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The space between two large plane surfac...

The space between two large plane surfaces is 2.5 cm and it is filled with glycerine.
What force is required to drag a very thin plate `1 m^2` in area, between the surfaces at a speed of 1m/s ? (The plate is at a distance of 1 cm from one of the surfaces. ) `(eta=0.75Ns//m^2)`

A

100 N

B

75 N

C

50N

D

125 N

Text Solution

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The correct Answer is:
To solve the problem, we need to calculate the force required to drag a thin plate of area \(1 \, m^2\) between two large plane surfaces filled with glycerine. The viscosity of glycerine is given as \( \eta = 0.75 \, Ns/m^2 \), and the distance between the surfaces is \(2.5 \, cm\) (or \(0.025 \, m\)). The plate is positioned \(1 \, cm\) (or \(0.01 \, m\)) from one surface, which means it is \(1.5 \, cm\) (or \(0.015 \, m\)) from the other surface. The speed of the plate is \(1 \, m/s\). ### Step-by-Step Solution: 1. **Identify the parameters:** - Area of the plate, \( A = 1 \, m^2 \) - Viscosity of glycerine, \( \eta = 0.75 \, Ns/m^2 \) - Speed of the plate, \( v = 1 \, m/s \) - Distance from one surface, \( z_1 = 0.01 \, m \) - Distance from the other surface, \( z_2 = 0.025 \, m - 0.01 \, m = 0.015 \, m \) 2. **Calculate the velocity gradient \( \frac{dv}{dz} \) for both surfaces:** - For the upper surface (distance \( z_1 \)): \[ \frac{dv}{dz_1} = \frac{v}{z_1} = \frac{1 \, m/s}{0.01 \, m} = 100 \, s^{-1} \] - For the lower surface (distance \( z_2 \)): \[ \frac{dv}{dz_2} = \frac{v}{z_2} = \frac{1 \, m/s}{0.015 \, m} = 66.67 \, s^{-1} \] 3. **Calculate the viscous force \( F_1 \) acting on the upper surface:** \[ F_1 = -\eta A \left(\frac{dv}{dz_1}\right) = -0.75 \, Ns/m^2 \times 1 \, m^2 \times 100 \, s^{-1} = -75 \, N \] 4. **Calculate the viscous force \( F_2 \) acting on the lower surface:** \[ F_2 = -\eta A \left(\frac{dv}{dz_2}\right) = -0.75 \, Ns/m^2 \times 1 \, m^2 \times 66.67 \, s^{-1} = -50 \, N \] 5. **Calculate the net force \( F \):** \[ F = F_1 + F_2 = -75 \, N - 50 \, N = -125 \, N \] The negative sign indicates that the viscous force acts in the direction opposite to the motion. 6. **Final Result:** The magnitude of the force required to drag the plate is: \[ F = 125 \, N \]
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