A metal plate has the dimensions `10cm xx 10cm xx 1mm`. One of its faces having larger area is fixed and a tangential force is applied to the opposite larger face. If the lateral displacement between the two surfaces is `1.2 xx 10^(-3) mm`, and the modulus of rigidity of the metal is `5 xx 10^(10) N//m^(2)`, then the tangential force is

Explanation

Data : A= 0.1 m` xx ` 0.1 m = ` 10^(-2) m^(2)`
y= 1mm =- `10^(-3) m, Deltax= 1.2 xx 10^(-6) m, n= 5 xx 10^(10) N//m^(2)`
Let F be the magnitude of each applied tangential force.
(i) The shear strain ` theta = (Delta x)/y= (1.2 xx 10^(-6))/(10^(-3)) = 1.2 xx 10^(-3)`
(ii) Modulus ofo rigidity
` n = (" shear stress")/("shear strain") = (F//A)/theta`
The tangential force ` F = A eta theta`
` (10^(-2)) ( 5 xx 10^(10)) (1.2 xx 10^(-3))`
` 6 xx 10^(5) N`