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?

Civil Engineering Hydraulics Difficulty: Easy
Choose an option
  • A
    MLT-2
  • B
    M-1LT2
  • C
    ML-2T
  • D
    M-1L2T-2
  • E
    none of these.

Answer

Correct Answer: MLT-2

Explanation

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:

Acceleration a → L T^-2.Mass m → M.Force F = m a → M * (L T^-2) = M L T^-2.

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

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