Dimensional analysis—dimensions of force in MLT system What is the correct dimensional formula for force (Newton’s second law) in terms of mass M, length L, and time T?

Introduction / Context:
Dimensional homogeneity is the backbone of fluid mechanics and all of engineering analysis. Identifying the correct dimensions of fundamental quantities helps in validating equations and building non-dimensional groups (e.g., Reynolds, Froude, Weber numbers).

Given Data / Assumptions:

• Force F is related to mass m and acceleration a by F = m * a.
• Base dimensions: mass M, length L, and time T.

Concept / Approach:

Acceleration a has dimensions of length per time squared, and mass is simply M. Multiplying these gives the well-known MLT^-2 for force. This is consistent with SI, where the Newton (N) is kg·m/s^2.

Step-by-Step Solution:

Verification / Alternative check:

Energy (work) W = F * L gives dimensions M L^2 T^-2, consistent with Joule, further validating F = M L T^-2.

Why Other Options Are Wrong:

(b), (c), and (d) mismatch exponents; (e) is unnecessary since a correct option exists.

Common Pitfalls:

Dropping negative signs on time exponents; confusing derived units (N) with base dimensions; mixing CGS and SI inconsistently.

Final Answer:

MLT-2