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A block of mass M is resting on an inclined plane shown in fig. The inclination of the plane to the horizontal is gradually increased . It is found that when the angle of inclination is $\theta$ the block just begins to the slide down the plane. What is the minimum force F applied parallel to the plane that would just make the block move up the plane ?

$\begin{array}{1 1} Mg\;\sin \theta \\ Mg\;cos \theta \\ 2\;Mg\;\cos \theta \\ 2\;Mg\;\sin \theta \end{array} $

1 Answer

Answer : $ 2\;Mg\;\sin \theta$
The component of weight $Mg$ of the block along the inclined plane $=Mg \sin \theta$.
The minimum frictional force to be overcome is also $Mg \sin \theta$
To make the block just move up the plane the minimum force applied must overcome the component $Mg \sin \theta$ of gravitational force as well as the frictional force $Mg \sin \theta = 2 \;Mg \sin \theta$.
answered Aug 26, 2014 by meena.p

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