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Confused about "motion to a vertical plane" questions

Whenever there is a question about "motion to the horizontal" I understand that its sine for vertical and cosine for horizontal components of motion, but get it mixed up for vertical.

I drew a diagram: https://postimg.cc/kVsqXr5x

So for example: "A ball is thrown horizontally from the vertical plane", would its horizontal component of velocity be Ucos and vertical component of velocity be Usin along the angle I have illustrated?
(edited 4 months ago)
Reply 1
Original post by MonoAno555
Whenever there is a question about "motion to the horizontal" I understand that its sine for vertical and cosine for horizontal components of motion, but get it mixed up for vertical.

I drew a diagram: https://postimg.cc/kVsqXr5x

So for example: "A ball is thrown horizontally from the vertical plane", would its horizontal component of velocity be Ucos and vertical component of velocity be Usin along the angle I have illustrated?

In your diagram the sin and cos are the wrong way round. Id always (until youre sure) draw a small right triangle with the velocity/force as the hypotenuse and the two directions (vert/horiz) as the legs and the relvant angle (theta) marked on. Fairly obv, the vertical would be cos and the horizontal would be sin in your diagram as vert is adj and horiz is opp.

Usually theta is relative to the horizontal though (convention) so the complement of what youve drawn, then (drawing a small right triangle) horizontal would be cos and vertical sin.
(edited 4 months ago)
Reply 2
Original post by mqb2766
In your diagram the sin and cos are the wrong way round. Id always (until youre sure) draw a small right triangle with the velocity/force as the hypotenuse and the two directions (vert/horiz) as the legs and the relvant angle (theta) marked on. Fairly obv, the vertical would be cos and the horizontal would be sin in your diagram as vert is adj and horiz is opp.

Usually theta is relative to the horizontal though (convention) so the complement of what youve drawn, then (drawing a small right triangle) horizontal would be cos and vertical sin.


Wow you are right (had to brush up on my sohcahtoa). Im confused at what i even wrote now.

Had to re-draw it (and with projections to horizontal aswell): https://postimg.cc/TywVmnnL

Thanks
Reply 3
Original post by MonoAno555
Wow you are right (had to brush up on my sohcahtoa). Im confused at what i even wrote now.

Had to re-draw it (and with projections to horizontal aswell): https://postimg.cc/TywVmnnL

Thanks


Looking better, though perhaps a bit laborious (not a bad thing if youre unsure). Id probably do something like
https://emedia.rmit.edu.au/learninglab/content/pf12-forces-slopes
(obv make the dashed triangle a bit smaller and note that here its para/perp rather than horiz/vert) so put a small resolved forces/velocity right triangle on the original diagram.

Note that as u (velocity) or f (force) is the hypotenuse, then the two legs must be u*cos(theta) and u*sin(theta) where cos is next to theta and sin opposite. Both the legs are smaller than the hypotenuse, so theyre cos/sin multipled by u or f. There isnt really a need to work out the trig / you should be able write down the legs for the resolved forces/velocity "without thinking". If youre having to resort to sohcahtoa, then maybe do more examples.
(edited 4 months ago)
Reply 4
Original post by mqb2766
Looking better, though perhaps a bit laborious (not a bad thing if youre unsure). Id probably do something like
https://emedia.rmit.edu.au/learninglab/content/pf12-forces-slopes
(obv make the dashed triangle a bit smaller and note that here its para/perp rather than horiz/vert) so put a small resolved forces/velocity right triangle on the original diagram.

Note that as u (velocity) or f (force) is the hypotenuse, then the two legs must be u*cos(theta) and u*sin(theta) where cos is next to theta and sin opposite. Both the legs are smaller than the hypotenuse, so theyre cos/sin multipled by u or f. There isnt really a need to work out the trig / you should be able write down the legs for the resolved forces/velocity "without thinking". If youre having to resort to sohcahtoa, then maybe do more examples.

Hi, I want to know if the same principle within the page you linked is used to solve this ball along a semi circular ramp problem:
Obviously not the same, but would the work done along the ramp at the angle be mgrsin(45) from point P to Q/ Q to P?
Screenshot_20231013-085127_Drive.jpg
(edited 4 months ago)
Reply 5
Original post by MonoAno555
Hi, I want to know if the same principle within the page you linked is used to solve this ball along a semi circular ramp problem:
Obviously not the same, but would the work done along the ramp at the angle be mgrsin(45) from point P to Q/ Q to P?
Screenshot_20231013-085127_Drive.jpg

Not really sure how youre answering it, but for me Id note that loss in GPE from 0 to Q is mgrcos(45) so that is the work done by friction from 0 to Q so from 0 to P is 2/3 of that which gives the KE at P.
Reply 6
Original post by mqb2766
Not really sure how youre answering it, but for me Id note that loss in GPE from 0 to Q is mgrcos(45) so that is the work done by friction from 0 to Q so from 0 to P is 2/3 of that which gives the KE at P.

Oop I believe i misread the question as it "first" passes P, as I thought you had to calculate it from when its at point Q to when it returns to P.

Anyways is the reason its "mgrcos(45)" and not tan is because the gpe loss it equates to is vertical motion?
Reply 7
Original post by MonoAno555
Oop I believe i misread the question as it "first" passes P, as I thought you had to calculate it from when its at point Q to when it returns to P.

Anyways is the reason its "mgrcos(45)" and not tan is because the gpe loss it equates to is vertical motion?

The vertical height of the loss is
r*cos(45)
just by drawing the leg which corresponds to a horizontal line from Q to the "y" axis. Then r is the hypotenuse and rcos(45) is the adjacent vertical side, so the loss in GPE or work done against friction.
(edited 4 months ago)

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