Physics Problem!

Let $L$ be the length of the string and $H$ be the height difference between the pulleys.

Let $x$ be the length of the horizontal piece of string and $y$ be the length of the vertical piece of string.

The slanted piece of string has length $L-x-y = \sqrt{x^2+H^2}$. Thus, $(L-x-y)^2 = x^2+H^2$.

Differentiation yields: $2(L-x-y)\left(-\dfrac{dx}{dt}-\dfrac{dy}{dt}\right) = 2x \dfrac{dx}{dt}$.

By definition, $\dfrac{dx}{dt} = -v_2$ and $\dfrac{dy}{dt} = v_1$. Using trigonometry $x = H\tan \theta$ and $L-x-y = H\sec \theta$.

Therefore, $2H\sec\theta \cdot (v_2 - v_1) = 2H\tan\theta \cdot (-v_2)$. Solving yields $v_1 = v_2(1+\sin \theta)$.

i think $v_1$=$v_2$$cos\theta$ Not pretty sure but still..i think..

use the concept that length on the thread remains constant. :)

Does $\theta$ stay constant? My guess is no, but I can't tell for certain from the diagram.

won't it also be in terms of the masses of the two objects? (determining their acceleration from the tension in the thread and gravity)

is the velocity v1 with respect to wedge or with respect to ground?

$v_2(1+\sin\theta)=v_1\sin\theta$

v1/(1+sinθ)=v2 I think so!!!

concept behind the problem is use of constraint motion

xdx=ldL sinθ=dL/dx sinθ+1=(dL+dx)/dx sinθ+1=V1/V2

constraint relations

Another way to derive the relation is by using the fact that work done by tension is zero (by newton's third law). We get $\sum$T.s=0 . Differentiating we get, $\sum$T.v=0

using the virtual work method..-T(1+sin(thetha))x2 + Tx1=0. Differentiating this constraint equation we get the relation v2(1+sin(thetha))=v1. (x2 is wedge's displacement and x1 is the block's displacement)

My guess... it's "v1 = v2 + v2*sin(theta)". Is that right?

by constrain relation,v2=v1tan theta