Pratham Ashish
Last Activity: 15 Years ago
hi,
at any time ,
let the length of string between the wedge & pully(which is stationary ) = x2
length of string between the block & pully (to which it is hanged ) = x1
length of string between the both pullys = x2/ cos(90-θ) = x2 / sinθ
since the length of strig would be const. so the sum of upper mentioned lenght would be const.
x1 + x2 + x2/ sinθ = const.
d/dt x1 + d/dt x2 + d/dt ( x2/ sinθ ) = 0
v1 - v2 - v2 /sin θ - x2 cosθ /sin^θ (d/dt θ) =0 .............(1)
let the height of the wedge be l,then
tan (90 -θ) = l / x2 x2 cot θ = l cotθ d/dt x2 + x2 d/dt cot θ = 0 -cotθ v2 - x2 cosec^2 θ (d/dt θ) =0 cosθ sinθ v2 + x2 (d/dt θ) = 0 (d/dt θ) = - cosθ sinθ v2 /x2 ..............put in eq . (1) v1 - v2 - v2 /sin θ - x2 cosθ /sin^θ (d/dt θ) =0 v1 - v2 - v2 /sin θ + x2 cosθ /sin^θ cosθ sinθ v2 /x2 =0
v1 - v2 - v2 /sin θ + v2 cos^θ / sinθ =0
v1- v2 - v2 /sinθ( 1 - cos^2θ ) = 0
v1 -v2 - v2 sinθ =0
v1 = v2 (1 + sinθ ) (d )