See Fig. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle theta=15^@with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?

See Fig. A mass of 4 kg rests on a horizontal plane. The plane is grad

1.	See Fig. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle theta=15^@with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?
Answer» Solution :The forces ACTING on a block of mass m at rest on an inclined PLANE are (i) the weight mg acting vertically downwards (ii) the normal FORCE N of the plane on the block, and (iii) the static frictional force fs opposing the impending motion. In EQUILIBRIUM, the resultant of these forces must be zero.Resolving the weight mg along the two directions shown, we have `mg SIN theta = f_s , mg cos theta = N` As `theta `increases , the self - adjusting frictional force `f_s` increases unit at ` theta = theta_(max), f_s` achieves its maximum value `(f_s)_(max) = mu_s N` . Therefore ,`tan theta_(max) = mu_s "or" theta_(max) = tan^(-1) mu_s` When` theta`becomes just a little more than ` theta_( max)` , there is a small net force on the block and it begins to slide.Note that`theta_( max)`depends only on µs and isindependent of the mass of the block. for`theta_(max) = 15^@` `mu_s = tan 15^@` = 0.27