Calculating Tension in Mass on a Slope with Standing Waves

[roam]

A thin wire is attached to a mass M = 3.67 kg, and hung over a pulley. The mass rests on a slope of angle θ = 35 °. The linear mass density of the wire is μ = 7.7 g/m.

Standing waves are excited in the vertical section of the wire:

The height of this vertical section is h = 2.0 m.
Take g = 9.8 ms−2.



(a) What is the tension of the wire (you may assume the wire is massless)?



I am confused because and I really don't know how to do this question;
I can't use the formula: $$v = \sqrt{\frac{T}{\mu}}$$ and then solve like $$T = \mu v^2$$ because the velocity is not given and I can't think of any way of calculating it..
The correct answer should be 20.6. I don't know what formula to use etc..

Thanks.

 

[Dick]

Now I'm confused. Is the mass hanging from a wire or is it sitting on a slope? I would think you would calculate the tension using force balance, but I cannot figure out what the geometry of your problem is?

 

[roam]

Is the mass hanging from a wire or is it sitting on a slope?

Actually it's both! Here's the diagram:

http://img87.imageshack.us/img87/4437/imagego2.gif [Broken]

It's very confusing, there's nothing on the equation sheet that works.

 

[Dick]

Ooohhh. I see. I think you are supposed to assume that the wire is tied to the ground 2m below the pulley. Use force balance to find the tension in the wire assuming it's massless.

 

[roam]

What do you mean by using "force balance"? - (?)

Could you please show me exactly what I need to do?

You mean, I use $$\Sigma_{y} = T - m g$$ => $$T = m_{block} g$$
3.67 kg × 9.8 ms^−2 = 36
But the correct answer is 20.6

(we were supposed to assume it's massless...)

 

[Dick]

You would only need 36N to hold the block up if the ramp were vertical. It's not. Just do the usual thing of splitting the tension into tangential and normal components according to the angle of the ramp.

 

[hayowazzup]

o, it's the assignment ;), just use the forces diagram

 

[Dick]

You would only need 36N to hold the block up if the ramp were vertical. It's not. Just do the usual thing of splitting the tension into tangential and normal components according to the angle of the ramp.

I meant 'splitting the WEIGHT into tangential and normal components'. Sorry.