A block of mass m is sliding on an inclined right angle through as shown in Fig. 7.46(a) and (b) If mu is the coefficient of kinetic friction, find the acceleration of the block .

A block of mass m is sliding on an inclined right angle through as sho

1.	A block of mass m is sliding on an inclined right angle through as shown in Fig. 7.46(a) and (b) If mu is the coefficient of kinetic friction, find the acceleration of the block .
Answer» Solution :As here reactions are from two surfaces and the body is a cube so `R_(1) = R_(2) = R `. The resultant of these reactions will be perpendicular to LINE AB and will be given by `R' = 2 R "cos" 45^(@) = sqrt2 R"" ….. `(i) Now for equilibrium in ADIRECTION perpendicular to AB , `R' = "MG cos" theta` which in the light of Eqn. (i) gives `R = (1 //sqrt2) "mg cos" theta "" .......... (ii)` And for the motion along the trough , mg sin `theta - muR_(1) - muR_(2) = ma` or `"" a = "G sin " theta - (2 muR//m) "" ["as" R_(1) = R_(2) = R]` Substituting R from Eqn . (ii) in the above , `a = g "sin" theta - (2mu)/(m) xx ("mg cos" theta )/(sqrt2)` i.e, `"" a = g["sin" theta - sqrt2 , mu "cos" theta ]`