Recent content by sojsail

Is it possible that your sketch is a misinterpretation of the problem description? If p is on the orbit, when the particle comes around to the location of p, the force would become infinite (with n being a negative exponent). I am assuming that r is the distance to p. And in that case, I can not...

In graphs of a thrown ball, assume that time starts when the ball leaves the hand. So would the acceleration of the throwing motion be part of the graph?

If you are confused by what is said in any of our posts, likely any of us that are connected could explain/reword the post for you. We don't explain how to get the answer, but if necessary we can explain posts that have come in for you.

I think your solution is equivalent to one of a list of perhaps 2 or 3 methods that would get full credit. One quibble about your worksheet: in one line you gave g as the unit where I think you knew, from how you converted the numerical part, is kg. That could cost you. There were several...

Could it be that you intended to ask about acceleration?

I found this to be an interesting discussion. I can not tell if the worry has been eliminated, so I would like to add my way of seeing that. Current would be flowing down the schematic thru R1. Let the potential of the "horizontal wire" be Vh. What if some current tried to continue going down...

Where did the 64 come from? Certainly not the 64 N in frame #8. In which direction are you now trying to work out the total force -- total force exerted on the ground?

Another approach: You know that the ammeter reading is to be zero. So assume that is the case and simplify your equations accordingly.

You are asked to find moment of this Force about point A. Don't assume the bracket is going to rotate to a new position. This would be an analysis of the twisting torque on the bracket (perhaps to see how strong the bracket needs to be considering what will be applied to it in use). Doc Al said...

I am back. I see that Orodruin caught me misreading the problem. He swims at 4 kph, so the minimum time to the opposite shore would be 540 s. So the book's 624.3 s may be right Edit: since the landing point is to be exactly opposite. Picture yourself at the bank of the river about to jump in...

I need to sign off for tonight. I hope someone else will take up monitoring and aiding your progress.

Doc Al said Doc Al's "at an angle" means not perpendicular. Your vertical and horizontal sides would be perpendicular, would they not?

OK, please describe your triangle.

Here is a nudge: Review Doc Al's post #8.

Yes, there is an error somewhere. Before we knew that the goal was to go straight across the river, I worked it out assuming all of his speed would be dedicated to getting across the river quickly. That would yield the minimum time considering his speed and width of the river. That time was 1080...