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

jeet jeetu, 11 years ago

Grade:11

6 Answers

Anirban Mukherjee

47 Points

11 years ago

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

Approved

Aman Bansal

592 Points

11 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 1^st 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 2^nd 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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Aman Bansal

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Chris Chan

33 Points

11 years ago

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

G Ashwin -

24 Points

11 years ago

you c

Neeraj nani

15 Points

4 years ago

MgBecause all components depend on

ankit singh

askIITians Faculty 614 Points

3 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 1^st 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