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Can somebody help me understanding why when free falling the force on the vertical axis is F= -mg-kv, where k is the constant of air resistance and v the velocity?

Suppose the vertical axis is positive in the upwards direction. Then the acceleration is negative, so this is okay. The air resistance should be positive thou, cause it should oppose to the acceleration. Am I right or wrong? Cause I don't understand how the air resistance and the acceleration could ever have the same sign. Indeed when opening a parachute, all we do is enhancing the air resistance and therefore balancing the two forces.

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  • $\begingroup$ You're right. Gravity and air resistance should oppose each other. It would be odd to have the constant of air resistance itself to be negative. $\endgroup$ – MaxW Feb 21 '16 at 19:01
  • $\begingroup$ Your equation is wrong. The sings must be opposite. $\endgroup$ – Yashas Feb 21 '16 at 19:01
  • $\begingroup$ So how come that in EVERY textbook, online pdf or mathematical model, -mg and -kv have the same sign? $\endgroup$ – Euler_Salter Feb 21 '16 at 19:08
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A general model of restive forces that depend on velocity can be given by the following equation,

$F = C_1v + C_2v^2$

We choose either the force to be dependent on either of the terms or both depending on the magnitude of velocity.

In your case, since we are talking about bodies falling we can assume the velocity to be small and ignore the $v^2$ term.

The resistance force always oppose the motion of a body.

The body you are talking about is falling down due to acceleration due to gravity. Since restive forces always tend to bring the body in motion to rest, the restive forces will act upwards in an attempt to slow down the object.

So we have two forces, gravity which pulls the object down and the air resistance which tries to slow down the object.

Writing the equation of motion for the body using your sign convention (upwards as positive), we get

$F = C_1v - mg$

The vector form of the equation would be

$F = -C_1\vec{v} - m\vec{g}$

Note that the velocity and acceleration are in opposite directions hence the additional negative sign.

The signs of the two forces are in opposite direction which is in agreement with out day to day experiences and intuition.

An object is said to be in free-fall if its motion is caused by gravity only. Your case where air provides resistance, the object in motion cannot be said to be in a free-fall.

(Video-Lecture) 12: Resistive Forces | 8.01 Classical Mechanics, Fall 1999 (Walter Lewin)

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  • $\begingroup$ according to you, what is the sign of c1? $\endgroup$ – Mrigank Feb 21 '16 at 22:36
  • $\begingroup$ The constant is positive. $\endgroup$ – Yashas Feb 22 '16 at 5:05
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    $\begingroup$ you do know that in this v is negative ?, hence this is the exact confusion op started with where both forces are in same direction $\endgroup$ – Mrigank Feb 22 '16 at 6:54
  • $\begingroup$ Yea, it must be speed instead of velocity. $\endgroup$ – Yashas Feb 22 '16 at 7:37
  • $\begingroup$ I have already stated that the force depends on the magnitude of velocity at the beginning. Will edit and make the answer clearer. $\endgroup$ – Yashas Feb 22 '16 at 7:40
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The force F is taken to be positive if it is upward and negative if it is downward. So -mg means that the gravitational force is downward. The term -kv means that, if the body is moving upward (positive v) the drag force is downward, and if the body is moving downward (negative v), the drag force is upward. This is completely consistent with our expectations.

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    $\begingroup$ The original equation should have been accompanied by the very important "WHERES". In this case, "where up is positive for forces and velocities, g is a positive number , and k is a positive number" $\endgroup$ – DJohnM Feb 23 '16 at 6:54
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You agree that in coordinate system that is generally used, up is "+" and down in "-" , so mg is "-" , on to your problem,

{ Frictional/Drag } FORCE DOES NOT HAVE A PREDFINED DIRECTION

rather it is opposite to direction of velocity at any point {or relative velocity in case of friction}, which is exactly what the equation says, no matter what coordinate system we would have used, air drag force would have always been "-Kv" in no way shape or form, it will ever support the velocity. where k is positive

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protected by Qmechanic Feb 21 '16 at 20:12

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