a sphere of radius r moving with
velocity v through a fluid experiences a retarding
force f given by F=6r^yv^zn^x where n is the
viscosity of the liquid. use the method of dimensions
to find the values of x,y,and z.
They're all 1
Answer
To find the values of x, y, and z using the method of dimensions, we need to equate the dimensions of the given equation on both sides. Let's break down the dimensions for each term:
On the left-hand side (LHS):
F = Force = [M L T^-2] (mass times length divided by time squared)
On the right-hand side (RHS):
6 = dimensionless constant
r^y = [L^y] (length raised to the power of y)
v^z = [L^z T^-z] (velocity raised to the power of z)
n^x = [M^x L^-x T^-x] (viscosity raised to the power of x)
By comparing the dimensions on both sides, we can equate them as follows:
[M L T^-2] = 6 [L^y] [L^z T^-z] [M^x L^-x T^-x]
Simplifying the equation, we can sort the terms based on their dimensions:
1. Mass: [M^1] = [M^x] ⇒ x = 1
2. Length: [L^1 T^-2] = 6 [L^y+z] [M^-x]
Equating the exponents, we get:
1 = y + z - x
1 = y + z - 1
y + z = 2
y = 2 - z
3. Time: [T^-2] = 6 [T^-z] [L^-x T^-x]
Equating the exponents, we get:
-2 = -z - x
-2 = -z - 1
z = -1
Therefore, the values of x, y, and z are:
x = 1
y = 2 - z = 2 - (-1) = 3
z = -1
So, the values of x, y, and z in the equation F = 6r^yv^zn^x are x = 1, y = 3, and z = -1.
A sphere of radius a moving through a fluid of density ρ with a velocity v experiences a
retarding force F which is given by F = k a
x ρ
yv
z where k is non dimensional coefficient. Using
method of dimension, find the value of x,y and z.