Question

A plank 2.00 cm thick and 13.0 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 68.4 kg is forced to stand on the very end. If the end of the board drops by 5.20 cm because of the man's weight, find the shear modulus of the wood.

Answer 1

Answer:

9.93 MPa

Explanation:

Given:

 - mass of the man = 68.4 kg

 - Deflection dx = 5.2 cm

 - thickness of plank t = 2.0 cm

 - width of plank w = 13.0 cm

 - Length subtended L = 2.0 m

Find:

Shear Modulus of Elasticity S :

                 S = shear stress / shear strain

                        Shear stress = F / A

                        Shear stress = 68.4*9.81 / 0.02*0.13

                        Shear stress = 258078.4615 Pa

                        Shear strain = dx / L

                        Shear Strain = 0.052 / 2

                        Shear Strain = 0.026  

Hence,

                  S = 258078.4615 / 0.026

                  S = 9.93 MPa