Friday, May 15, 2020

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 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 36.87= 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 will vary depending upon the values of ∑V and  ∑H.
When,
∑V is positive, 0a <1800  
∑V is negative, 180a <3600 
Again when,
∑V is positive,  0a <900  or 270a <3600 
∑V is negative, 90a <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 20125 cos 35 
                                                                    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 2044.72 cos 35 
                                              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 20 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 30+ 1958.4 sin 35 
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 30+ 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 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 forc...