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)]You need to login to perform this action.
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