question_answer

A couple is of moment G and the force forming the couple is P. If P is turned through a right angle, the moment of the couple thus formed is H. If instead, the forces P is turned through an angle a, then the moment of couple becomes     AIEEE  Solved  Paper-2003

A)                         \[G\sin \alpha -H\cos \alpha \]                 

B) \[H\cos \alpha +G\sin \alpha \]

C)       \[G\cos \alpha +H\sin \alpha \]

D)       \[H\sin \alpha -G\cos \alpha \]

Correct Answer: C

Solution :

Let \[AB=d\] AM is the perpendicular distance from A to B.                 In \[\Delta ABM\],                                 \[\sin \theta =\frac{AM}{AB}\]                                 \[AM=d\sin \theta \] Moment of couple, G = P . AM = P \[d\sin \theta \]                  ... (i) When, P is turned a right angle, then Moment of couple. \[H=P\,.\,\,d\sin \,({{90}^{o}}+\theta )=P\,d\cos \theta \]           ... (ii) when P is turned through an angle a, then Moment of couple                 \[=P\,.\,d\sin (\alpha +\theta )\]                 \[=P\,.\,d(\sin \alpha \cos \alpha +\cos \alpha \sin \theta )\]                 \[=(P\,.\,d\cos \theta )\sin \alpha +(P.\,\,d\sin \theta )\cos \alpha \] \[=G\cos \alpha +H\sin \alpha \]               [from Eqs. (i) and (ii)]