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# A block of mass m is placed on a smooth inclined plane of inclination q with the horizontal. The forceexerted by the plane on the block has a magnitude.(a) mg (b) mg/cos q (c) mg cos q (d) mg tan q

Anirban Mukherjee
47 Points
8 years ago

From FBD of the situations we get R= mg cos A

Aman Bansal
592 Points
8 years ago

Dea Jeet,

 Consider a block of mass "m" placed on an inclined plane, which makes an angle q 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:wcosq and wsinqother 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. For latest information , free computer courses and high impact notes visit : www.citycollegiate.com Fx = 0f – wsinq = 0 -------(1) Fy = 0R– wcosq = 0 -------(2) Since there is no motion in the direction perpendicular to the inclined plane, therefore wcosq is balanced by R i.e. R = wcosq If block slides down with an acceleration equal to ''a'', then the resultant force is equal to ''ma'' and the force on block will be: wsinq – f According to 2nd law of motion wsinq – f = ma If the force of friction is negligible, then wsinq = ma but w = mg This expression shows that if friction is negligible the acceleration of a body on an inclined plane is independent of mass but is directly proportional to sinq. WHEN FRICTION IS NOT NEGLIGIBLE (i) If block moves upward f - wsinq = ma (ii) If block moves downward wsinq – f = ma

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Thanks

Aman Bansal

Chris Chan
33 Points
8 years ago

mg cos q - because the plane is horizontal...

G Ashwin -
24 Points
8 years ago
you c
Neeraj nani
15 Points
2 years ago
MgBecause all components depend on
ankit singh