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`)

Answer

0.16 mm

Explanation

Here `A = 50cmxx 10cm = 500 cm^2`
` = 500 xx 10^(-4) m^2 = 0.05 m^2 ,`
` L = 50 cm = 0.50 m, Delta L = ? ,`
`G = 5.6 xx10^9 N m^(-2), F = 9.0 xx 10^4 N`
As `G = (F)/(A)xx(L)/(Delta L) or Delta L = (F)/(A)xx(L)/(G)`
` = ((9.0xx10^4)xx0.50)/(0.05xx5.6 xx 10^9) = 1.6 xx 10^(-4)m` =0.16 mm