Problem 16: Viscous Drag on Metal Sheet
1. Viscous Force Equation
According to Newton’s law of viscosity, the viscous drag force $F_v$ is given by:
$$ F_v = \eta A \frac{dv}{dy} $$Since the oil layer is very thin ($h$) and the velocity profile is linear, the velocity gradient is $\frac{v}{h}$.
As the sheet moves by a distance $x$, the area of the sheet in contact with the oil decreases. If the sheet is square with side $l$, the contact area $A$ at any instant is:
$$ A = l \times (l – x) $$Substituting this into the force equation:
$$ F_v = \eta \cdot l(l-x) \cdot \frac{v}{h} $$2. Equation of Motion
The sheet is “almost inertia-less” and pulled with a constant force $F$. This implies the applied force is always balanced by the viscous drag force ($F_{applied} = F_v$).
$$ F = \frac{\eta l (l-x)}{h} v $$Substituting $v = \frac{dx}{dt}$:
$$ F = \frac{\eta l (l-x)}{h} \frac{dx}{dt} $$3. Integration
Rearrange the terms to integrate time $t$ and displacement $x$. We need to find the time to pull half the sheet off the table, so limits for $x$ are $0$ to $l/2$.
$$ \frac{F h}{\eta l} dt = (l – x) dx $$ $$ \frac{F h}{\eta l} \int_0^t dt = \int_0^{l/2} (l – x) dx $$ $$ \frac{F h}{\eta l} [t] = \left[ lx – \frac{x^2}{2} \right]_0^{l/2} $$Evaluating the limits:
$$ \frac{F h}{\eta l} t = \left( l(\frac{l}{2}) – \frac{(l/2)^2}{2} \right) – 0 $$ $$ \frac{F h}{\eta l} t = \frac{l^2}{2} – \frac{l^2}{8} = \frac{3l^2}{8} $$4. Solving for Time
$$ t = \frac{3l^2}{8} \times \frac{\eta l}{F h} = \frac{3 \eta l^3}{8 F h} $$Now, substitute the given values:
- $\eta = 0.2 \, \text{N}\cdot\text{s}/\text{m}^2$
- $l = 1.0 \, \text{m}$
- $F = 15 \, \text{N}$
- $h = 1.0 \, \text{mm} = 10^{-3} \, \text{m}$