Recent content by jonny997

Ahhh okay, thank you. For some reason it didn’t click that I could solve for V using ## T = \frac{2a}{R}\left(\frac{1-\frac{b}{V}}{b}\right) ## and then sub that into the other equation 😬 I’ve got it now

Hey. Thanks for the response. Do you have any idea how I would go about obtaining the expression my lecturer has provided? Namely, $$ P_{inv} = \left[\frac{2a}{b^2} - \frac{RT}{b}\right]e^{\frac{1}{2}-\frac{a}{RTb}}$$ If I am to rewrite ## T = \frac{2a}{R}\left(\frac{1-\frac{b}{V}}{b}\right) ##...

The notes my lecturer has provided state that the maximum temperature can be found taking p = 0 in the inversion curve formula, given as: I’m not sure how to obtain this?? These are the formulas: This is my attempt at a solution : Not sure if this approach is right?

Thanks, I appreciate all the help

So in that case we would be taking the surface of the fluid to be the origin and measuring downwards?

Ohhh okay, that makes everything a lot clearer... So dp is the difference in pressure between the top and bottom portions of the slap, therefore p_top = p_bottom + dp So the force from the pressure at the top of slab acts downwards and is given by (p+dp)A

I think its mainly the (p+dp)A force I'm not understanding. Specifically the dp part, I'm not sure exactly what that is supposed to represent or why its acting downwards. I thought there would be more pressure at the bottom?

dw is the weight of the slab of fluid and the negative sign implies that it acts downwards. (p+dp)A is the force on the slab due to the fluid above it, it also acts downwards, hence the negative sign. pA is the force acting upwards on the slab and so its positive... The net force acting on...

No, not exactly. :/

I'm still kinda confused... :cry: I'm not sure where the (p+dp)A came from... How do you work out that p+dp is less than p from the drawing?

v

Thanks for help haha. I appreciate it

Ahhh okay, I'll keep that in mind. Thank you!

Thanks. It's definitely more intuitive for me to think of the distance as d-x. In this case x-d would be a negative value for m2 to the right of string A, am I right in saying that..? I think that's why I was getting so confused by the signs towards the end...

Oh wait, no I've made a really dumb mistake... The block musr definitely be to the left of string A for the tension in string B to equal 0. I've just got the signs for distance wrong... 😭 the way it is right now (x-d) is inverted which is why the direction of the torque is as well... Is that...