It is possible to estimate the percentage of body fat by measuring the resistance of the upper leg instead of the upper arm; the calculation is similar. A person’s leg measures 40 from knee to hip, with an average leg diameter (excluding bone and other poorly conducting tissue) of 12. A potential difference of 0.80 causes a current of 1.6 µm. What are the muscle and fat fractions in the leg?
It seems that there is not enough information. For instance :
– what are the units used for each value?
– what resistivity values should be used for muscle and fat?
– what is the average diameter of the legs?
If you repost with the complete/correct information, you might get replies.
Alright, so some background info: masteringphysics hope you know that the resistivities of muscle and fat are 13 ohms*m and 25 ohms*m respectively (they provide a graph in the book that matches homework, yes that’s not is not provided at home work).
As for the question when it said “calculation is similar” (between the upper leg not the arm) they were referring to an integrative example which was also inside the manual.
Background: When we think of the upper arm, or upper leg, or appendages, we might think of the most basic shape – a cylinder. inside this cylinder are bones, muscles and fat (although we ignore bones in terms of calculations). one important thing to note is that *muscle and fat are connected in parallel* – for example, one cylinder splitting into two half-circle cylinders consisting of fat and the other one muscle rather than a splitting in half which would translate to two cylinders (in this case indicative of a series connection).
it is important because
a) when potential is applied, the current will split into “muscle resistance” and “fat resistance”
b) the nature of the equivalent resistances connected in parallel
There are two ways (or so I can see) to answer this problem: using currents or using resistors – although they both result in the same equation.
non menclature: _T (like VT, IT, RT) will denote the total of this amount; _m to describe the muscle; _f for fat; r for rho/resistivity; L for length; pi to pi; x for muscle fraction; 1-x for the fat fraction (since muscle + fat must equal 1 = you could also have used x + y = 1, but you would need a common variable and assign y = 1-x anyway) ; everything has been converted to SI units
====1st Via: Via Correntes====
A) it’s a parallel circuit, the current is divided, so IT = Im +If
B) it’s parallel, so each branch has the same VT
1) Im = VT/Rm
2) If = VT/Rf
*C) we said earlier that we should think of fat and muscle as partial circular cylinders, so the resistance formulas come out in such a way that the percentage is counted in the area. that’s to say
1) Rf = (rf*L)/[(1-x)A]
2) Rm = (rm*L)/(xA)
D) Combining AC and we get
IT = (VT*x*A)/(rm*L) + (VT*A*[1-x])/(rf*L)
E) from here is a bunch of algebra to isolate x (and sub into a formula for the area of the circle) and finally we get…
*Configuration: x = [(4rm*rf*L*IT)/(VT*pi*d^2) – rm]*[(rf-rm)^-1]
Plug & Chug: x = [(4*13*25*.40*.0016)/(.80*pi*.12^2)-13]*[(25-13)^-1]
YouGet: x=0.832 while 1-x=0.168 for muscle and fat respectively
====2nd channel: via resistors====
A) it is a parallel circuit, so 1/RT = 1/Rm + 1/Rf
You can rewrite as RT = (Rm^-1 + Rf^-1)^-1
B) VT = IT*RT for all circuits
Can rewrite as RT = VT/IT
C) (see 1st way C for the derivation of the formula)
1) Rm = rm*L / (xA)
2) Rf = rf*L/ [(1-x)A]
D) We combine AC to obtain
VT/IT = [ (1-x)A/(rf*L) + xA/(rm*L) ]^-1
E) go through some algebra, rewrite some formulas, sub in (pi*d^2)/4 for A and we get the exact same formula we derived the 1st way, i.e.
x = [(4rm*rf*L*IT)/(VT*pi*d^2) – rm]*[(rf-rm)^-1]
So three tips:
1) homework must really say what is given and what is not
2) appendages can be considered in circuits as parallel
3) the “fractions” of muscle and fat are due to different slices of the area/cylinder (the percentages are attributed to the areas of the resistors, not the resistances of the resistors themselves, i.e. x *RT or x*r*L /A vs r*L/(x*A) with the first being wrong and the last being right.