When solving certain questions , I noticed that in some places we take component of gravity along the string while in others, we take component of tension along gravity, I think that both of these methods should give the same result but they dont. Some people also told me that assume you have a car being pulled by two ropes, will acceleration be double? But why is tension force so special? What am i doing wrong by taking component of tension along an other force? Thanks in advance.+
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Without explicit examples, it is impossible to understand what you are puzzled about. – mike stone Jul 17 '21 at 12:59
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https://imgur.com/a/XmBZymk sort of like in this one – Akshat Jangra Jul 17 '21 at 13:03
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What is $u$ in your diagram? You have it pointing in three different directions. – garyp Jul 17 '21 at 13:06
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@garyp It is the velocity with which string is being pulled. – Akshat Jangra Jul 17 '21 at 13:09
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This may also help: https://physics.stackexchange.com/a/648788/305718 – ACB Jul 17 '21 at 14:01
2 Answers
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If the distance between the pulleys is $2x$ and the weight is a distance $y$ below the pulleys then the distance $z$ from either pulley satisfies
$z^2 = x^2 + y^2$
Differentiating and using the fact that $x$ is constant gives
$\displaystyle 2z \frac {dz}{dt} = 2y \frac{dy}{dt} \\ \displaystyle \Rightarrow \frac {dy}{dt} = \frac z y \frac {dz}{dt} = \sec \theta \frac {dz}{dt}$
Your method assumes that $y = z \cos \theta$ implies $\frac {dy}{dt} = \frac {dz}{dt}\cos \theta$ but this would only be true if $\theta$ were constant, which it is not.
gandalf61
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The force which a tension exerts on an object is a vector like any other force and can be broken into components. If you are getting a conflict, you are probably doing something wrong. (I'm not seeing a diagram.)
R.W. Bird
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