BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer The resultant of two forces P and Q is of magnitude p. If P be doubled, the resultant will be inclined to Q at an angle

    A) \[0{}^\circ \]                                      

    B) \[30{}^\circ \]

    C) \[60{}^\circ \]                   

    D)        \[90{}^\circ \]

    Correct Answer: D

    Solution :

    Resultant of P and Q, \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\] Here,\[R=P\] \[\therefore \]  \[P=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\] \[{{P}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta \]           \[{{Q}^{2}}+2PQ\cos \theta =0\] \[Q+2P\cos \theta =0\] \[(\because \,P\ne 0)\]                 Required angle                                 \[\beta ={{\tan }^{-1}}\left[ \frac{p}{Q+p\cos \theta } \right]\]                 \[={{\tan }^{-1}}\left( \frac{2P}{Q+2P\cos \theta } \right)\]          \[(\because \,P=2P)\]                                 \[={{\tan }^{-1}}\left( \frac{2P}{0} \right)\]                                 \[={{\tan }^{-1}}(\infty )={{90}^{o}}\]

adversite


You need to login to perform this action.
You will be redirected in 3 sec spinner