A question about linear drag force

AI Thread Summary
The discussion clarifies the relationship between velocity (v) and position (r) in the context of linear drag force in classical mechanics. It emphasizes that while velocity is defined as the derivative of position, it does not depend on position in a functional sense for the equations of motion being considered. The drag force is expressed as f = -bv, leading to a first-order differential equation that simplifies the analysis. The key takeaway is that velocity is treated as a separate variable that influences motion but does not directly depend on position in this context. Understanding this distinction is crucial for solving problems involving drag forces in projectile motion.
shanname
Messages
8
Reaction score
0
My classical mechanics textbook says that, for a projectile, the linear drag force is given by f = -bv and the second law is written as m\ddot{r} = mg - bv (a second order differential equation) which can be rewritten as m\dot{v} = mg - bv (a first order differential equation) because the forces depend only on v and not on r. But I can't figure out why this is the case. Doesn't v depend on r?
 
Physics news on Phys.org
No. Velocity doesn't depend on location as such, only to the extent that location may effect some of the parameters.
 
##\mathbf v ## doesn't "depend" on ##\mathbf r## in the sense that it is some (unknown) function of ##\mathbf r## and probably some other variables as well.

The point is that ##\mathbf v## is just another name for ##\mathbf{\dot r}##, (that's what "velocity" means!) and differentiating, ##\mathbf{\dot v}## is identically equal to ##\mathbf{\ddot r}##.
 
Thank you, AlephZero. I believe I understand. I mean, I know that v is just another name for \dot{r} and the like, I just thought it could be rewritten in terms of v for that reason. No one ever explained that this is true specfically because v did not "depend" on r... is it ever the case that v does depend on r?
 
sorry, didn't realize I wasn't bolding \dot{r}.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top