Question

A mass of 4 kg rests on a horizontal plane . the plane gradually inclined until an angle 15 with the horizontal , the mass just begins to slide . what is the coefficient of static friction between the block and the surface

Answer 1

Answer:

0.27.

Explanation:

The frictional force is given as μN = fs which is the static friction. Hence, we know that the N is mgcos15. Where 15 is the angle of inclination. The Fs from the free body diagram is mgsin15. The body will slide freely along the inclined plane if there was no friction

So, on equating both of the equation we will get the coeffiicient of the friction as μ=tan15 which is equal to the 0.27.

Answer 2

mass of body , m = 4kg rests on a horizontal plane. now the plane gradually inclined untill angle 15° with the horizontal.

The mass just begins to slide (i.e., downwards) so, static friction applied just opposite direction of it and hence, static friction acts upwards along plane.

at that instant of time.

weight of body along plane = static friction force

or, mgsinФ = μ N , where N is normal reaction acting on the body.

N = component of weight of body perpendicular to plane = mgcosΦ

now, mgsinΦ = μmgcosΦ

⇒ tanФ = μ

now putting the value of Φ = 15°

so, μ = tan15° = 2 - √3

hence, coefficient of static friction between the block and the surface is (2-√3)