Question

A horizontal force of 0.5 N is required to move a metal plate of area \( 10^{-2} \, {m}^2 \) with a velocity of \( 3 \times 10^{-2} \, {m/s} \), when it rests on \( 0.5 \times 10^{-3} \, {m} \) thick layer of glycerin. Find the viscosity of glycerin.

Hint:

Viscosity measures a fluid's resistance to flow and is calculated using the applied force, fluid thickness, area, and velocity.

Answer 1

The viscosity \( \eta \) of a fluid is expressed by the formula: \[ \eta = \frac{F \cdot d}{A \cdot v} \] where \( F = 0.5 \, {N} \) is the applied force, \( d = 0.5 \times 10^{-3} \, {m} \) is the thickness of the fluid layer, \( A = 10^{-2} \, {m}^2 \) is the cross-sectional area, and \( v = 3 \times 10^{-2} \, {m/s} \) is the velocity. Substituting the given values: \[ \eta = \frac{0.5 \times 0.5 \times 10^{-3}}{10^{-2} \times 3 \times 10^{-2}} = 0.833 \, {N} \cdot {s/m}^2 \] \bigskip