What happens to the coefficient of friction, when the normal reaction is halved

The correct answer is No change.

To understand why, we must first look at the fundamental definition of the coefficient of friction (μ).

The coefficient of friction is defined by the relationship between the frictional force (f) and the normal reaction force (N):

\mu =\frac{f}{N}

Or equivalently:

f=\mu ·N

Here, μ is not a variable that depends on f or N. It is a constant of proportionality — a material property that depends only on the nature of the two surfaces in contact (their texture, composition, roughness, etc.).

Step-by-step reasoning:

Step 1: Suppose the original normal reaction is N and the original frictional force is f.
Then: \mu =\frac{f}{N}

Step 2: Now, the normal reaction is halved, so the new normal reaction becomes N/2.
Since f = μN, the new frictional force also adjusts proportionally: f_{new}=\mu ·\frac{N}{2}=\frac{f}{2}

Step 3: Now compute the new coefficient of friction: {\mu }_{new}=\frac{f_{new}}{N_{new}}=\frac{f/2}{N/2}=\frac{f}{N}=\mu

Conclusion: The coefficient of friction remains unchanged. Both the frictional force and the normal reaction are halved, so their ratio stays exactly the same. μ is an intrinsic property of the surface pair — it does not depend on how much force is applied or how heavy the object is. Halving (or doubling) the normal reaction only changes the magnitude of friction, never the coefficient itself.