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A block of mass m is placed on a smooth inclined plane of inclination q with the horizontal. The force exerted by the plane on the block has a magnitude. (a) mg (b) mg/cos q (c) mg cos q (d) mg tan q A block of mass m is placed on a smooth inclined plane of inclination q with the horizontal. The force exerted by the plane on the block has a magnitude. (a) mg (b) mg/cos q (c) mg cos q (d) mg tan q
A block of mass m is placed on a smooth inclined plane of inclination q with the horizontal. The force
exerted by the plane on the block has a magnitude.
(a) mg (b) mg/cos q (c) mg cos q (d) mg tan q
From FBD of the situations we get R= mg cos A
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mg cos q - because the plane is horizontal...
you c
MgBecause all components depend on
other forces acting on the block are:(i) Surface reaction (R) which is perpendicular to the plane(ii) Force of friction (f) acting opposite to the direction of motion of block.Let us take x-axis perpendicular to the inclined plane. If the block is at rest, then wsinq acting down the plane balances the opposing frictional force.Applying 1st condition of equilibrium.q and wsinq with the horizontal plane. The weight "W" of the block is acting vertically downward. The weight of the block can be resolved into two rectangular components:wcosqConsider a block of mass "m" placed on an inclined plane, which makes an angle
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