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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


Grade:11

6 Answers

Anirban Mukherjee
47 Points
9 years ago

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

Aman Bansal
592 Points
9 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 wsinq
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.
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Fx = 0
f – wsinq = 0 -------(1)
Fy = 0
R– 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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Chris Chan
33 Points
9 years ago

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

G Ashwin -
24 Points
9 years ago
you c
Neeraj nani
15 Points
3 years ago
MgBecause all components depend on
ankit singh
askIITians Faculty 614 Points
2 years ago
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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