BLOCK towards right.             For the condition of equilibrium             \[mg\sin \theta =ma\COS \theta \] Þ \[a=\frac{g\sin \theta }{\cos \theta }\]             \[\THEREFORE \] Force exerted by the WEDGE on the block             \[R=mg\cos \theta +ma\sin \theta \]    R\[=mg\cos \theta +m\left( \frac{g\sin \theta }{\cos \theta } \right)\sin \theta \]\[=\frac{mg({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}{\cos \theta }\]             R\[=\frac{mg}{\cos \theta }\]

"> BLOCK towards right.             For the condition of equilibrium             \[mg\sin \theta =ma\COS \theta \] Þ \[a=\frac{g\sin \theta }{\cos \theta }\]             \[\THEREFORE \] Force exerted by the WEDGE on the block             \[R=mg\cos \theta +ma\sin \theta \]    R\[=mg\cos \theta +m\left( \frac{g\sin \theta }{\cos \theta } \right)\sin \theta \]\[=\frac{mg({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}{\cos \theta }\]             R\[=\frac{mg}{\cos \theta }\]

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A block of mass m is placed on a smooth wedge of inclination \[\theta \]. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be                               [CBSE PMT 2004]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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When the whole system is accelerated towards left then pseudo force (ma) works on a BLOCK towards right.             For the condition of equilibrium             \[mg\sin \theta =ma\COS \theta \] Þ \[a=\frac{g\sin \theta }{\cos \theta }\]             \[\THEREFORE \] Force exerted by the WEDGE on the block             \[R=mg\cos \theta +ma\sin \theta \]    R\[=mg\cos \theta +m\left( \frac{g\sin \theta }{\cos \theta } \right)\sin \theta \]\[=\frac{mg({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}{\cos \theta }\]             R\[=\frac{mg}{\cos \theta }\]

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