Mechanics (1 Viewer)

[kwabon]

in my Indian brother's book (S.K. Patel)
there was this example on pg 253

it goes
a particle of mass m falls under gravity from rest in a medium with the resistive force given by R(v) = kv. discuss the motion

my answer (part of it at least), goes like

ma = mg - kv
then i divide by m and then integrate and so on.

but when i look at the answer, it goes
ma = mg - mkv

i do not think that is right, what do you guys think about the answer to the example??

your help is very much appreciated.

 

[Timothy.Siu]

it doesn't really matter because k is a constant, mk is still constant

 

[Drongoski]

kwabon

u have a valid point.

If Patel says: ma = mg - mkv

it implies that 'kv' is "acceleraration" part rather than 'force' as in resistive force R(v) = kv.

I think he just overlooked. Authors/editors are human.

 

[kwabon]

kwabon

u have a valid point.

If Patel says: ma = mg - mkv

it implies that 'kv' is "acceleraration" part rather than 'force' as in resistive force R(v) = kv.

I think he just overlooked. Authors/editors are human.

hahaha, just making sure that is all

it doesn't really matter because k is a constant, mk is still constant

if that is true, then
ma = mg - kv
a = g -(k/m)v

let that (k/m) = x
a = g - xv

not sure if that is what you meant, but yeh if you do this, the answer will still be the same

 

[study-freak]

If the book is right about saying R(v)=kv is the resistive FORCE, the answer is wrong as you pointed out.

But usually, R(v)=kv^n refers to acceleration, I think.

 

[kwabon]

If the book is right about saying R(v)=kv is the resistive FORCE, the answer is wrong as you pointed out.

But usually, R(v)=kv^n refers to acceleration, I think.

should that be divided by m, and then it can be referred to acceleration
not sure whether i am right, so here goes
F=ma
R(v) = F
ma = kv^n
a = (k/m)v^n

 

[study-freak]

should that be divided by m, and then it can be referred to acceleration
not sure whether i am right, so here goes
F=ma
R(v) = F
ma = kv^n
a = (k/m)v^n

You are right if and only if Force=kv^n
but I'm saying that in many cases in textbooks, acceleration=kv^n, not force.

 

[kwabon]

You are right if and only if Force=kv^n
but I'm saying that in many cases in textbooks, acceleration=kv^n, not force.

oh right. but what i thought was only put m into the resistive force when they specifically say that R(v) = mkv, or say that the resistance is kv^n per unit squared.
thats what i thought at least, whats your thoughts on that, study-freak????

btw study-freak i get what u mean, but the thing i said above, is that how you figure out whether m should be placed or not into the acceleration.

LOL, i myself dont know what i am ranting on about, yeh but please try to answer the question as best as you can lol.

 

[study-freak]

oh right. but what i thought was only put m into the resistive force when they specifically say that R(v) = mkv, or say that the resistance is kv^n per unit squared.
thats what i thought at least, whats your thoughts on that, study-freak????

btw study-freak i get what u mean, but the thing i said above, is that how you figure out whether m should be placed or not into the acceleration.

I was confused greatly about this too when I first studied mechanics (Frankly speaking, I still am).

But textbooks make it unclear. For example, Fitzpatrick one varies from question to question in the exercise and cambridge exercises always include m in resistive force although it sometimes says "resistance is proportional to kv^n" and another time "resistance is mk times square of the speed,"etc.
The tutor I had last year told me to put m into force only if the question says "in a resistive medium" or whatever but that doesn't work in some questions, too.

I think your rule is right in most cases though.
And exam questions should make it clear whether m is included in resistive force.

 

[riseek]

TIP: just read the question (in exam), do random stuff, and hope for the best.
i shall get back to you after further enquiring.

 

[kwabon]

TIP: just read the question (in exam), do random stuff, and hope for the best.
i shall get back to you after further enquiring.

that helps a lot riseek.

 

[riseek]

as i stated i shall GET BACK AFTER FURTHER ENQUIRING

 

[vds700]

in my Indian brother's book (S.K. Patel)
there was this example on pg 253

it goes
a particle of mass m falls under gravity from rest in a medium with the resistive force given by R(v) = kv. discuss the motion

my answer (part of it at least), goes like

ma = mg - kv
then i divide by m and then integrate and so on.

but when i look at the answer, it goes
ma = mg - mkv

i do not think that is right, what do you guys think about the answer to the example??

your help is very much appreciated.

yeah this used to annoy me last year. I think in most circumstances, the question will make it clear whether to put the m in or not.

If it doesnt say, look in the later parts of the question, it might say prove v (or x)= ..., if it has no m, put m in front of the resistive force.

If still not sure, id say put it in, they cant mark u wrong either way, and with the m, its easier to integrate things

 

[David.Lai]

Hey, the reason for ma = mg -mkv,
since every resisted near the earth, that is; where mg is applied,
then m times R (R= resisted force), is required as all resisted motions near the earth
is affected by mass. that is the reason for using mkv