Retarding Force proportional to the velocity falling particl

AI Thread Summary
The discussion focuses on solving a classical mechanics problem involving a particle in vertical motion with a retarding force proportional to its velocity. The equation of motion is established as F = m(dv/dt) = -mg - kmv, leading to a differential equation that is solved for velocity and displacement. Participants clarify the interpretation of forces, noting that the retarding force acts opposite to gravity, and discuss the mathematical manipulation of the exponential term in the solution. The conversation emphasizes the importance of correctly interpreting the signs of velocity and forces in the context of the problem. Overall, the thread aims to clarify the mathematical steps and physical concepts involved in the problem.
Decypher
Messages
2
Reaction score
0
Example problem from a Classical Mechanics book.
Find the displacement and velocity of a particle undergoing vertical motion in a medium having a retarding force proportional to the velocity.

F = m(dv/dt) = -mg-kmv

dv/(kv + g) = -dt

(1/k)ln(kv + g) = -t + c

kv + g = e^(-kt+kc)

v = (dz/dt) = (-g/k) + ((k(Vo) + g)/k)e^(-kt)

I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
I know a property of exponents allows you to separate the exponents powers when they are being added, then each one becomes the power of an exponent which is where e^-kt comes from, but I am not sure about where the other half of the exponent goes.
 
Physics news on Phys.org
Decypher said:
F = m(dv/dt) = -mg-kmv
Hi Decypher:

I think y ou have a sigh error. mg is the force of gravity pulling the mass towards rthe Earth, and kmv is a retarding force, slowing down the pull. The direction of these forces are opposite.

Hope this is helpful.

Regards,
Buzz
 
I spent a great amount of time on this part you are mentioning. The book does not provide a great description on this statement.
"Where -kmv represents a positive upward force since we take z and v = z(dot) to be positive upward, and the motion is downward-that is, v<0, so that -kmv > 0."
I think they are saying that because we are treating a positive velocity as one that is going higher on the z axis, whe have to treat -kmv as kmv because when a negative velocity is plugged in, kmv becomes an opposing force once again. It makes more sense to me to have it be g-kv just like a force opposing gravity should be.
 
Decypher said:
Find the displacement and velocity of a particle undergoing vertical motion
Decypher said:
we are treating a positive velocity as one that is going higher on the z axis
Hi Decypher:

OK. I think the first quoted sentence would have been clearer if the word "upward" had preceded "vertical".

Decypher said:
I don't see how the e^(-kt + kc) is turning into ((k(Vo) + g)/k
The above quote is not what you want to do. I suggest you take the equation
Decypher said:
kv + g = e^(-kt+kc)
and fill in the missing step: subtract g from both sides, and divide both sides by k. Then make the substitution v = dz/dt.

Then you should be able to make a substitution for c (which is an arbitrary constant of integration) which will give you what you want.

Hope this helps.

Regards,
Buzz
 
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