significance of parallelogram law of force s
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Answer:
The law of parallelogram of forces:
The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram drawn from the same point .
Magnitude and Direction of Resultant:
Draw a perpendicular QN to OP produced.
And let us assume that OP=A, OS= PQ= B, OQ=R and angle SOP= angle QPN = θ.
Now considering this if we proceed further , in the case of triangle law of vector addition , the magnitude and direction of resultant vector will be given by
R= sqrt of A^2 + B^2 + 2 AB cosθ
tan B = B sinθ/ A+B cosθ
Special cases:
(1) When two vectors are acting in the same direction , then θ= 0 , cosθ=1 and sinθ= 0
R= sqrt of A^2 + B^2 + 2 AB
=sqrt of (A+B)^2 = A + B
tan Beta = B X 0/ A+B = 0
Beta = 0
Thus for two vectors acting in the same direction the magnitude of the resultant vector is equal to the sum of the magnitudes of two vectors and act along the direction of A and B.
(2) When two vectors are acting in opposite directions , then θ= 180 , cos θ= -1 and sinθ= 0
R= sqrt of A^2+ B^2+ 2 AB (-1)
= sqrt of (A-B) or (B-A)
tan beta = B X 0/ A+ B (-1)= 0
Beta = 0 or 180.
Thus for two vectors acting in opposite directions, the magnitude of the resultant vector is equal to the difference of the magnitudes of the two vectors and acts in the direction of bigger vector.
(3) When two vectors act at right angle to each other θ = 90 , sinθ = 1 and cosθ = 0
R= sqrt of A^2+B^2 + 2 AB (0)
= sqrt of A^2+B^2
tan beta = B(1)/A+B(0)= B/A
or, Beta = tan^-1 B/A
