HISTORY of SCI & TECH MATH STUDIES LEARN FRENCH STUDY GUIDE PHOTO GALLERY
ATOMIC and NUCLEAR ELECTRICITY and MAGNETISM MEASUREMENTS MECHANICS OPTICS PRACTICAL WORK QUESTIONS RELATIVITY THERMAL PHYSICS WAVES
MECHANICS
Google
Custom Search
Conditions for the Equilibrium of Three Non-Parallel Forces
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 clock-wise moments about any point must have the same magnitude as the sum of all the anti-clockwise 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 clock-wise turning effect about this point and the weight, mg, has an anti-clock-wise 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  
SITE MAP
WHAT'S NEW
ABOUT
PRIVACY
COPYRIGHT
SPONSORSHIP
DONATIONS
ADVERTISING
 

© The Open Door Team
2016
Any questions or
problems regarding
this site should be
addressed to
the webmaster

David Hoult 2017

Hosted By
Web Hosting by HostCentric

 
SiteLock
 
 
Mechanics Index Page