A rectangular plate 12 m by 0.4 m weighing 970 Newton slides down at a 45degree 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 45degree 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

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, $g$

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

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 Pascalseconds (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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