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Physics Problem!

Find the relation b/w v1 and v2 in terms of v1,v2, and theta

Note by Ajitesh Mishra
4 years ago

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

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@Jimmy Kariznov Awesome Solution Dude...!!! Ajitesh Mishra · 3 years, 12 months ago

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@Jimmy Kariznov Wow! Good job Jimmy! Ivan Sekovanić · 3 years, 12 months ago

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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) 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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@Jord W nope ..masses aren't needed.. Ajitesh Mishra · 3 years, 12 months ago

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

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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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@Ritvik Choudhary no .. it isn't the right ans Ajitesh Mishra · 3 years, 12 months ago

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

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@Lokesh Sharma is it a guess or something else?? Ajitesh Mishra · 3 years, 12 months ago

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@Ajitesh Mishra More than a guess. I have used logic but might be wrong. Lokesh Sharma · 3 years, 12 months ago

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

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