Forum Klix.ba

kukuruz_kokicar wrote:A to uze ne klize nikako? Ako je tako, onda je ugao nebitan - ono nece klizati i da je na ravnom.

kukuruz_kokicar wrote:A to uze ne klize nikako? Ako je tako, onda je ugao nebitan - ono nece klizati i da je na ravnom.

kukuruz_kokicar wrote:Ugao (izmedju h i l) mora biti manji od 45 stepeni. OK?

nellington wrote:

kukuruz_kokicar wrote:Ugao (izmedju h i l) mora biti manji od 45 stepeni. OK?

Moj račun kaže da ne mora.

Melanholik wrote:

nellington wrote:

kukuruz_kokicar wrote:Ugao (izmedju h i l) mora biti manji od 45 stepeni. OK?

Moj račun kaže da ne mora.

ma je li ovdje uopšte stvar u vrhu?

kukuruz_kokicar wrote:Sada sam ja pogrijesio, ugao izmedju h i l treba biti manji ili jednak 45 da uze ne sklizne, u suprotnom klize prema gore i ... spada.

nellington wrote:

kukuruz_kokicar wrote:Sada sam ja pogrijesio, ugao izmedju h i l treba biti manji ili jednak 45 da uze ne sklizne, u suprotnom klize prema gore i ... spada.

Ne bih te na tu sitnicu lovio, ne brini Ugao može biti i veći, značajno veći.

David Hmelnicki wrote:The answer: The critical angle A between the generatrix of the cone and the vertical (see Fig.(a) below) is 30o.

The solution:

The solution is given by a simple lemma, consisting in the following: Consider a rope on a convex surface with two equal weights at its ends. It is obvious that the length must be the shortest possible, otherwise the potential energy of the weights would not be minimum. So, according to the lemma, a rope on a convex surface lies along a geodesic line, i.e. the shortest curve between two points lying wholly on the surface.

Now, if we cut our cone along the generatrix, passing through the knot of the loop (green line on Fig.(a) below), and then roll it out on the plane, we will get a sector, which will look either like Fig.(b) or Fig.(c) below depending on the angle A. Green lines on the resulting sectors are just the line along which the cone was cut. Here the thick lines are the projections of the loop on the plane. A geodesic on a plane is a straight line. So, the straight line is exactly what we need. Moreover, we can see that the loop does not slip off the top of the mountain only if the angle B at the top of the sector does not exceed 180o as depicted on Fig.(b). Otherwise (see Fig.(c)), the line lies outside the sector and the loop has no equilibrium on the cone.

To find the critical angle A of the cone, we note that if L is the generatrix of the cone, then the length of the arc, which subtends the angle B, equals the circumference of the base of the cone 2{pi}r, where r is the radius of the base: B L = 2 {pi} r. As r = L sin(A) and the critical angle B = {pi}, sin(A) = 0.5, and therefore A= 30o.

kukuruz_kokicar wrote:

nellington wrote:

kukuruz_kokicar wrote:Sada sam ja pogrijesio, ugao izmedju h i l treba biti manji ili jednak 45 da uze ne sklizne, u suprotnom klize prema gore i ... spada.

Ne bih te na tu sitnicu lovio, ne brini Ugao može biti i veći, značajno veći.

E sada mi fakat nista nije jasno. Kako 30 vece od 45.

nellington wrote:

David Hmelnicki wrote:Green lines on the resulting sectors are just the line along which the cone was cut. Here the thick lines are the projections of the loop on the plane. A geodesic on a plane is a straight line. So, the straight line is exactly what we need.

nellington wrote: Ja govorim o nagibu, ti govoriš o uglu između h i l. Totalno suprotne stvari, kod mene taj ugao može ići do 60°.

EDIT: Sad vidim tvoj edit, sve ok

kukuruz_kokicar wrote:

nellington wrote: Ja govorim o nagibu, ti govoriš o uglu između h i l. Totalno suprotne stvari, kod mene taj ugao može ići do 60°.

EDIT: Sad vidim tvoj edit, sve ok

Nista nije OK, dok ja ne kazem da je ok .

Sala mala, no u rjesenju koje si prikazao - kriticni ugao A nije nagib vec isto sto i ugao l/h i on je 30 - dakle je to manje od 45 stepeni. To vec ima smisla. Za rjesenje koje tvrdi da je moguc ugao veci od 45 bi bio spreman rizikovati i zivot ... tvoj.

nellington wrote:Po mom računu, minimalni nagib je 30°. Koliko je atko dobio?

atko wrote:

nellington wrote:Po mom računu, minimalni nagib je 30°. Koliko je atko dobio?

toliko moRe u matematiKi ... u masinstvu je rezultat drugaciji moze i malo vise i malo manje jer zavisi od trenja i/ili velicine ugla izmedju uzeta i kose ravni kupe ... sto je kod tebe nula predpostavljam ...