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)

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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## 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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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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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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constraint relations

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

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concept behind the problem is use of constraint motion

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v1/(1+sinθ)=v2 I think so!!!

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

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

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

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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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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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