Question

A mass of 4 kg rest on a horizontal plane . The plane is gradually inclined until an angle `0 = 15^(@)` with the horizontal and the mass just begins to slide . What is the coefficient of static friction between the block and the surface ?

Answer 1

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 `f_s` opposing the impending motion. In equilibrium , the resulatant of these forces must be zero. Resolving the weight mg along the two directions shown, we have
m g sin `theta` = `f_s` , m g `cos theta` = N
As `theta` increases , the self-adjusting frictional force `f_s` increases until at `theta` = `theta_max`, `f_s` achieves its maximum value , (`f_s`)max = `mu_s`N.
Therefore , `tan lttheta_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 `mu_s` and is independent of the mass of the block.
For `theta _max` = `15^(@)`
`mu_s` = tan `15^(@)`
= 0.27