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
Equilibrium of Forces: Two Examples
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 m1 = 2kg and m2 = 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 Fp 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 = anti-clockwise 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) T1 and T2 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 T1 and T2  
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 T1 to be about 12.73N and that of T2 to be about 7.35N.  
   
Looking back on all this... it would appear that this last method is the easiest, in this case.  
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