A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force (on its narrow face) of `9xx10^4N`. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear modulus of lead`=5.6xx10^9Nm^-2`)

Explanation

Here, `L-50 cm = 50 xx 10^(-2)m` ,
`G=5.6 xx 10^(9)Pa, F=9.0 xx 10^(4)N`.
area of the face on which force is applied ,
`A=50 xx 10 =500 sq cm =0.05m^(2)`
If `Delta L` is the displacement of the upper edge of the slab due to tangential force F applied, then
`G=(F//A)/(Delta L//L)`
or `Delta L=(FL)/(GA) = (9 xx 10^(4)xx50 xx 10^(-2))/(5.6 xx 10^(9) xx 0.05)`
`=1.6 xx 10^(4)m`.