A block of mass m is at rest on an inclined plane having coefficient of friction less than tan theta
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Weight of the block, Mg, in vertically downward direction. This force is not due to inclined plane , but it is due to gravitation.
Normal force ,N. This force is due to inclined plane. Notice that for the equilibrium in the direction perpendicular to the inclined plane, N= Mg cos (theta). Here Mg cos(theta) is component of Mg in the direction perpendicular to the inclined plane.
Now, the block is also in equilibrium in the direction parallel to inclined plane. Therefore , the frictional force exerted by inclined plane, (mu) N= (mu) Mg cos ( theta) balances the Mg sin( theta) component of Mg.
Thus, we see that there are two forces exerted on the block BY inclined plane. These are N and ( mu) Mg cos(theta). These forces are perpendicular to each other.
Therefore , the total force on the block due to inclined plane is [N^2 + ( mu)^2 N^2]^1/2 = N[1 +(mu)^2]^1/2=Mg cos(theta) [1+(mu)^2]^1/2.
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If the reaction is N then
Frictional force=u N (u is coefficient of friction).
Since block is placed on the inclined plane, so
N=uN Cos (theta)
Here N and uN both the forces are applied on the block by the inclined plane and both are acting at right angle between them, therefore total force applied on the block by the inclined plane is
F (net)=[(uN)^2+N^2]^(1/2)
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The basic concept regarding this question is to know what is friction.
Now , Friction = coefficient of force(mew)*Normal reaction force acting upwards on the body. Here ,the weight of body is Mg acting vertically downwards. If we consider the X-Y plane on the Normal reaction ,then it will be equal to Mg cos(theta).
Therefore friction=mew* Mgcos(theta).
hope it helps you dear