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Sagot :
To determine the formula for the viscous force [tex]\( f \)[/tex] acting on a solid sphere moving through a liquid, we use dimensional analysis. The viscous force is found to depend on the coefficient of viscosity [tex]\( \eta \)[/tex], the radius [tex]\( M \)[/tex] of the sphere, and its speed [tex]\( U \)[/tex].
We will express [tex]\( f \)[/tex] as a function of [tex]\( \eta \)[/tex], [tex]\( M \)[/tex], [tex]\( U \)[/tex], and a proportionality constant [tex]\( k \)[/tex]:
[tex]\[ f = k \cdot \eta^a \cdot M^b \cdot U^c \][/tex]
### Step 1: Determine Dimensions of Each Quantity
1. Viscous Force [tex]\( f \)[/tex]:
- Dimension: [tex]\([f] = \text{MLT}^{-2}\)[/tex]
2. Coefficient of Viscosity [tex]\( \eta \)[/tex]:
- Dimension: [tex]\([\eta] = \text{ML}^{-1}\text{T}^{-1}\)[/tex]
3. Radius [tex]\( M \)[/tex]:
- Dimension: [tex]\([M] = \text{L}\)[/tex]
4. Speed [tex]\( U \)[/tex]:
- Dimension: [tex]\([U] = \text{LT}^{-1}\)[/tex]
### Step 2: Set Up Dimensional Equation
We need the dimensions on both sides of the equation [tex]\( f = k \eta^a M^b U^c \)[/tex] to be consistent. Thus,
[tex]\[ [f] = [k \eta^a M^b U^c] \][/tex]
Substituting the dimensions, we get:
[tex]\[ \text{MLT}^{-2} = [k] \cdot (\text{ML}^{-1}\text{T}^{-1})^a \cdot (\text{L})^b \cdot (\text{LT}^{-1})^c \][/tex]
### Step 3: Break Down the Dimensions
Combining the dimensions, we get:
[tex]\[ \text{MLT}^{-2} = \text{M}^a \cdot \text{L}^{-a} \cdot \text{T}^{-a} \cdot \text{L}^b \cdot \text{L}^c \cdot \text{T}^{-c} \][/tex]
Simplifying the dimensions, we have:
[tex]\[ \text{MLT}^{-2} = \text{M}^a \cdot \text{L}^{-a+b+c} \cdot \text{T}^{-a-c} \][/tex]
### Step 4: Equate the Exponents
Equating the exponents for [tex]\( \text{M} \)[/tex], [tex]\( \text{L} \)[/tex], and [tex]\( \text{T} \)[/tex] on both sides, we get three equations:
1. For mass (M):
[tex]\[ 1 = a \][/tex]
2. For length (L):
[tex]\[ 1 = -a + b + c \][/tex]
3. For time (T):
[tex]\[ -2 = -a - c \][/tex]
### Step 5: Solve the System of Equations
From the first equation:
[tex]\[ a = 1 \][/tex]
Substituting [tex]\( a = 1 \)[/tex] in the third equation:
[tex]\[ -2 = -1 - c \][/tex]
[tex]\[ -1 - c = -2 \][/tex]
[tex]\[ c = 1 \][/tex]
Substituting [tex]\( a = 1 \)[/tex] and [tex]\( c = 1 \)[/tex] in the second equation:
[tex]\[ 1 = -1 + b + 1 \][/tex]
[tex]\[ 1 = b \][/tex]
### Step 6: Write the Final Formula
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex] into our original expression [tex]\( f = k \eta^a M^b U^c \)[/tex], we obtain:
[tex]\[ f = k \eta^1 M^1 U^1 \][/tex]
Therefore, the formula for the viscous force [tex]\( f \)[/tex] is:
[tex]\[ f = k \eta M U \][/tex]
where [tex]\( k \)[/tex] is a dimensionless proportionality constant.
We will express [tex]\( f \)[/tex] as a function of [tex]\( \eta \)[/tex], [tex]\( M \)[/tex], [tex]\( U \)[/tex], and a proportionality constant [tex]\( k \)[/tex]:
[tex]\[ f = k \cdot \eta^a \cdot M^b \cdot U^c \][/tex]
### Step 1: Determine Dimensions of Each Quantity
1. Viscous Force [tex]\( f \)[/tex]:
- Dimension: [tex]\([f] = \text{MLT}^{-2}\)[/tex]
2. Coefficient of Viscosity [tex]\( \eta \)[/tex]:
- Dimension: [tex]\([\eta] = \text{ML}^{-1}\text{T}^{-1}\)[/tex]
3. Radius [tex]\( M \)[/tex]:
- Dimension: [tex]\([M] = \text{L}\)[/tex]
4. Speed [tex]\( U \)[/tex]:
- Dimension: [tex]\([U] = \text{LT}^{-1}\)[/tex]
### Step 2: Set Up Dimensional Equation
We need the dimensions on both sides of the equation [tex]\( f = k \eta^a M^b U^c \)[/tex] to be consistent. Thus,
[tex]\[ [f] = [k \eta^a M^b U^c] \][/tex]
Substituting the dimensions, we get:
[tex]\[ \text{MLT}^{-2} = [k] \cdot (\text{ML}^{-1}\text{T}^{-1})^a \cdot (\text{L})^b \cdot (\text{LT}^{-1})^c \][/tex]
### Step 3: Break Down the Dimensions
Combining the dimensions, we get:
[tex]\[ \text{MLT}^{-2} = \text{M}^a \cdot \text{L}^{-a} \cdot \text{T}^{-a} \cdot \text{L}^b \cdot \text{L}^c \cdot \text{T}^{-c} \][/tex]
Simplifying the dimensions, we have:
[tex]\[ \text{MLT}^{-2} = \text{M}^a \cdot \text{L}^{-a+b+c} \cdot \text{T}^{-a-c} \][/tex]
### Step 4: Equate the Exponents
Equating the exponents for [tex]\( \text{M} \)[/tex], [tex]\( \text{L} \)[/tex], and [tex]\( \text{T} \)[/tex] on both sides, we get three equations:
1. For mass (M):
[tex]\[ 1 = a \][/tex]
2. For length (L):
[tex]\[ 1 = -a + b + c \][/tex]
3. For time (T):
[tex]\[ -2 = -a - c \][/tex]
### Step 5: Solve the System of Equations
From the first equation:
[tex]\[ a = 1 \][/tex]
Substituting [tex]\( a = 1 \)[/tex] in the third equation:
[tex]\[ -2 = -1 - c \][/tex]
[tex]\[ -1 - c = -2 \][/tex]
[tex]\[ c = 1 \][/tex]
Substituting [tex]\( a = 1 \)[/tex] and [tex]\( c = 1 \)[/tex] in the second equation:
[tex]\[ 1 = -1 + b + 1 \][/tex]
[tex]\[ 1 = b \][/tex]
### Step 6: Write the Final Formula
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex] into our original expression [tex]\( f = k \eta^a M^b U^c \)[/tex], we obtain:
[tex]\[ f = k \eta^1 M^1 U^1 \][/tex]
Therefore, the formula for the viscous force [tex]\( f \)[/tex] is:
[tex]\[ f = k \eta M U \][/tex]
where [tex]\( k \)[/tex] is a dimensionless proportionality constant.
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