Physics Problem!

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

Note by Ajitesh Mishra
6 years, 1 month ago

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20 votes

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Let LL be the length of the string and HH be the height difference between the pulleys.

Let xx be the length of the horizontal piece of string and yy be the length of the vertical piece of string.

The slanted piece of string has length Lxy=x2+H2L-x-y = \sqrt{x^2+H^2}. Thus, (Lxy)2=x2+H2(L-x-y)^2 = x^2+H^2.

Differentiation yields: 2(Lxy)(dxdtdydt)=2xdxdt2(L-x-y)\left(-\dfrac{dx}{dt}-\dfrac{dy}{dt}\right) = 2x \dfrac{dx}{dt}.

By definition, dxdt=v2\dfrac{dx}{dt} = -v_2 and dydt=v1\dfrac{dy}{dt} = v_1. Using trigonometry x=Htanθx = H\tan \theta and Lxy=HsecθL-x-y = H\sec \theta.

Therefore, 2Hsecθ(v2v1)=2Htanθ(v2)2H\sec\theta \cdot (v_2 - v_1) = 2H\tan\theta \cdot (-v_2). Solving yields v1=v2(1+sinθ)v_1 = v_2(1+\sin \theta).

Jimmy Kariznov - 6 years, 1 month ago

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Awesome Solution Dude...!!!

Ajitesh Mishra - 6 years, 1 month ago

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Wow! Good job Jimmy!

Ivan Sekovanić - 6 years, 1 month ago

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i think v1v_1=v2v_2cosθcos\theta Not pretty sure but still..i think..

Ritvik Choudhary - 6 years, 1 month ago

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no .. it isn't the right ans

Ajitesh Mishra - 6 years, 1 month ago

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

Advitiya Brijesh - 6 years, 1 month ago

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

Jimmy Kariznov - 6 years, 1 month ago

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

Priyansh Sangule - 6 years, 1 month 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 - 6 years, 1 month ago

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nope ..masses aren't needed..

Ajitesh Mishra - 6 years, 1 month ago

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

Rahul Nahata - 6 years, 1 month ago

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v2(1+sinθ)=v1sinθv_2(1+\sin\theta)=v_1\sin\theta

Arnab Animesh Das - 6 years, 1 month ago

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v1/(1+sinθ)=v2 I think so!!!

chetan Vibhandik - 6 years, 1 month ago

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concept behind the problem is use of constraint motion

Deepankar Jangid - 6 years, 1 month ago

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

akella ravitej - 6 years, 1 month ago

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constraint relations

Goutam Narayan - 6 years, 1 month 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 \sumT.s=0 . Differentiating we get, \sumT.v=0

Sambit Senapati - 6 years, 1 month 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 - 6 years, 1 month ago

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

Lokesh Sharma - 6 years, 1 month ago

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is it a guess or something else??

Ajitesh Mishra - 6 years, 1 month ago

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

Lokesh Sharma - 6 years, 1 month ago

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by constrain relation,v2=v1tan theta

Nayan Pathak - 6 years, 1 month ago

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