Recent content by sgholami

Ah! I see! I just find it counter-intuitive that A need not have any relationship to ##y = \sqrt x##. I now understand that my original question should have been how the integral we're asked to evaluate, ##\int (A \cdot ds)##, is relevant to the function ##y##. They just seemed like separate...

Right; but I'm not sure how A relates to the function, ##y = \sqrt x##. It seems that the function gives us our path, but then what is A used for?

Thank you, haruspex. I was able to find a solution to this problem. Like you said, the goal is to integrate a vector function (A = x2ˆx + y2ˆy + z2ˆz) dotted with the derivative of the position vector (ds), over a specified path [i.e. from (0, 0) to (2, √2)]. Even though the vector function A...

Aah, thank you again. So, what I understand is that the centripetal force is just a result of the real, applied forces (namely Fn and mg*cos(θ)). If it's a result of the two...then is it true that mα = mg*cos(θ) - Fn ? Then, said another way, mα is used to represent the sum of all forces in...

Hello. I'm new to physics, and the problem I have seems so basic, mathematically speaking. I'm just failing to grasp exactly what is being asked. If I can find that, I believe I can find the answer. Here it is: 1. Homework Statement Let A = x2ˆx + y2ˆy + z2ˆz Consider the parabolic path y2 =...

Thanks for that pointer. I think this is where my confusion lies. I figure that, in the radial direction, you have three different forces: Fn in the outward (negative) direction, and mα + mg*cos(θ) in the centripetal (positive) direction. But something about that feels problematic... Maybe my...

Hello! I am just stuck on one part of this question and would be grateful for any help. Question A small block of ice slides from rest from the top of an inverted frictionless bowl of radius R (above right). How far below the top x does the ice lose contact with the bowl? Equations mgx =...