Understanding the parallelogram law of forces through problems : Part-I

It is impossible to make a concept gullible without applying it. In other words no concept has ever evolved; in someone's brain, without he or she encountering a problem at some point in their life and squeezing the brain to find a way out of the problem. The famous story of Sir Issac Newton challenging an apple's falling, which eventually turns into the concept of Gravitational force is the best example related to it..

In this piece of blog (Part-I and Part-II), we will see practical problems from some of the best books available for Engineering Mechanics and will solve them applying our learning from previous blogs.         

Before, going into the problems there are few concepts needed to be understood. And they are in fact handy and very useful while solving the problems. And these are :

• The Triangular Law of Forces
• The Polygon Law of Forces
• The Resolution of Forces  

Laws of The Resultant Force

• The Triangular Law of Forces
• The Polygon Law of Forces

The Triangular Law of Forces :

"If two forces acting simultaneously in a particle, be represented in magnitude and direction by the two sides of a triangle, taken in order; their resultant may be represented in magnitude and direction by the third side of the triangle, taken in opposite order".  

Figure-1 | Triangular Law of Forces

Basically, when two forces are arranged in such a way that they are any two sides of a triangle, and their directions are same. Then their resultant will be the third side of the same triangle in magnitude, and the direction will be opposite to the other forces. The concept of Triangular forces are directly imported from Vector Algebra.

This law along with Parallelogram law of forces can be useful tools to solve the problems graphically. That means a geometry box is sufficient to determine resultant from a force system. And also no equation, calculation and remembrance of formula is required. Though the answer may not be accurate all the time rather they will be close to the answers of analytical solutions due to many factors like accuracy of drawing and precision of the instruments of geometry box. The process is also time consuming for solving any problem. In spite of that, such learning enhances our understanding of the fact that, how these laws works!     

To understand the practical aspect of the above discussion, we can solve a problem graphically to show the readers, how the concept works!

(Q1): A disable automobile is pulled by means of ropes subjected to the two forces as shown. Determine graphically, the magnitude and direction of their resulting using (a) the parallelogram law (b) the triangle rule. 

Figure-2 

Solution, 
To approach the problem for a graphical solution, we need a sharpened pencil, a scale, a protractor and a compass from geometry box.

Let,
1 mm length in the geometry scale is representing 1 KN of force shown in Fig-2

(a) Solution with Parallelogram Law   

Steps

• We draw a horizontal straight line of 4 mm which represents the 4 KN force
• Then we draw an another co-planer and concurrent force line of 2 mm, and inclined to 55⁰ (i.e. 30⁰+25⁰) with the horizontal force. Which represents the 2 KN force from Fig-2.
• Their directions are shown as given in the Fig-2.
• Now we make a parallelogram by connecting the parallel lines of those we have drawn, with the help of compass.  
• Now we have to join the two points of the distant corners of the parallelogram thus drawing its major diagonal.
• The length of the Diagonal comes out to be 5.4 mm in geometry scale. From which we can say the magnitude of the resultant is 5.4 KN 

Figure-2A

(b) Solution with Triangular Law
 Steps

• First we draw a horizontal line of 4 mm representing the 4 KN force.
• Then we draw an another co-planer concurrent force line of 2 mm length, inclined to the horizontal at 125⁰ (i.e. 180⁰- 55⁰). Which represents the 2 KN force.
• Their directions are shown as per Fig-2    
• Now we join the open parts of the figure we have drawn, thus making it a triangle. As this line is representing the resultant force, so as per Triangular law, the direction should be opposite to the other forces.
• The length of this line represents the magnitude of resultant force. By measuring the length we can say the magnitude of resultant force is 5.3 KN     

Figure-2B

Now let us compare the results with the analytical (trigonometric) solution of the same problem

Figure-2C | Trigonometric solution of problem (Q1)

In analytical solution the magnitude of resultant force comes out to be s 5.276 KN, which is satisfactory and all the values of the answer are near to each other     

Answers

(a) 5.4 KN

(b) 5.3 KN

(c) Trigonometric (Analytical) solution = 5.276 KN

The Polygon Law of Forces :

This law is an extension of Triangle Law of Forces for more than two forces, and it states that :

"If a number of forces acting simultaneously on a particle, be represented in magnitude and direction, by the sides of a polygon taken in order; then the resultant of all these forces may be represented, in magnitude and direction, by the closing side of the polygon, taken in opposite order".

Let us take six forces and arrange them as they represents the sides of a polygon. Their directions are taken as similar to each other. 

Now, the Fig-3 shows, how using the Triangular law, we can find the ultimate resultant and also validate the statement of Polygon Law of forces,

Figure-3|Polygon Law of Forces

Where  

Resolution of Forces : 

The process of splitting up the given force into a number of components, without changing its effect is known as Resolution of force.

The principle of resolution may be precisely stated as "The algebraic sum of the resolved parts of a number of forces, in a given direction, is equal to the resolved part of their resultant in the same direction"

Generally forces are resolved in vertical and horizontal directions.

For better understanding of the resolution of force, let's solve a problem

(Q2): A machine component of 1.5 m long and weight 1000 N is suspended by two ropes AB and CD as shown in Fig. 4 given below.

Figure-4

Calculate the tensions T1 and T2 in the ropes AB and CD.

Solution, 

In this problem, first we can check visually. There are 3 forces acting in the system. The tension force T1 and T2 with their respective angle and the weight of the machine part, which is given as 1000N.

Now, let us resolve the forces horizontally; that means we are equating the horizontal forces of opposite directions :

T1 cos 60⁰ = T2 cos 45⁰

    

T1 = 1.414 X T2............(i) 

Now, let us resolve the forces horizontally,

T2 sin 45⁰ + T1 sin 60⁰ = 1000

T2 sin 45⁰ + (1.414 X T2) X sin 60⁰ = 1000

0.707 X T2 + (1.414 X T2) X 0.707  = 1000

               

1.93 X T2 = 1000..........(ii)

So,   

T2 = 518.1 N

      

T1 = 1.141 X 518.1 = 732.6 N

Hence, our required answers are calculated above.

We will continue our journey of solving problems in the next piece of blog. We will also enjoy learning the method of resolution for the resultant force. The readers must be assured that their author will be working hard to make the coming blogs exciting and satisfying too. 

Readers are also encouraged to comment on the content and suggest ways of making the presentation better and also requested to share the blog.