Recent content by jna

Not the height *ratio* ?

Maybe, I guess @AmirWG should comment, and I do not mean to diminish your effort answering what you believe was the OP's confusion. All I can say with absolute certainty is that the highlighted wording in the textbook is irritatingly confusing to me.

I think that the OP's intended question has not been answered by any of the above (albeit informative) responses. I think his/her confusion stems from the inaccurate wording in the textbook. The text claims that the height is not affected by the diameter of the tube. Clearly, that is incorrect...

Thank you!

Thank you!

So, as the solution for F_fr I am getting 80/3 (approx. 26.7) N. Unfortunately, that does not match any of the values in the multiple choice: {A. 99 N, B. 12 N, C. 63 N, D. 75 N}. I am ok with that as long as I know this result is correct - are you getting the same? Thanks for all your help again.

Let me try (forces acting to the left +, to the right -): For Block A: (net force A) = (acting force F) - (friction/drag from from block B) For Block B: (net force B) = (friction force from A) where (friction force from A) = - (friction/drag from from block B) --- so: m_A * a_A = F - F_fr m_B...

Yes, I agree that 75N/30kg was not the right thing to solve this (original attempt). That's why I posted the revised attempt (#3) which does not go into determining the acceleration of the block A at all, just considers forces.

I see. Re the first part, fF=63N causing an acceleration of 2.625 m/s^2 was still fine to me because I only assumed a relative reduction of acceleration of B by 0.5 m/s^2 (the 0.5 m/s^2 is specified as "to the right relative to block A). I do understand that the force 63N makes B accelerate to...

It would be helpful if you could point out where I went wrong with the 63N reasoning. I am just unsure what to do with the two hints.

Ah sorry i actually meant to say the same (misread your question). But can you please help me correct my answer?

I'd say yes.

Hmm, yes I am pretty sure I broke something. So a more careful but still blurry look: Block A exerts force through friction on block B in the direction <--. Obviously, that force exceeded the opposing max friction force of B on A pointing -->, since block B is sliding and accelerating 0.5 m/s^2...

Homework Statement Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. What is the friction force of block A on block B if the block B has a mass of 24 kg and is accelerating at 0.50 m/s2 to the right relative to the block...

I am trying to imagine the action: suppose there is a drag force acting exactly at the perigee. This results in lowering the altitude of every point of the orbit except the perigee, making both \dot e and \dot a negative. I can't explain to myself how any amount of drag applied at perigee could...