Forces

// The link below has an incredible video on BASE (Building Antennae Span Earth) jumping with wingsuits, very much related to terminal velocity. They call the cliffs they jump from "terminal cliffs" since the cliffs are high enough to reach terminal velocity. The wingsuits slow their terminal velocity but they're still traveling at about 100 mph. Check it out!

[|Terminal Velocity Video] //

__DEFINITION:__ A force is a push or a pull which exerts an acceleration on a mass. =**NEWTON'S LAWS OF MOTION**=

__Newton's first law__- Basically states that a body in motion (or lack thereof) will remain in said motion so long as no net force is applied.

__Newton's second law__- Pretty much states that: , or that Force is the change in momentum with respect to time. When there is constant mass, this becomes the more familiar:
 * F=ma**. (sort of important)

__Newton's third law__- Every action generates an equal and opposite reaction. Or, if you think of the universe (or anything, really) and everything inside it as a whole, the sum of the internal forces is zero. Important for friction problems, tension, springs, but not really anything else.

__Notes__: In general:
 * F=ma**. (m is the mass of the object receiving the force, a is the induced acceleration)

For springs: , with k the spring constant of the spring, and x the displacement in meters.

For gravity: , with G the gravitational constant (=6.67x10^-11), m and M the masses acting on each other through gravity, and r the distance between them.

For frictional force: , with µ the coefficient of static friction and Fn the normal force on the object from the surface exerting the friction.

For centripetal force: , or equivalently: , with M the mass being acted on, Ac the centripetal acceleration, which is equal to v^2/r, with v the velocity of the object and r the radius of the mass from the center of the circle being followed.

For drag force: Suppose drag force is directly proportional to velocity- i.e. there exists some b such that: Now, we originally wonder as to what velocity an object dropped from sufficiently high up in the sky reaches as a maximum- i.e. what the terminal velocity is. Well, since terminal velocity means acceleration down is 0, or the sum of the forces in the y-direction is 0, so (taking up to be positive): Since that's rather unimpressive, we wonder as to what the equation for velocity at any time t is, so: sum of the forces F equals ma, and, (taking down to be positive) we already know what mg/b is (it's the terminal velocity, which we'll call VT) hence: so, to find C, we take t to be 0, and since we know v=v(0), so **C=v(0)-VT** hence, so **v = (v(0)-VT)*exp(-bt/m)+VT.** If we took v(0) to be 0 (i.e. we just dropped it instead of throwing it, etc.), factoring out VT: =//__**v**__//= //**VT * ( 1 - e^(-bt/m) ).**//=
 * Fd=-bv**.
 * Fd-Fg=0**
 * Fd=Fg**
 * bv=mg**
 * v=mg/b.**
 * Fg-Fd=mg-bv=ma,**
 * a=dv/dt,**
 * mg-bv=m(dv/dt),**
 * dv/(v-mg/b) = -(b/m) dt,**
 * ln ( v-VT ) = -bt/m + C***
 * v-VT=C*e^(-bt/m)** (note: C=e^C*)
 * v(0)-VT=C*e^0**
 * v-VT=(v(0)-VT)*exp(-bt/m)** (exp(x)=e^x, and it's just easier to write)
 * v = (-VT)*exp(-bt/m)+VT**



Don't forget good old fashioned FRICTION too!