METHODS OF RESOLUTION

RESOLUTION OF FORCE

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Ḟ  = F < Ө       ( F = Magnitude , Ө = Direction)

from ∆OAC , OA/OC = COS Ө , OA = OC COS Ө = F COS Ө

This is the y – component

PROVE

From ∆OAC , x²=(F COS Ө)²+(F SIN Ө)²

X² = F²(SIN² Ө+ COS² Ө)

X² = F²

X = F (Proved)

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Algebric sum of X – component

(  ͢+) Σf_x= +f₁cos Ө₁- f₂cos Ө₂- f₃cos Ө₃+ f₄cos Ө₄

Algebric sum of Y – component

(+↑)Σf_y = +f₁ sin Ө₁+ f₂ sin Ө₂- f₃ sin Ө₃- f₄ sin Ө₄

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Resultant force R² = Σf_x²+ Σf_y²

Magnitude of  Ṝ = (Σf_x²+ Σf_y²)^1/2

Direction α  =tan^-1(Y-component/X-component)

                  α  =tan^-1(Σf_x/ Σf_y)

                                  TRIANGLE LAW

STATEMENT

If two forces are acting simultaneously on a body may represent by the two adjacent sides of a triangle taken in order , then the third side of the triangle will represent the resultant force taken in opposite order.

EXPLANATION

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AB = F₁ , BC = F₂ , AC = R = resultant force in opposite order.

                                                         POLYGON LAW

STATEMENT

If a number of forces (more than two) are acting simultaneously on abody or a particle may be represented by the side of a polygon taken in order , then the closing side of polygon will represent the resultant force taken opposite order.

EXPLANATION

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Assuming F₁ ,F₂ , F₃ ,F₄ are the forces acting simultaneously on a body as soon in figure.

Then , according to polygon law R be the resultant force.