Understanding the parallelogram law of forces through problems : Part II

In this piece of writing, we will be understanding the way of using resolution of forces in order to determine the value of resultant force of complex force systems. As we know concept of resolution tells us to resolve any force into the directions of our favour (generally perpendicular). Thus we will break every force of a given system into the direction of our preference. As we choose the direction,  hence we know the angle between the resolved forces. Then we can simply calculate the net force in each direction and put their value in the formula of resultant and obtain the net resultant of the system. 

Apart from the above concept we will also travel through some of the problems related to the learning so far and their solutions to bolster the understanding of the resolution of forces.  

Resolution for the Resultant 

Basically it is an algebraic operation, where net horizontal and vertical forces  are determined through resolution; and a resultant force is computed. The real taste of understanding of this concept can be grasped by solving a problem.

Figure-1

The force system in Figure-1 is posing a question; to determine the magnitude of resultant force of the system. There is liberty to assume that 40 N force is acting in the positive X-direction and 50 N is in the positive Y-direction.      

To determine, the resultant,k we first need to resolve the 30 N force into the direction of other two forces in order to compute net forces. But before that we need to determine the q in Figure-1:

Now, we are free to resolve the 30 N force in vertical and horizontal direction as well. And that is precisely shown in Figure-2 below.

Figure-2

After resolution; the components of 30 N force in the horizontal and vertical directions are :

Horizontal Component : 30 cos 36.87⁰ = 24 N ( Negative X Direction)
    Vertical Component : 30 sin 36.87⁰  = 18 N ( Negative Y Direction)

Now, we can have the net force in both the directions as :
Net force in Horizontal Direction ∑H = 40-24 =16 N (Positive X Direction)
   Net force in Vertical Direction ∑V =  50-18 =32 N (Positive Y Direction)

As we know the angle between both the net forces is 90⁰. Hence the equation for net resultant force can be written as :

Therefore the magnitude obtained for resultant force is about 36 N. Apart from that we also need to find the direction of the resultant force, as it is a vector quantity. The direction of the resultant can be indicated by showing its inclination angle with any of the axis, as the constituent  net forces are taken through X and Y axis. If we chose to show the angle (say a) with the horizontal, then we can write :

The problem and the solution teaches us about how to approach such questions! How to identify;  which force needs resolution in what direction! and both magnitude and direction can be found. 

So, we now can make a list of general tasks to be remembered and applied in such problems. 

Methods of Resolution for the Resultant Force

• Resolve all the forces horizontally and find the the algebraic sum of all the horizontal components (i.e.,∑H).      
• Resolve all the forces vertically and find the the algebraic sum of all the vertical components (i.e.,∑V).   
• The resultant R of the given forces will be given by the equation:               
  
• The resultant force will be inclined at an angle  a, (if) with the horizontal, such that : 
  

The value of the angle a will vary depending upon the values of ∑V and  ∑H.

When,

∑V is positive, 0⁰ < a <180⁰  

∑V is negative, 180⁰ < a <360⁰ 
Again when,
∑V is positive,  0⁰ < a <90⁰  or 270⁰ < a <360⁰ 
∑V is negative, 90⁰ < a <270⁰ 

Above conclusions can be tested by randomly putting the values of ∑V and  ∑H for a

Let us now move on to some problems as promised, and their solutions to consolidate our understanding of the application of our learnings. The problems are chosen from exercise of the book named "Vector Mechanics for Engineers" by Beer, Johnson and others. These are simple problems with the taste of real life situations. Going through them will indeed be a pleasure of the readers hopefully.

(Q-1) 
Case-1 : Two forces P and Q are applied as shown in Figure-3 at point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine the magnitude of resultant and its direction using simple trigonometry.

Figure-3

Case-2 : For the hook support, knowing that the magnitude of P is 75 N, determine by trigonometry :
a) The required magnitude of force Q if the resultant R of the two forces applied at A is to be vertical.
b) The corresponding magnitude of R.

Solution for Case-1
Let us first draw the corresponding free body diagram (Figure-3A) with the conditions of Case-1

Figure-3A

As the Case-1 does not state either any horizontal movement or inclination of the hook. Therefore we can consider the horizontal components of P and Q, are cancelling themselves and thus making the system horizontally steady.

Hence the required resultant force is (as indicated in the figure)

                                          R = 75 cos 20⁰ + 125 cos 35 ⁰ 

                                                                    R = 172.87 N

Solution for Case-2
(a) The Case-2 talks about the resultant which is vertical. Therefore let us draw an another free body diagram with the condition of Case-2 (Figure-3B).

Figure-3B

Now equating the horizontal components, we get :

                                           Q x sin 35⁰ =  75 sin 20⁰ 
                                                                             Q = 44.72 N

(b) Now from the Figure-3B we can, compute the resultant as :
                                        R = 75 cos 20⁰ + 44.72 cos 35 ⁰ 
                                              R = 107.12 N
Hence all the required answers of Q-1 is obtained.

(Q-2) Case-1 : A steel tank is to be positioned in an excavation. Knowing that a = 20 ⁰ determine by trigonometry :
(a) The required magnitude of force P if the resultant R of the two forces applied at A is to be vertical.
(b) The corresponding magnitude of R.

Figure-4

Case-2 : Knowing that the magnitude of P is 3000 N, determine by trigonometry.
(a) The required angle if the resultant R of the two forces applied at A is to be vertical.
(b)  The corresponding magnitude of R.

Solution for Case-1
(a) Let us first draw the free body diagram with the conditions of Case-1

Figure-4A

As there is no horizontal motion, therefore we can equate the horizontal components of both the forces in order to find the magnitude of P.

            P x cos 20⁰ = 2125  cos 30⁰ 

P = 1958.4 N

(b) Since we have computed the magnitude of P. Therefore the magnitude of resultant can be obtained from the equation indicated in Figure-4A.

R = 2125 sin 30⁰ + 1958.4 sin 35 ⁰ 

R = 1732.31 N

Solution for Case-2

A free body diagram with the conditions of Case-2 is drawn below.

Figure-4B

(a) As there is no horizontal motion, therefore we can equate the horizontal components of both the forces in order to find the magnitude of a. And we get :

(b) As we have obtained the value of a, therefore from Figure-4B, we can get the magnitude of resultant indicated in the figure.

R = 2125 sin 30⁰ + 3000 sin 69.26 ⁰

R = 3868.09 N  

Hence all the required answers of Q-2 is obtained.

Here is the end of this blog, the next one will be fully problem oriented and the solutions will be discussed with all efforts. Readers are requested to comment their opinions and also any correction in the content will be addressed with care.

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.  

The Force systems and the Principles of Mechanics : Parallelogram Law of Forces

Understanding the Principles of Mechanics 

To understand the principles of the elementary mechanics, it is necessary  to know about the various kinds of orientations of forces or the system of forces. 

"When two or more forces act on a body, they are called the system of forces".  

Following systems of force are defined below to address the need of their references and understanding in future topics. 

1. Coplanner Forces :  The forces, whose lines of action lies on the same plane, are known as co planner forces.
2. Co-linear Forces : The forces, whose lines of action lies on the same line, are known as co linear forces. Co-linear forces may not be on the same plane but they will be linear in acting all the time 
3. Concurrent Forces : The forces, Which meet at one point, are known as concurrent forces. (Irrespective to their lines and planes.)  
4. Coplanner concurrent Forces : The forces, which meet at one point and their lines of action also lie on the same plane, are known as co-planner concurrent forces.
5. Coplanner non-concurrent Forces : The forces, which don not meet at one point, but their lines of action lie on the same plane, are known as coplanner non-concurrent forces. 
6. Non-coplanner concurrent Forces : The forces, which meet at one point but their lines of action do not lie on the same plane, are known as Non-coplanner concurrent forces.
7. Non-coplanner non-concurrent Forces : The forces, which don not meet at one point, and their lines of action do not lie on the same plane, are known as Non-coplanner non-concurrent forces. 

Resultant of A Force : 

The idea of resultant is very important in mechanics as it accommodates the effect of all the acting forces and turn them together into a single representative force. A resultant force properly can be defined as, "If a number of forces acting simultaneously on a particle, then it is possible to find out a single force which would produce the same effect as produced by all the given forces. This single force is called the resultant force" 

POINTS

• The number of given forces are known as component forces.
• The process of finding out the resultant force, of a number of given forces, is called composition of forces.

As mentioned in my earlier blog, we are now ready to learn the the first foundational Principle of elementary mechanics, which is the "Parallelogram Law of Forces". The above written ideas and their understanding is important for moving further into the coming topic.

Parallelogram Law of Forces : 

This law essentially gives the clear idea of a resultant and it shows the way of calculating in the force systems. The statement of the law can be written as :

"If two forces acting simultaneously on a particle, be represented in magnitude and direction by two adjacent sides of a parallelogram; their resultant may be represented in magnitude and direction by the diagonal of the parallelogram, which passes through their intersection".   

        Figure - 1

Now, referring to figure-1, 

Let, F1, F2 be the forces whose resultant R is required to be found out and q is the angle between the forces F1 & F2.

Let, a be the angle which the resultant force makes with one of the forces (Say F1)  

Then, as per Vector Algebra we can write 

 

Now, If the angle (a) which the resultant force makes with the other force F2.

Some important corollaries from Eqn(1)  

• If q=0⁰, when the forces act along the same line and direction, then                  R = F1 + F2                                   (Since cos 0⁰= 1)
• If q=90⁰, when the forces act at right angles                                                       
  
                       (Since cos 90⁰= 0)                                                                                                                                                                                                                                                              
• If q=180⁰, when the forces act along the same line and opposite direction, then,                                                                                                      R = F1 - F2                                          (Since cos 180⁰= -1)                        In this case resultant forces will act in the direction of the greater force       
• If the two forces are equal i.e., when F1 = F2= F then                               
  
  In the next piece of blog we will continue to travel through the learning of this blog and also explore its application in some exciting problems.