Recent content by omegasquared

Thank you both for your help, in combination I believe I understand this much better now but I'll continue to ponder this idea of approximations (in terms of power series) over. Much appreciated!

Expanding previously, Fr = (2kL+2kx)y / (2L+x), so if x is small then 2kx ≈ 0 and 2L + x ≈ 2L? Which would leave 2kLy/2L = ky but then surely the y on the top would make the whole thing ≈ 0?

I'm still unsure how to get from there to simply ky however.

Ah, good point. So it should be: Fr = 2k(L+x)sinθ Which would then be: = 2ky(L+x) / (2L+x) Agreed?

Additional extension beyond what it was (L) at y=0.

The x is to do with the extension of the spring which will occur when you displace the ball by y upwards.

Homework Statement Mass M is supported by a smooth table and connected by two light horizontal springs to two fixed blocks. Each spring is of natural length L. Both springs are initially extended by L to get a total width between blocks of 4L. The spring constant of both springs is k. When...

Thanks, that pointed me in the right direction! Much appreciated!

Sorry, noticed that almost straight away. Thanks for your help there. I'm still unsure how it is possible to resolve to get a value for cosθ which doesn't involve R though.

Like in the modified diagram attached?

If it's a normal force, what is it normal to? The surface/edge of the hoop?

I've attached the provided diagram of the system. I assume the action force would make the same angle θ as it's been pulled in that direction by the tension. Resolving parallel to the reaction & tension would yield a result still involving R though.

Homework Statement A ring of mass m slides on a smooth circular hoop with radius r in the vertical plane. The ring is connected to the top of the hoop by a spring with natural length r and spring constant k. By resolving in one direction only show that in static equilibrium the angle the...