A slightly loose fit widnow is balanced by two strings which are connected to weights w//2 each. The stringes pass over the frictionless pulleys as shown in the figure. The stringes are tied almost at the corner of the window accelerates downward. If coefficient of friction between the window and the side supports is mu then calculate the acceleration of the window in terms of mu, a, b and g. whereas width and b is the lengths of he window. .

A slightly loose fit widnow is balanced by two strings which are conne

1.	A slightly loose fit widnow is balanced by two strings which are connected to weights w//2 each. The stringes pass over the frictionless pulleys as shown in the figure. The stringes are tied almost at the corner of the window accelerates downward. If coefficient of friction between the window and the side supports is mu then calculate the acceleration of the window in terms of mu, a, b and g. whereas width and b is the lengths of he window. .
Answer» Solution :Translational motion of window `w - T - 2f = (w)/(G) A` `w - T - 2 mu N = (w)/(g) A`…(1) Translational motion of weight `T - (w)/(2) = (w)/(2 g) A rArr T = (w)/(2) (1 + (A)/(g))` ….(2) TAKING torque about `C.M` `F xx (a)/(2) + T xx(a)/(2) -2N xx (b)/(2) -FXX(a)/(2) = 0` `T = 2N (b)/(a)`...(3) From (2) and (3) `(w)/(2) ((g + A))/(g) = (2Nb)/(a)` `N = (wa(g+A))/(4 bg)`...(4) From (1),(2) and (4) `w -(w)/(2) ((g+A))/(g) - (2 mu wa(g +A))/(4 bg) = (w)/(g) A` `1-(1)/(2) ((g-A))/(g) - ((2mu ag +A))/(4bg) = (A)/(g)` `1 -(1)/(2) -(mu a)/(2b) = (A)/(g) + (A)/(2g) + (mu aA)/(2 bg)` `((b - mua))/(2b) = ((3b + mua)/(2bg)) A rArr A = ((b - mu a)g)/((3b + mu a))`.