Q.A metal block of area $0.10\ \text{m}^{2}$ is connected to a $0.010\ \text{kg}$ mass via a string that passes over an ideal pulley (considered massless and frictionless). A liquid with a film thickness of $0.30\ \text{mm}$ is placed between the block and the table. When released the block moves to the right with a constant speed of $0.085\ \text{m s}^{-1}$. Find the coefficient of viscosity of the liquid.

Viscous Force Balance: From Intuition to Precision

Imagine you're pushing a heavy box across a rough floor. The harder you push, the faster it moves — but there's a constant resistance from the floor trying to slow it down. If you push with a steady force, the box eventually moves at a constant speed. At that moment, your pushing force exactly equals the friction force. The box is in force balance.

Now replace the box with a tiny sphere falling through honey. The honey resists the motion — that resistance is a viscous force. As the sphere speeds up, the viscous force grows. Eventually, it becomes large enough to exactly balance the weight pulling the sphere down. The sphere then falls at a constant speed (terminal velocity). That's viscous force balance in action.

The Core Idea

This is just Newton's first law applied to a fluid environment: if the sum of forces is zero, the object moves with constant velocity — it doesn't speed up or slow down.

The Precise Statement

For an object moving through a viscous fluid, the equation of motion is:

where:

• $F_{\text{other}}$ is the sum of all non-viscous forces (gravity, buoyancy, applied forces, etc.)
• $F_{\text{viscous}}$ is the drag force from the fluid

Viscous force balance is the condition:

which gives $\frac{dv}{dt} = 0$, i.e., constant velocity.

The Two Common Forms of Viscous Force

The exact expression for $F_{\text{viscous}}$ depends on the flow regime:

Here $\eta$ is fluid viscosity, $r$ is object radius, $\rho$ is fluid density, $A$ is cross-sectional area, $C_d$ is drag coefficient.

A Concrete Example: The Falling Sphere

Consider a sphere of mass $m$ and radius $r$ falling through a viscous fluid of density $\rho_f$. The forces are:

• Weight downward: $mg$
• Buoyancy upward: $\frac{4}{3}\pi r^3 \rho_f g$
• Viscous drag upward (Stokes' law): $6\pi\eta r v$

The net downward force is:

At balance, $F_{\text{net}} = 0$, so:

Solving for the terminal velocity:

Why This Matters

Viscous force balance is not just a textbook concept — it's the principle behind:

• Sedimentation: particles settling in a liquid (used in water treatment, blood tests) …