Recent content by link223

why is the left one not permitted? What I mean is that the normal force calculated doesn't come out correctly, tension dfor is correct however

Haii, I don't understand why I need to choose my n-t components in the direction of a circular motion and can't just use them with the n-axis along the rope and the binormal perpendicular to the surface.

How in the world I am supposed to start with this problem? No clue, so can't provide HW solution by any means. regards.

sorry didn't see your comment when you posted it. ( I kind of said the same tjhing in the intial part.)oh okay that is interesting. So the problem gave it, but i wanted to see how I could've determined it on my own. But euh- why is that difference by factor of 4?

no but that is what is not the case on this one, because the forces themselves are calculated by the constraint equation F = ##mu * N##. This means that I can get only two equations that useful from Newton's laws. And then I can either say that the rope is masseless such that the accelerations...

By the way I did forget the normal force of A on B and of B on A in the analysis of equilibirum but is of no real consequence to the conclusion that 'd be drawn hand, at least I think ofc.

That's why I say Newton's laws for the static situation and see in which dir. it would move when there was no friction and then just oppose that motion. So to say Mass A goes up, thus friction towards the left.

even if I knew the friction forces, that doesn't tell me nothing, because there's no direction I can associate with them right?

well no, but I determine now what the direction of motion would be such that I can say that the friction opposes the motioin, isn't that a good way to do this.?

don't care about those right now? or do we? I mean they are let me calc real quick...

Can I now think of this as to have A at rest, the rope needs to exert a 490.5N force, but at that point, mass B will already be moving such that it just goes upwards.

sorry oki, equilibrium B with friction neglected. ##T_b = m_b*g/2 = 245.3N## Eq. A with friction neglected: ##T_a = m_a*g/2 = 49.05N## This is the place where there is actually no friction yet, just to correct miself.

I did the analysis where they are in equilibrium then the tension force in A is way less than B, but how could this conclude that B goes down whilst A goes upP?

I am not very sure how I would be approach this. Obviously it is stated in which direction it's going where we see that mass A goes to the right, but how do I determine this stuff analytically.

sure thing. for B ##F_b = 42.48n## for A ##F_a = 22.1n## and this concludes that it will push on B and not pull it gadaimm