JEE Main & Advanced Mathematics Statistics Parallelogram Law of Forces

If two forces, acting at a point, be represented in magnitude and direction by the two sides of a parallelogram drawn from one of its angular points, their resultant is represented both in magnitude and direction of the parallelogram drawn through that point.

If \[OA\] and \[OB\] represent the forces \[P\] and \[Q\] acting at a point \[O\] and inclined to each other at an angle\[\alpha \]. If \[R\] is the resultant of these forces represented by the diagonal \[OC\] of the parallelogram \[OACB\] and \[R\] makes an angle \[\theta \]  with \[P\].

i.e., \[\angle COA=\theta \], then \[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \alpha \] and \[\tan \theta =\frac{Q\sin \alpha }{P+Q\cos \alpha }\]

The angle \[{{\theta }_{1}}\] which the resultant \[R\] makes with the direction of the force \[Q\] is given by \[{{\theta }_{1}}={{\tan }^{-1}}\left( \frac{P\sin \alpha }{Q+P\cos \alpha } \right)\]

Case (i) : If \[P=Q\]

\[\therefore R=2P\cos \,\left( \frac{\alpha }{2} \right)\] and \[\tan \theta =\tan \left( \frac{\alpha }{2} \right)\] or \[\theta =\frac{\alpha }{2}\]

Case (ii) : If \[\alpha =90{}^\circ \], i.e. forces are perpendicular

\[\therefore R=\sqrt{{{P}^{2}}+{{Q}^{2}}}\]and \[\tan \theta =\frac{Q}{P}\]

Case (iii) : If \[\alpha =0{}^\circ \], i.e. forces act in the same direction

\[\therefore {{R}_{\max }}=P+Q\]

Case (iv) : If \[\alpha =180{}^\circ \], i.e. forces act in opposite direction

\[\therefore {{R}_{\min }}=P-Q\]

• The resultant of two forces is closer to the larger force.

• The resultant of two equal forces of magnitude \[P\] acting at an angle \[\alpha \] is \[2P\cos \frac{\alpha }{2}\] and it bisects the angle between the forces.

• If the resultant R of two forces \[P\] and \[Q\] acting at an angle \[\alpha \] makes an angle \[\theta \]  with the direction of \[P,\] then  \[\sin \theta =\frac{Q\sin \alpha }{R}\]and \[\cos \theta =\frac{P+Q\cos \alpha }{R}\]

• If the resultant \[R\] of the forces \[P\] and \[Q\] acting at an angle \[\alpha \] makes an angle \[\theta \] with the direction of the force \[Q,\] then \[\sin \theta =\frac{P\sin \alpha }{R}\] and \[\cos \theta =\frac{Q+P\sin \alpha }{R}\]

Component of a force in two directions : The component of a force \[R\] in two directions making angles \[\alpha \]  and \[\beta \]  with the line of action of \[R\] on and opposite sides of it are

\[{{F}_{1}}=\frac{OC.\sin \beta }{\sin (\alpha +\beta )}=\frac{R\sin \beta }{\sin (\alpha +\beta )}\]

and          \[{{F}_{2}}=\frac{OC.\sin \alpha }{\sin (\alpha +\beta )}=\frac{R.\sin \alpha }{\sin (\alpha +\beta )}\]

\[\mathbf{\lambda r\mu }\] theorem : The resultant of two forces acting at a point \[O\] in directions \[OA\] and \[OB\]represented in magnitudes by \[\lambda .OA\]and \[\mu .OB\] respectively is represented by \[(\lambda +\mu )OC\], where \[C\] is a point in \[AB\] such that \[\lambda .CA=\mu .CB\].