Question

Relation for which block doesnot slide on an incident plane

Answer 1

❤️$\huge\boxed{\fcolorbox{blue}{pink}{Explanation}}$❤️

Look at the fig. A block of mass m is placed on an inclined plane of inclination (theta).

The forces acting on the block are:

(1) The weight mg of the block.

(2) Normal force due to incline, N

(3) Limiting frictional force uN. Here, u is coefficient of friction between block - plane surfaces in contact.

See that we have taken two components of mg. These are mgcos(theta) perpendicular to the plane and mg sin( theta) parallel (downward ) to the plane.

For the block to remain on the block, N must be reaction to mg cos (theta) or

mg cos (theta) = N……………..(1)

The equation of motion of the block will be:

ma=mg sin (theta) - uN =mg sin (theta) - umg cos (theta). Here, we have used eq.(1) for N.

Cancelling m throughout ,

a=g[ sin (theta)- u cos (theta)]………………….(2)

Note that acceleration of the block depends on angle of inclined plane and coefficient of friction.

Starting from small value of (theta), we initially find that u cos( theta) > sin (theta) , then a is negative ,which is not possible in the present case, hence , the block does not move. In this situation the static frictional force is less than limiting frictional force and is equal to mg sin of angle at that time.

As we slowly go on increase the value of ( theta), a stage is reached, when

u cos (theta)=sin (theta) , then the block just starts sliding or tends to slide.

Or

u=tan(theta)…………………(3). The (theta) satisfying eq.(3) is called angle of angle of repose.

Here, in our problem angle of repose is 60 degree . Therefore .

coefficient of static friction ,u=tan (60) degree= sqrt(3).