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Extremely basic question [mechanics] (2 Viewers)

Revacious

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"A particle of mass m is projected vertically upwards with a velocity of u m/s, with air resistance proportional to its velocity"

So naturally, i assumed the forces working on the body were

F = mg
F = kv

net F = - (mg + kv)

and hence

a = - (mg + kv)/m


BUT the first line of the solution is:

R = mkv
F= mg

net F = - (mg + mkv)

which would seem to imply resistance is acceleration.


So the question is, which is correct?

(the very next question's solution makes it a force, so its either something i completely don't understand, or some bullshit inconsistency)

Also, can you help me clarify between resistance and retardation? If there's a difference?
 
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clintmyster

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this was sorta bugging me for a bit as well. It seems that air resistance has a component of mass in there so what I normally do is this:

F = ma = -mg - mkv
therefore a = -(g+kv)

For some question however that say F(r) = kv, then you don't treat it as mkv.

About the resistance/retardation thing, I haven't seen retardation used much so can't really remember. Can you find an example please?
 

Affinity

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doesn't matter in cases where your mass in constant.
 

annabackwards

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From what i've researched it is as affinity said - it doesn't matter whether you let R = -mkv or -kv because k is a constant which would include the mass of the object if you didn't.

I always use R = -mkv because calculations are easier unless they tell me what Resistance is... they can't mark you down whichever way you do it right?
 

clintmyster

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From what i've researched it is as affinity said - it doesn't matter whether you let R = -mkv or -kv because k is a constant which would include the mass of the object if you didn't.

I always use R = -mkv because calculations are easier unless they tell me what Resistance is... they can't mark you down whichever way you do it right?
really i just think it matters what your answer is for the acceleration. If that's wrong then everything else is wrong.
 

Revacious

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so basically, whether or not i integrate
a = - (mg + kv)/m

or

a = - (g+kv)

the answer should be the same (except k differs?)


And from people i ask, retardation generally refers to negative acceleration, not a force.

(im still pretty confused)
 

annabackwards

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really i just think it matters what your answer is for the acceleration. If that's wrong then everything else is wrong.
Ah, thanks ^^

so basically, whether or not i integrate
a = - (mg + kv)/m

or

a = - (g+kv)

the answer should be the same (except k differs?)


And from people i ask, retardation generally refers to negative acceleration, not a force.

(im still pretty confused)
Pretty much, try both ways and you'll see ;)

Retardation is negative acceleration. But if you think about it logically, any acceleration on an object with a mass will act as a force on that object (F = ma). I hope that clears things up :)
 

Dumbledore

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since ur treating k like it literally can be any constant:
if u mix it up between acceleration and retardation, can u still get full marks by having the same answer except k is a negative constant

not that i would try to mix it up, just incase.
 

clintmyster

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so basically, whether or not i integrate
a = - (mg + kv)/m

or

a = - (g+kv)

the answer should be the same (except k differs?)
I don't think thats right because it should be - (mg + mkv)/m and that simplifies down to a = - (g+kv). I'm not all that sure about this but from all the textbook questions I've done, I don't recall seeing one where you have to find the value of k. In general, they make you integrate and then they might say given that k = this and U is this, find the value of H.
 

Dumbledore

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lol bos should have stuck to saying acceleration is ...etc, it would stop all the confusion and any 4u student should be able to apply newtons 2nd law(at least i hope so)
 

Revacious

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I don't think thats right because it should be - (mg + mkv)/m and that simplifies down to a = - (g+kv). I'm not all that sure about this but from all the textbook questions I've done, I don't recall seeing one where you have to find the value of k. In general, they make you integrate and then they might say given that k = this and U is this, find the value of H.
hmmm i see what theyre saying though, basically, since k varies, it doesnt matter; granted it might be more convenient to choose one way or another, but it should be right seeing as k can actually be anything (unless stated otherwise).

its only problematic when they ask you to prove the height above ground is something (for instance). Then your final answer will differ, becasue if their acceleration was based off having no m's in it, and you have ms in it, its not going to match up unless you just erase all the m's.

so prehaps its just educated guessing to some extent. thanks guys.
 

yibbon

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"A particle of mass m is projected vertically upwards with a velocity of u m/s, with air resistance proportional to its velocity"
Iv'e always taken the air resistance as mkv in this case, that's at least what our teacher showed us. But there are questions that have the m's in there during the integration, so never crossed them off unless you can.
 

Mingiee

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oh i think i understand it now! i dont believe that it can be both!
Ok like someone had previously said retardation is negative acceleration.
And if anyone has done fitzpatrick questions... the solutions use R=mkv when it says "Retardation due to resistance"... and only R=kv when its just "resistance"

I need to use a proper example to explain
"A particle falls from rest in a medium. If the retardation due to resistance in the medium varies as the square of the velocity... etc"
So its the retardation that equals kv^2 NOT resistance!
Resistance is the force... and includes retardation!
Resistance is F=ma. But a=retardation(which is acceleration)=kv^2

So when we're talking about forces acting on the object... its R= m.kv^2

I hope this theory of mine applies to other questions!! Good luck everyone
 

Drongoski

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Here's my 2 cents' worth of clarification or further confusion.

resistance is not of fixed direction unlike gravity (down to the centre of the earth)
If we take 'down' to be positive then acceleration due to gravity is g (where g is positive) and the corresponding force is mg. now the magnitude of this resistance is: a constant x v. This constant can be written simply as 'k' or 'mk'; if 'mk' then this k is 1/m of the earlier k.

Now whilst the direction of gravity is down, the direction of resistance is always opposite to the direction the particle happens to be travelling. Take the common problem of a particle mass m being projected upwards with initial velocity u say and subject to only the forces of gravity and of resistance which is as posted: proportional to the velocity. Lets use the 2nd option for magnitude of resistance = mkv where k is a positive constant. The problem is usually broken into 2 parts; the upward leg and the downward leg.

For upward leg, where we treat displacement up to be positive: the net force working on the particle = -mg -mkv

For the downward leg (taking displacement downwards to be positive now): the net force on the particle = mg - mkv

In each case resistance works against the prevailing motion (that's why it is called resistance). Thus in the upward leg, the particle is being slowed down by gravity and is abetted in this by the resistance (they are partners in slowing down the particle). In the downward leg, the particle is being speeded up by gravity; the resistance now acts to slow it down (resistance always try to slow the particle down, whichever direction it travels). If you think of it the resistance in the upward leg is opposite in direction to the resistance (same magnitude) in the downward leg.

Hopefully I have not added to the confusion.
 
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eriito

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From what i've researched it is as affinity said - it doesn't matter whether you let R = -mkv or -kv because k is a constant which would include the mass of the object if you didn't.

I always use R = -mkv because calculations are easier unless they tell me what Resistance is... they can't mark you down whichever way you do it right?
thanks. i did not know that :eek:
 

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