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M3 Elastic Strings

A particle of mass m is supported by 2 light elastic strings, each of natural length a and modulus of elasticity 15mg16 \frac{15mg}{16}. the other ends of the strings are attached to 2 fixed points A and B where A and B are i the same horizontal line with AB=2a. when the particle hangs at rest in equilibrium below AB, each string makes and angle θ\theta with the vertical.
Verify that cosθ=45\theta = \frac{4}{5}

okay seems simple enough, but i'm not sure...
ive got T=15mgx16a T=\frac{15mgx}{16a} and 2Tcosθ=mg 2Tcos\theta = mg

so 12cosθ=15x16a\frac{1}{2cos\theta}=\frac{15x}{16a} but i don't see how to get rid of a and x
apocalipse117
A particle of mass m is supported by 2 light elastic strings, each of natural length a and modulus of elasticity 15mg16 \frac{15mg}{16}. the other ends of the strings are attached to 2 fixed points A and B where A and B are i the same horizontal line with AB=2a. when the particle hangs at rest in equilibrium below AB, each string makes and angle θ\theta with the vertical.
Verify that cosθ=45\theta = \frac{4}{5}

okay seems simple enough, but i'm not sure...
ive got T=15mgx16a T=\frac{15mgx}{16a} and 2Tcosθ=mg 2Tcos\theta = mg

so 12cosθ=15x16a\frac{1}{2cos\theta}=\frac{15x}{16a} but i don't see how to get rid of a and x


You need to be able to express x as a function of a and theta. To do this consider one of the triangles consisting of half the overall diagram.

However, I have not been able to work it through. I get a quartic in cos theta, of which 4/5 is the only valid solution, but I am unable to prove it. So I am missing a trick somewhere.

Edit: The only other option I can see, is energy considerations. Is that something you've covered? I've not tried to work it through by that method yet.
I don't get elastic strings/springs!!! :'(
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rbnphlp
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apocalipse117
A particle of mass m is supported by 2 light elastic strings, each of natural length a and modulus of elasticity 15mg16 \frac{15mg}{16}. the other ends of the strings are attached to 2 fixed points A and B where A and B are i the same horizontal line with AB=2a. when the particle hangs at rest in equilibrium below AB, each string makes and angle θ\theta with the vertical.
Verify that cosθ=45\theta = \frac{4}{5}

okay seems simple enough, but i'm not sure...
ive got T=15mgx16a T=\frac{15mgx}{16a} and 2Tcosθ=mg 2Tcos\theta = mg

so 12cosθ=15x16a\frac{1}{2cos\theta}=\frac{15x}{16a} but i don't see how to get rid of a and x

iis a diagram provided for this?or do you have to sketch one?
because from the info given I get a triangle shape with the strings and the mass hanging somewhere between the strings.
Hey Ghostwalker, can you explain how to get the quartic? I'd know how to do it from there. Much appreciated :smile:
Original post by karlpopper
Hey Ghostwalker, can you explain how to get the quartic? I'd know how to do it from there. Much appreciated :smile:



Draw a labelled diagram if you haven't already. And follow the advice in my previous post.

I suspect however, this is not the intended method. Since it says verify cos theta = 4/5, it may be enough to set cos theta to 4/5 and verify that the system is in eqilibrium at that angle. I can't imagine you being expected to solve a quartic.
(edited 8 years ago)

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