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A weight "m" is suspended from the center of a light, taut and originally horizontal rope. After suspending the weight "m", what angle must the rope make with the horizontal if the tension in the rope is to equal the weight "m"?

Here is what I have so far. enter image description here

I'm not sure what to do from here on out. I'm assuming T1 and T2 will be equal and i'm also assuming that T1 and T2 have about half the mass of M. So is the idea to solve for M then substitute I'm guessing but when I did that i ended up with something like 2sin(theta)=sin(theta).

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  • $\begingroup$ Does the length of the rope change? I assume it doesn't and that the initial information is useless. Because if the length of the rope changes then you need a lot more information (like what is the tension in the rope as it stretches). It seems to me that you are being given that the $T_1 = T_2 = mg$. In this case it's a simple matter of equating $2mg\sin(\theta) = mg\rightarrow \sin(\theta) = \frac{1}{2} \rightarrow \theta = 30^\circ$. $\endgroup$ – Jared Mar 23 '15 at 6:32
  • $\begingroup$ No it does not. Gah, now I get it. Both tensions are equal and their force combined is the force created by "m" which is just mass*gravity. Thank you Jared, if you post it as an answer I'll accept it. $\endgroup$ – Lefty Mar 23 '15 at 6:42
  • $\begingroup$ Physics stackexchange should strongly discourage such posts. $\endgroup$ – Jaswin Mar 23 '15 at 6:47
  • $\begingroup$ @Lefty In my opinion this is a very misleading question because the entire setup is irrelevant--but it's not if you want to deal in reality (and the last thing we should want to do is have physics students question reality when they are given theoretical questions)! It's just that they give you information (i.e. the tension) that negates any prior setup. $\endgroup$ – Jared Mar 23 '15 at 6:47
  • $\begingroup$ @Jaswin My comment is bad (since I gave the answer), but I don't think the question is necessarily to be discouraged. All I would say is to encourage the homework tag (as I added) when appropriate. $\endgroup$ – Jared Mar 23 '15 at 6:54
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I'm assuming T1 and T2 will be equal

What made you assume that? They may or maynot.

and i'm also assuming that T1 and T2 have about half the mass of M.

"about" when the string are almost equal, not always

So is the idea to solve for M then substitute I'm guessing but when I did that i ended up with something like 2sin(theta)=sin(theta).

that is $2\sin\theta=\sin\theta\implies\theta=0$? that doesn't make sense


You should do:

  • In equilibrium, in a suitable cordinate axes the forces (or their respective components along the axis in consideration) should balance each other.
  • Take in account weight of body, and the tension in two wires and try to make the triple point at equilibrium.
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