Question

A tangential force of 1500 N exerted upon the upper surface of a cube of 20 cm edge. Calulate the shear modulus of the cube matarial if the result displacement was 0.1cm

Answer 1

Answer:

The shear modulus of the cube material is  $7.5 \times 10^6 N/m^2$.

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 $A=a \times a=(20 \times 10^(^-^2^))^2=400 \times 10^(^-^4^) m$

shear strain = tan⁡ θ = $\frac {x}{h} = \frac {0.001}{0.2} =0.005$  

shearing stress = $\frac {F}{A} = \frac{1500}{0.04} = 37500 N$

modulus of rigidity η $= \frac{(shearing \ stress)}{(shearing \ strain)}$

$= \frac{37500}{0.005}=7.5 \times 10^6 N/m^2$