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Differential equation for grav, boyant and drag force
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[QUOTE="Agustin, post: 5258698, member: 574122"] [h2]Homework Statement [/h2] So there is a falling object, you have to take into account the boyant force, the pull of gravity and the drag force A time dependent distance equation is what we're looking for [h2]Homework Equations[/h2] F[SUB]d[/SUB]=C[SUB]d[/SUB]Ap[SUB]a[/SUB]v[SUP]2[/SUP]/2 Where F[SUB]d[/SUB] is the drag force C[SUB]d[/SUB] is the drag coefficient A is the area exposed to the fall p[SUB]a[/SUB] the air density v the immediate velocity F[SUB]b[/SUB]=mgp[SUB]a[/SUB]/p[SUB]c[/SUB] F[SUB]b[/SUB] is the boyant force mg is the weight of the object p[SUB]c[/SUB] is the object's density Note ( this equation is found from the original equation F[SUB]b[/SUB]=Volume submerged x air density x gravity; where the submerged volume is m/p[SUB]c[/SUB]) F[SUB]g[/SUB]=mg [h2]The Attempt at a Solution[/h2] ma=mg - mgp[SUB]a[/SUB]/p[SUB]c[/SUB] - C[SUB]d[/SUB]Ap[SUB]a[/SUB]v[SUP]2[/SUP]/2 Or dv/dt = A - Bv[SUP]2[/SUP] Where A=g( 1 - p[SUB]a[/SUB]/p[SUB]c[/SUB]) B=C[SUB]d[/SUB]Ap[SUB]a[/SUB]/2m I solve for v v (t) = (A/B)[SUP]1/2[/SUP] (1 + C[SUB]1[/SUB]e[SUP]-2(BA)[SUP]1/2[/SUP]t[/SUP])/( 1 - C[SUB]1[/SUB]e[SUP]-2(BA)[SUP]1/2[/SUP]t[/SUP]) Where C[SUB]1[/SUB] is some arbitrary constant Integrating we get the distance formula: X (t) = (A/B)[SUP]1/2[/SUP]t+(1/B)ln( 1 - C[SUB]1[/SUB]e[SUP]-2 (BA)[SUP]1/2[/SUP]t[/SUP]) + C[SUB]2[/SUB] I don't know wether it's correct or not. I've used techniques i found on the internet for the integration. http://www.freemathhelp.com/forum/threads/47073-integral-(1-(e-x-1))-dx-Using-Partial-Fractions [/QUOTE]
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Differential equation for grav, boyant and drag force
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