Answer:
The shear modulus of the cube material is [tex]7.5 \times 10^6 N/m^2[/tex].
Explanation:
Given that shearing force applied F = 1500 N
Displacement produced x = 0.1 cm=0.001 m
side of the cube =20 cm = 0.2 m
Since the object is a cube the upper surface is a square and it is on this surface the shearing
force is applied
area of the upper surface [tex]A=a \times a=(20 \times 10^(^-^2^))^2=400 \times 10^(^-^4^) m[/tex]
shear strain = tan θ = [tex]\frac {x}{h} = \frac {0.001}{0.2} =0.005[/tex]
shearing stress = [tex]\frac {F}{A} = \frac{1500}{0.04} = 37500 N[/tex]
modulus of rigidity η [tex]= \frac{(shearing \ stress)}{(shearing \ strain)}[/tex]
[tex]= \frac{37500}{0.005}=7.5 \times 10^6 N/m^2[/tex]
