Floating beam analogy – maximum shear force in a floating rectangular log with a central load A rectangular wooden log is floating in still water and a 100 N load is applied at its midspan (center). Treating buoyant support as continuous, what is the maximum shear force developed in the log?

Introduction / Context:A floating log subjected to a concentrated load behaves, in a first approximation, like a simply supported beam with equal reactions provided by buoyancy on either side of the load. Recognizing the reaction distribution allows quick identification of shear force magnitudes.

Given Data / Assumptions:

• Applied point load W = 100 N at the midspan.
• Hydrostatic buoyancy adjusts to provide static equilibrium.
• Neglect wave action and dynamic effects; treat as quasi-static.

Concept / Approach:For a simply supported beam with a central point load W, the end reactions are equal: R_A = R_B = W/2. The maximum positive/negative shear magnitude equals the reaction magnitude and occurs just to the left/right of the midspan or at the supports in the shear force diagram.

Step-by-Step Solution:W = 100 N at midspan → reactions R_A = R_B = W/2 = 50 N.Shear diagram jumps from +50 N near the left support to −50 N near the right support at the load location.Maximum absolute shear magnitude = 50 N at each support region.

Verification / Alternative check:Equilibrium: ΣV = R_A + R_B − W = 50 + 50 − 100 = 0; ΣM about any support confirms R = W/2.

Why Other Options Are Wrong:

• (b) Shear at the centre is zero immediately under a central point load for an idealized SFD, not 50 N.
• (c) 100 N at centre violates equilibrium distribution.
• (d) Not applicable; a correct option exists.

Common Pitfalls:Confusing support shear with bending moment; mixing total load with reaction magnitudes.

Final Answer:50 N at each end