A rectangular plate 12 m by 0.4 m weighing 970 Newton slides down at a 45-degree incline at a uniform velocity of 2.25 m per second. If the 2 mm gap between the plate and the incli... A rectangular plate 12 m by 0.4 m weighing 970 Newton slides down at a 45-degree incline at a uniform velocity of 2.25 m per second. If the 2 mm gap between the plate and the inclined surface is filled with oil, calculate the viscosity.
Understand the Problem
The question is asking to calculate the viscosity of oil given a rectangular plate sliding down an inclined surface at a uniform velocity under the influence of gravity. We need to apply principles from fluid dynamics and use the dimensions provided to determine the viscosity.
Answer
$$ \eta = \frac{\rho \cdot (L \cdot W \cdot H) \cdot g \cdot \sin(\theta) \cdot d}{L \cdot W \cdot v} $$
Answer for screen readers
$$ \eta = \frac{\rho \cdot (L \cdot W \cdot H) \cdot g \cdot \sin(\theta) \cdot d}{L \cdot W \cdot v} $$
Steps to Solve
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Identify the given values
List the dimensions of the rectangular plate, the angle of inclination, and any other necessary constants, such as the gravitational acceleration. For example, let’s assume:
- Length of the plate, $L$
- Width of the plate, $W$
- Height of the plate, $H$
- Angle of inclination, $\theta$
- Gravitational acceleration, $g$
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Calculate the gravitational force acting on the plate
The weight of the plate $W_{plate}$ is calculated using:
$$ W_{plate} = \text{Density} \times \text{Volume} \times g = \rho \cdot (L \cdot W \cdot H) \cdot g $$ -
Determine the forces acting along the incline
The component of the gravitational force pulling the plate down the inclination is:
$$ F_{gravity} = W_{plate} \cdot \sin(\theta) $$ -
Calculate the viscous force opposing the motion
The viscous force $F_{viscosity}$ is given by:
$$ F_{viscosity} = \eta \cdot \frac{A \cdot v}{d} $$
Where:
- $\eta$ is the viscosity
- $A$ is the area of the plate in contact with the fluid (which can be $L \cdot W$)
- $v$ is the velocity of the plate
- $d$ is the thickness of the fluid layer
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Set the forces equal for uniform velocity
Since the plate moves with uniform velocity, we can set the gravitational force equal to the viscous force:
$$ F_{gravity} = F_{viscosity} $$
$$ W_{plate} \cdot \sin(\theta) = \eta \cdot \frac{A \cdot v}{d} $$ -
Solve for viscosity $\eta$
Now, rearranging the equation, we can express viscosity as:
$$ \eta = \frac{W_{plate} \cdot \sin(\theta) \cdot d}{A \cdot v} $$
Substituting in the previously calculated $W_{plate}$ will give the formula for viscosity.
$$ \eta = \frac{\rho \cdot (L \cdot W \cdot H) \cdot g \cdot \sin(\theta) \cdot d}{L \cdot W \cdot v} $$
More Information
The formula derived for viscosity is critical in fluid dynamics and helps understand how fluids flow under different forces. The units of viscosity are typically in Pascal-seconds (Pa·s) in the SI system.
Tips
- Forgetting to convert units: Ensure all measurements are in the same units before calculations.
- Confusing forces: Remember that for uniform motion, the net force must equal zero; thus, forces must balance.
- Improper area calculations: Make sure to use the correct area ($A$) for the fluid–plate interaction.
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