Okay, yeah, that makes sense, so the ##F_x = Fcosθ = 2k(\sqrt(x^2+y^2+(z - S)^2) - l_0)cos(arcsin(\frac{S}{\sqrt(x^2+y^2+(z - S)^2)}))## but that... that seems like too much trigonometry to be correct, plus I don't think there's a trig identity that simplifies cos(arcsin(whatever))? Could be...
ohh, so it's just trigonometry now right? yea... yea I think that makes sense, so then
$$ V = k*\sqrt(x^2+y^2+(z - S)^2) \Rightarrow F_x = -\frac{1}{2}k*2x = -kx $$
But that's even further from the expression we're trying to get?? Plus, how will I be able to get a general expression for the...
Oh, yea, that makes sense. So I don't even have to derive an expression for V in terms of ##\phi##? That's a relief.
How far is it from the anchor point? Well it's a length z - S from the anchor point, because it's a distance z from the origin, and the origin is a distance S away from the...
What do you mean by this? Why don't the springs change length when the mass is being displaced from equilibrium?
It can be at any length ##l_0## from the spring if it is relaxed (and the second one doesn't break, but we're not given any condition for breaking so I'll just assume that's not a...
So at first I tried to express the potential energy as a function of x, y and z, but since I'm not quite sure about the geometry of the situation, I decided to separate out the potential energy into three components: ##V_x, V_y, V_z## (I'm pretty sure this is valid because in the partial...
If I'm understanding this correctly, the constraints represent the angle ##\theta## when the mass slips off, correct? Or are you simply rewriting everything in Cartesian coordinates?
As to answer your question of "what is this," it is my attempt at applying the chain rule in a place where it...
[Rewriting this as per the suggestions. Thanks once again.]
I won't be using the Lagrangian because it was never explicitly stated that I have to so I'll just use conservation of energy.
$$ T = \frac{1}{2}mv^2 = \frac{1}{2}m(R\dot{\theta})^2 = \frac{1}{2}mR^2\dot{\theta}^2 $$
$$ V = mgy =...
Okay, so I figured it out, I was using the wrong equation for ΔU_isobaric, it's still ##nC_VΔT##, I got that formula confused with the one for heat transfer in an isobaric process, ##Q = nC_PΔT##.
My problem isn't exactly with calculating the actual changes in internal energy, I'll put those values below. My problem is that I can't get the values to add up to 0, and I don't understand why since for cyclic processes, by definition, ΔU must equal 0.
$$ΔU_{AB} = ΔU_{isothermal} = 0$$...