A box is placed on a plank One end of the plank is gradually

A box is placed on a plank. One end of the plank is gradually raised. When the angle of inclination with the horizontal reaches 30 degrees, the box starts to slip and it then slides 2.5 m down the plank in 4.5 seconds. Find (a) the coeffiencent of static friction, and (b) coefficient of kinetic friction between the box and the plank

Solution

(a) Just before sliding, static friction balances the component of weight along the incline.

So, µ_smgcos = mgsin

=> coefficient of static friction, µ_s = tan = tan30^o = 0.577

(b) Force equation:

mgsin - µ_kmgcos = ma

where a = 2s/t² = 2 * 2.5 / 4.5² = 0.247 m/s²

So, gsin - µ_kgcos = 0.247

=> coefficient of kinetic friction, µ_k = tan - 0.247/gcos = tan30^o - 0.247/(9.81*cos30^o) = 0.548