Haryana PMT Haryana PMT Solved Paper-2010

  • question_answer
    A body of mass 10 kg slides along a rough horizontal surface. The coefficient of friction is I\[1/\sqrt{3}\]. Taking g = \[10m/{{s}^{2}},\] the least force which acts at an angle of 30° to the horizontal is

    A)  25 N                                     

    B)  100 N

    C)  50 N                                     

    D) \[\frac{50}{\sqrt{3}}N\]  

    Correct Answer: C

    Solution :

                    Let P force is acting at an angle \[{{30}^{o}}\] with the horizontal for the condition of motion, \[F=\mu R\] \[P\,\,\cos \,{{30}^{o}}=\mu (mg-P\,\,\sin {{30}^{o}})\]                 \[\Rightarrow \]               \[P\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}\left( 100-P\times \frac{1}{2} \right)\]                 \[\Rightarrow \]               \[\frac{3P}{2}=\left( 100-\frac{P}{2} \right)\] \[\Rightarrow \]               \[2P=100\]  \[\Rightarrow \]  \[P=50N\]


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