InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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» 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`. |
|