Net Force Equation for Object in Free Fall with Gravity and Drag Forces

[Theorγ]

Given a situation where an object is dropped at a certain height, how should the net force equation be written considering the fact that only gravity and drag acts on it? I'm stuck on whether the equation is written like this:

\Sigma F = F_{g} - F_{d}
Or like this because g is negative?:
\Sigma F = -F_{g} + F_{d}

 

[rock.freak667]

It depends on the sign convention you've chosen.

If you choose down as positive, then it would be the first.

If down is negative then it's the second.

In most cases people find the equation based on the resultant.

So if your object is falling then gravity is more than drag and the object is moving downward such that the convention would give the first equation.

 

[D H]

Neither.

\vec F_{\text{net}} = \vec F_g + \vec F_d
Edit: WTF? Why is the latex stuff not printing as latex stuff?
Whoa! The magic still works, but only after you wait for a while.

 

[Theorγ]

Why neither? And by the way, I'm asking this question primarily for integrating and finding an equation for the velocity of an object at a certain time given that gravity and drag are affecting it. I thought the signs of those respective forces were somewhat relevant.

 

[HallsofIvy]

DH is writing the forces as vectors. The direction is included in the vector rather than as a sign.

However, if the motion is in a straight line, there are only two possible directions which we treat as "+" and "-". As rockfreak667 said, the sign depends upon your convention. Most common is to choose "+" upward. In that case the gravitational force, downward, would be negative. Drag is opposite to the direction of motion so if you take drag proportional to speed, it would be -k dx/dt, with k a positive coefficient of proportionality. In some cases (mainly very small objects) it is more accurate to say the drag is proportional to the square of the velocity. Since a square is always positive we "lose" the sign and it becomes a bit harder to model.

 

[brainpushups]

Sorry to revive a thread that is a month old, but this was exactly my question.

Take the case for linear drag with down being positive. Assume that the object is dropped from rest.

m dv/dt = mg - bv

After separating variables and integrating with the appropriate limits the equation for velocity as a function of time is

v(t) = (m g)/b -(m g)/b Exp[-b/m t]

As expected, as time passes the object approaches its terminal velocityNow assume the same situation but with up being positive

m dv/dt = -mg + bv

The equation for the velocity is now v(t) = (m g)/b - (m g)/b Exp[b/m t]

A similar problem occurs if you assume that the object is shot upward with an initial velocity. If positive y is also measured up then the math works out, but if you define positive y to be downward the argument of the exponential will be positive.It seems fairly obvious what is happening. In the case where down is positive, the velocity is also positive so when the difference is taken (mg - bv) gets smaller until it reaches terminal velocity. In the case where down is negative the velocity is also negative so the sum (-mg + (-bv)) gets larger in magnitude corresponding to a higher downward acceleration.

So it seems like choosing the sign of what is positive or negative for the direction of the vectors is more important than other situations I’ve encountered. One thought is that since the assumption is the drag force has the form -f(v) \hat{v} the sign of the acceleration and drag force should always be opposite.

I guess my question is: is the fact that Fdrag = -f(v) \hat{v} the reason that one must choose the signs of the acceleration and drag force to be opposite?

Thoughts and corrections welcomed.

 

[rock.freak667]

Sorry to revive a thread that is a month old, but this was exactly my question.

Take the case for linear drag with down being positive. Assume that the object is dropped from rest.

m dv/dt = mg - bv

In this case the resultant force is downwards, so m*dv/dt is positive

Now assume the same situation but with up being positive

m dv/dt = -mg + bv

The resultant here is still downward, so 'mdv/dt' should be

-m*dv/dt = -mg +bv

or else your equation would show the mass moving upwards while falling.