Yes! Exactly, yeah I finally got it. This was satisfying to figure out because at first glance there seems to be so much crucial info missing. Thank you for the help.
Ok right so F = m*a. There is acceleration "to the left" and thus, the sum of the forces in this situation should be in that direction as well. m is known, a is currently not but should be equal to v^2 / ρ (also unknown), and F for the box, I believe should be the combination of the friction...
Ok so since the truck has acceleration towards the curve's centre, what happens in regards to the box is that the "flooring" beneath it is going left, and this should be why the friction has to prevent the box going right on the truck bed, i think. Maybe this is what is actually happening rather...
Ok so it should be something like this I suppose. The truck is in a circular motion with the centre of the curve to the left. Although both the velocity of the truck and radius of curve is unknown. EDIT: Wait I didn't see this message ok hold on separate fbds ok got it
Ok so, I don't really know how I should go about depicting P. Since the truck is taking a curve, the box is supposed to be sliding to the right on the truck, were it not for F counteracting it. That is what P is meant to represent. I'm honestly not even sure if it's an actual force, but it is...
So the problem describes a truck carrying a box taking a curve (which is also slanted), It takes the curve at such speed that the box would start sliding up to the right in the picture if it were not for the friction force F. So P is a centripetal force that F counteracts. (I hope that it's...
The problem that I immediately ran into was how I would calculate N without knowing Fmax. I didn't think the y-component of N would simply be the same magnitude as mg. After being stuck for a good while I even tested if it was, by dividing the magnitude of mg with cosθ, which of course ended up...
OHHHHHHHHHHH! Pythagoras that's right! Got the right answer now. Thank you so much! Feels kind of good that that's what I missed, at least understood what formulas to use.
After 3,32 seconds, vt should have varied by 0,695*3,32. I have done a previous exercise where you only needed to calculate the radial acceleration in this scenario. There, I took the vt after the given time, squared it and then divided with the radius. I remember clearing that one, so in this...