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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Siddharth Shah
·
3 years, 12 months ago

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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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Sambit Senapati
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3 years, 12 months ago

xdx=ldL
sinθ=dL/dx
sinθ+1=(dL+dx)/dx
sinθ+1=V1/V2
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Akella Ravitej
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3 years, 12 months ago

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concept behind the problem is use of constraint motion
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Deepankar Jangid
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3 years, 12 months ago

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v1/(1+sinθ)=v2
I think so!!!
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Chetan Vibhandik
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3 years, 12 months ago

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\(v_2(1+\sin\theta)=v_1\sin\theta\)
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Arnab Animesh Das
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3 years, 12 months ago

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is the velocity v1 with respect to wedge or with respect to ground?
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Rahul Nahata
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3 years, 12 months ago

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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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Jord W
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3 years, 12 months ago

Does \(\theta\) stay constant? My guess is no, but I can't tell for certain from the diagram.
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Jimmy Kariznov
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3 years, 12 months ago

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@Jimmy Kariznov
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Yes , \( \theta \) will definitely not remain the same . \( \theta \) would gradually decrease .
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Priyansh Sangule
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3 years, 12 months ago

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use the concept that length on the thread remains constant. :)
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Advitiya Brijesh
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3 years, 12 months ago

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i think \(v_1\)=\(v_2\)\(cos\theta\)
Not pretty sure but still..i think..
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Ritvik Choudhary
·
3 years, 12 months ago

## 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 \(v_1 = v_2(1+\sin \theta)\). – Jimmy Kariznov · 3 years, 12 months ago

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– Ajitesh Mishra · 3 years, 12 months ago

Awesome Solution Dude...!!!Log in to reply

– Ivan Sekovanić · 3 years, 12 months ago

Wow! Good job Jimmy!Log in to reply

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) – Siddharth Shah · 3 years, 12 months ago

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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 – Sambit Senapati · 3 years, 12 months ago

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constraint relations – Goutam Narayan · 3 years, 12 months ago

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xdx=ldL sinθ=dL/dx sinθ+1=(dL+dx)/dx sinθ+1=V1/V2 – Akella Ravitej · 3 years, 12 months ago

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concept behind the problem is use of constraint motion – Deepankar Jangid · 3 years, 12 months ago

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v1/(1+sinθ)=v2 I think so!!! – Chetan Vibhandik · 3 years, 12 months ago

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\(v_2(1+\sin\theta)=v_1\sin\theta\) – Arnab Animesh Das · 3 years, 12 months ago

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is the velocity v1 with respect to wedge or with respect to ground? – Rahul Nahata · 3 years, 12 months ago

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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) – Jord W · 3 years, 12 months ago

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– Ajitesh Mishra · 3 years, 12 months ago

nope ..masses aren't needed..Log in to reply

Does \(\theta\) stay constant? My guess is no, but I can't tell for certain from the diagram. – Jimmy Kariznov · 3 years, 12 months ago

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– Priyansh Sangule · 3 years, 12 months ago

Yes , \( \theta \) will definitely not remain the same . \( \theta \) would gradually decrease .Log in to reply

use the concept that length on the thread remains constant. :) – Advitiya Brijesh · 3 years, 12 months ago

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i think \(v_1\)=\(v_2\)\(cos\theta\) Not pretty sure but still..i think.. – Ritvik Choudhary · 3 years, 12 months ago

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– Ajitesh Mishra · 3 years, 12 months ago

no .. it isn't the right ansLog in to reply

My guess... it's "v1 = v2 + v2*sin(theta)". Is that right? – Lokesh Sharma · 3 years, 12 months ago

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– Ajitesh Mishra · 3 years, 12 months ago

is it a guess or something else??Log in to reply

– Lokesh Sharma · 3 years, 12 months ago

More than a guess. I have used logic but might be wrong.Log in to reply

by constrain relation,v2=v1tan theta – Nayan Pathak · 3 years, 12 months ago

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