Question

A copper metal cube has each side of length 1 m. The bottom edge of cube is fixed and tangential force 4.2 x 10⁸ N is applied to top surface. Calculate lateral displacement of top surface, if modulus of rigidity of copper is 14 x 10¹⁰ N/m²). (Ans : 3 mm)

Answer 1

we know, separation between two faces of metal cube , l = side length = 1m

Cross sectional area of metal cube , A = (1m)² = 1m²

Force, F = 4.2 × 10^8 N

modulus of rigidity of copper, Y = 14 × 10^10

Now, use formula,
Young's modulus, Y = Fl/A∆l

14 × 10^10 = 4.2 × 10^8 × 1/(1 × ∆l)

∆l = 4.2 × 10^8/14 × 10^10 = 3 × 10^-3 m

So, lateral displacement of top surface, ∆l = 3 × 10^-3 m or 3mm

Answer 2

Answer:

we know, separation between two faces of metal cube , l = side length = 1m

Cross sectional area of metal cube , A = (1m)² = 1m²

Force, F = 4.2 × 10^8 N

modulus of rigidity of copper, Y = 14 × 10^10

Now, use formula,

Young's modulus, Y = Fl/A∆l

14 × 10^10 = 4.2 × 10^8 × 1/(1 × ∆l)

∆l = 4.2 × 10^8/14 × 10^10 = 3 × 10^-3 m

So, lateral displacement of top surface, ∆l = 3 × 10^-3 m or 3mm