A force p=3mg acts on a …
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Govind Singh 7 years, 11 months ago
Here the vertical component of the force is {tex}F\sin \theta = 3mg\;\sin \left( {30^\circ } \right) = 1.5mg{/tex}, and it will be in vertically upward. So the resultant force will be 1.5mg - mg = 0.5mg, and upwards. Since the resultant force is upwards, the normal reaction between ground and the block will be zero. So the frictional force will also be zero.
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