Recent content by Alex126

Oh, well, if there was no block then :D

I guess when there is something else pulling on the other side...? Something like friction on the table maybe?

Ok, thanks for the answers. That's kinda what was bothering me, because I was thinking of other kinds of exercises. One in particular had two "concentric overlapping discs of pulleys" (I'm sure there is a better term to indicate this, but I don't know it) with two blocks hanging on either side...

Yea, got it. I think. So if I drew the diagram only for the block, it would show these three forces: Weight, Normal force, and a Tension pointing to the right. For the pulley, there would be two forces: F pointing down, and T pointing to the left. The T from the block and the T from the pulley...

Homework Statement There is a block of known mass m on a horizontal surface. The block is connected to a pulley via a rope. The other end of the rope is pulled vertically, downwards, with a known force F. The pulley has a known moment of inertia I and radius r. Calculate the acceleration a of...

I have a general understanding of how torque works, at least for "simple" objects that can be drawn as a single "bar" under the effect of various forces. In this problem there is a slightly more "complex" object though, and I'd like to know if there is a way to solve it without doing what I did...

Alright, I think that's beyond my current understandings and knowledge. I'm going to pass on that last bit for now :D Thanks again everyone.

lol So if I'm reading this right you mean that after the object detaches it would start its parabolic motion, but it "hits its head" on the drum because the parabola trajectory "goes up" higher than the drum's circumference. Then, once the drum's circumference is "higher" than the parabola...

I think I get why μ must be infinite. With X centripetal axis, Y tangential: X axis: +N + Wx = m*a Y axis: +Friction - Wy = 0 => N*μ = Wy => N = Wy/μ Since N must be zero at the point of detachment, then N = Wy/μ => we need μ to be infinite to make that fraction -> 0. Just to be clear, you...

Everyone says it should be 0.85 (and I'm sufficiently convinced that's the right one), though the textbook says 0.91. Could just be an error on the textbook part, at this point that's my guess. I'd like to quote it exactly, but it's not in English. You can trust my translation though -- I've...

Yea, but I'm better at solving things when I understand them than when I have to accept an "insight" that falls out of the schemes I'm familiar with. I guess I try too hard to make things fit into a pre-made picture that I have of the problem, and thus find it hard to "adjust" to different...

Oh, so you mean W_x and W_y. Yea. So "tangential" here means "perpendicular to the centripetal axis". Yea. Ok, thanks (to you and everyone else). Alright. I still am not entirely convinced though; it kinda feels like we are making the most out of what we have, but we are missing a necessary...

Uhm...no, sorry, I just don't get it. I've always thought that the weight force is the same, so W can't possibly have any other value than m*g. The thing that would change is, with different angles, how much W_x and W_y we have, but W is the same. So how can gravity "provide more and more"...

Ehr...tangential component of gravity? If R is not perpendicular, then where is it? Centrifugal force is not enough anymore to counterbalance the Weight (in the second drawing I made, the W_x). Or, in other words, the W_x cannot provide a big-enough centripetal force. Ok, but how? I attempted...

It doesn't mention anything about friction, which was my very first problem (as I'm about to type). It is constant rotation, yes. By "frequency" it means rounds per second [Hz], so as I mentioned earlier I presume it wants me to find the acceleration, and from the acceleration I can then find...