Let ‘P’ be the given force represented in magnitude and direction by OB as shown in Fig. 1. Also let OX and OY be two given directions along which the components of P are to be found out.
Let < BOX = α and < BOY = β
From B, lines BA and BC are drawn parallel to OY and OX respectively. Then the required components of the given force P along OX and OY are represented in magnitude and direction by OA and OC respectively.
Since AB is parallel to OC,
<BAX = <AOC = α + β
< AOB = 1800 – (α + β)
Now, in Δ OAB
Determination of Resolved Parts of a Force
Resolved parts of a force mean components of the force along two mutually perpendicular directions. Let a force F represented in magnitude and direction by OC make an angle θ with OX. Line OY is drawn through O at right angles to OX as shown in Fig.2.
Through C, lines CA and CB are drawn parallel to OY and OX respectively. Then the resolved parts of the force F along OX and OY are represented in magnitude and direction by OA and OB respectively.
Now in the right angled Δ AOC,
i.e. OA = F cos θ
Since OA is parallel to BC,
<OCB = <AOC = θ
In the right angled Δ OBC,
i.e., OB = F sin θ
Thus, the resolved parts of F along OX and OY are respectively F cos θ and F sin θ.
Significance of the Resolved Parts of A Force
Similarly, if a force of 43.3kN is required to be applied to the body to lift it vertically upward, then the body will be lifted vertically upward if a force of 50kN is applied to the body at an angle of 60° with the horizontal, as the resolved part of 50kN along the vertical CE = 50 sin60° = 43.3kN.
Thus, the resolved part of a force in any direction represents the whole effect of the force in that direction.
Equilibriant
Equilibrant of a system of forces is a single force which will keep the given forces in equilibrium. Evidently, equilibrant is equal and opposite to the resultant of the given forces.
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