Recent content by omega5

I'm glad you pointed that out. Come to think of it, that has contributed to my mistakes on other problems. So, at the point just before the string leaves the hoop, the tension is acting upward on the hoop but downward on the point of string there?

Thank you for the clear explanation. One more, nitpicky question: how does the force provide upward acceleration and not just torque? I just want to understand everything that's going on.

Thanks for responding! I should have been clearer, I meant equilibrium along the y-axis.

I love free textbooks (especially since standard ones are so expensive), and I discovered this site, OpenCulture, which maintains a list of free textbooks, each authored by university professors or departments of education. Scroll down to "Physics" or "Mathematics" for the links: Hope this helps!

Homework Statement As shown in the figure, a string is wrapped several times around the rim of a small hoop with radius R and mass M. The free end of the string is pulled upward in just the right way so that the hoop does not move vertically as the string unwinds. Homework Equations...

Thank you! I understand now. Sounds dumb, but I neglected that angular velocity. Under my assumptions, the rings would have just fell straight down after they reached the ends of the rod. That defies inertia.

Homework Statement A uniform rod of mass 3.15×10−2kg and length 0.380m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.250kg , are mounted so that they can slide along the rod. They are initially held by catches...

Ah! Thank you very much. That was the missing piece.

I thought 3.66t would give the acceleration since the original equation gives the jerk.

Homework Statement A 4.50kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t)is applied to the end of the rope, and the height of the crate above its initial position is given by ##y(t) = (2.80m/s)t +(0.61m/s^3)t^3## What is the force at 4...

Oh! It just hit me. For some reason I had realized the lever arm was at the point the force acted at for FM but didn't realize that for FJ. I am very grateful!

The joint itself - I tried setting the axis as the point where FJ acts, but thought the lever arms would still have to be include x to define them from that frame of reference.

Thanks for responding! Do you mean FJ cos θ°? I didn't think it would give me the lever arm distance, just the horizontal force.

Homework Statement Hello and thank you in advance for your help! I am taking an introductory algebra-based physics class and am trying to solve this problem: Suppose an arm holds an 5.8-kg mass. The total mass of the arm is 3.3 kg. Gravity acts from the center of the arm, which is 24 cm away...