Law of Parallelogram of Forces: Definition, Derivative, Special Case

Hello friends, Today I am going to discuss the law of parallelogram of forces in which you will be aware of all the aspects related to the law of parallelogram of forces such as their definition, magnitude, and direction with derivative proof, along with also discuss some special cases.

So without wasting time let's get started.

What is the Law of Parallelogram of Forces?

According to this Law "If two coplanar forces, acting simultaneously on a point of the body, be represented in magnitude and direction by the two adjacent sides of a parallelogram then their resultant can be represented in magnitude and direction parallelogram." by the diagonal of the parallelogram".

This method is only useful for finding the resultant of only two coplanar concurrent forces.

Also Read: Law of Gearing

Magnitude and Direction Derivative with Proof

See in the figure,

Where,

OA =P  &  OB = Q 

Now from the figure (Corresponding angles)

 ∠AOB = ∠DAC = α

Now in △ADC 

As we know 

sinα = perpendicular/hypotenuse

cosα  = base/hypotenuse

So,

sinα = CD/AC 

 CD = AC sinα  

And

cosα = AD/AC

 AD = AC cosα

Now,

Now in △ODC,

OC²  = OD² + CD² = (OA+ AD)²+ CD² 

( since OD = OA + AD, from figure)

        = OA² + AD² + 2 OA ×AD + CD²

   = OA² + (AC cosα )² + 2 OA × AC cosα +(ACsinα)²

  = OA² + AC ² cos²α + 2 OA × AC cosα + (AC²

 sin²α)

= OA² + AC ²(sin²α + cos²α ) + 2 OA × ACcos α 

As We know,

sin²α + cos²α  = 1

Then

OC² = OA² + AC ²(1) + 2 OA × ACcos α 

OC² = OA² + AC ² + 2 OA × ACcos α 

Since,

OA = P

AC = AB = Q

OC = R

So,

R²= P ² + Q ² + 2P × Q cosα

Then,

Resultant 

R = √(P ² + Q ² + 2P × Q cosα)

Now for finding the direction of R, let the resultant make angle θ with force P.

 Now in △ODC,

tanθ = CD/OD 

tanθ = CD/(OA+AD)

CD = AC × sinα

AD = AC × cosα

tanθ = AC sinα/(OA + AC cosα)

tanθ = (Q sinα)/(P + Q cosα)

θ =  tan-1{(Q sinα)/(P + Q cosα)}

Special Cases in Law of Parallelogram of Forces

See various case studies in the figure which will be discussed in detail below.

Case 1:  When P and Q act along the same line (α = 0°)

R = √(P ² + Q ² + 2P × Q cosα)

R = √(P ² + Q ² + 2P × Q cos0°)

As we know,

Cos0° = 1

R = √(P ² + Q ² + 2P × Q )

As we know,

(a + b)²  = a²  + b²  + 2ab

So,

R = √(P ² + Q ²)

R = P + Q

Case 2:  When P and Q act along the same line but in opposite direction (α = 180°)

R = √(P ² + Q ² + 2P × Q cos180°)

R = √(P ² + Q ² + 2P × -Q )

As we know,

Cos180° = -1

So,

R = √(P ² + Q ² - 2P × Q )

As we know,

(a-b) ² = (a ² + b ² - 2ab)

So,

R = √(P ² - Q ²)

R = (P - Q)

Case 3 : When P and Q act at right angles (α = 90°)

R = √(P ² + Q ² + 2P × Q cosα)

R = √(P ² + Q ² + 2P × Q cos90°)

As we know,

cos90° = 0

So,

R = √(P ² + Q ²)

Case 4 : When P and Q are equal in magnitude, act along the same line but in opposite direction  ( α = 180 ° &  P = Q )

R = √(P ² + Q ² + 2P × Q cosα)

R = √(P ² + P ² + 2P × P cos180°)

R = √(P ² + Q ² - 2P × P) 

Since,

cos180° = -1

Hence,

R = √(P ² + P ² - 2P × P) 

R = √(2P ²  - 2P ² ) 

R = 0

Previous Related Article:

.Varignon's Theorem

.Moment of Force

.Theories of Failure

.Maximum Principle Stress Theory

.Principal Stress

So here I discussed the law of parallelogram of forces and their magnitude and direction.

I hope you all enjoy this topic.

Thank You.