Question

A block of mass 10kg is placed on an inclined plane.when the angle of inclination is 30 degree the block just begins to slide down the plane.the force of static friction

Answer 1

Answer:

plz check the question because value of coefficient of friction should be given to find the force of static friction.

if it is put the in last value and find the ans.

Explanation:

Draw F.B.D of block -

There is only component of gravity mgsin30°

N = mgcos30° = 10 x 10 x √3/2 = 50√3

the block just begins to move

therefore maximum static friction is applied

fsmax = in = u50√3 N

(where u is coefficient of friction)

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