Question

A uniform rod of length L and density ρ is being pulled along A smooth floor with a horizontal acceleration α what is the magnitude of the stress at the transverse cross-section through the midpoint of the rod? 

(A) (1/3) Lρ α 

(B) Lρ α 

(C) (2/3) Lρ ρ

(D) {(Lρ α) / 2}

Answer 1

The correct option is  (D) {(Lρ α) / 2}.

Explanation:

Mass of the first half of the rod = density × volume 

= ρ ∙ V = ρ ∙ Area × length 

= ρ ∙ A ∙ (L/2) 

Force F acting on the cross section through the midpoint of rod is 

F = mass of first half × acceleration 

F = (L/2) ρA × α 

Stress = (F/A) = {(ρAαL) / (2 ∙ A)} = {(ραL) / 2}