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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)

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestLet $L$ 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 $v_1 = v_2(1+\sin \theta)$.

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Awesome Solution Dude...!!!

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Wow! Good job Jimmy!

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

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no .. it isn't the right ans

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

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

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Yes , $\theta$ will definitely not remain the same . $\theta$ would gradually decrease .

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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)

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nope ..masses aren't needed..

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

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

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v1/(1+sinθ)=v2 I think so!!!

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concept behind the problem is use of constraint motion

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

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constraint relations

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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

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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)

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

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is it a guess or something else??

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More than a guess. I have used logic but might be wrong.

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by constrain relation,v2=v1tan theta

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