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 ?

Options

A. 0.5
B. 0.27
C. 0.85
D. 0.6

Answer

B

Explanation

The forces acting on a block of massmatrest on an inclined plane the weight mg acting vertically downwards (m) the normal force N of the plane on the block, and (mm) the static frictional force `f_(k)` 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 rArr tan theta =f_(s)/N`
As `theta` increases, the self-adjusting frictional forces increases until at `theta = theta_("max"), f_(b)` achieves its maximum value, `(f_(s))_("max") = mu_(s)N rArr (f_(u))_("max")//N = mu_(s)`
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 depends only on and is independent of the mass of the block. For `theta_("max") = 15^(@) rArr mu_(s) = tan 15^(@) = 0.27`