Given:
Side length of the cube, L=5 cm =0.05 m
Modulus of rigidity, η=105 N/m2
Force applied on the upper face, F=10 N
Area of the upper face, A=L2=(0.05)2=2.5×10−3 m2
Shear stress is given by:
Stress=AF=2.5×10−310=4000 N/m2
Shear strain is given by:
Strain=LΔx
Modulus of rigidity is defined as the ratio of shear stress to shear strain:
η=StrainStress=Δx/LF/A
Rearranging for displacement Δx:
Δx=A⋅ηF⋅L
Substituting the values:
Δx=2.5×10−3×10510×0.05
Δx=2500.5=0.002 m
Converting to millimeters:
Δx=0.002×1000=2 mm
Answer: 2