Question

A square lead slab of side $50\,cm$ and thickness $10\,cm$ is subjected to a shearing force (on its narrow face) of $9 \times {10^4}\,N$ . The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear modulus of lead $= 5.6 \times {10^{ - 9}}\,N{m^{ - 2}}$ )
A. $0.16\,mm$
B. $1.6\,mm$
C. $0.16\,cm$
D.$1.6\,cm$

Answer 1

There are two narrow sides of the square slab. One side is free and the other side is attached to the floor such that it does not have any effect on the applied force. The force is applied parallel to the narrow side of the slab so that it is displaced by a small length. The narrow face or side of the slab consists of its length and thickness. Therefore, the value of breadth is not needed here.

Complete step by step solution:
Whenever an object is subjected to an external force, it experiences deformation i.e. change in its dimensions. The force can be applied in a parallel direction or perpendicular direction to the plane of the object. When the force is applied parallel to the plane of the object, it gets deformed in that direction and such a force that causes the stress is called shear stress. Here the force is parallel to the cross-sectional area of the object. The SI unit of shear stress is $N/{m^2}$ . It can be defined as the ratio of force to the area of the cross-section.
Shear stress,
$\tau = \dfrac{F}{A}$
Shear Strain is defined as the ratio of the change in the dimension of an object due to stress to its original dimension. It is a dimensionless quantity.
Shear strain,
$\gamma = \dfrac{{\Delta l}}{l}$
The ratio of shear stress to shear strain is known as shear modulus. It tells about the rigidity of the material. It's S.I. unit is Pascal. If a material has a large shear modulus, then a large force is required to deform that material and if a material has a small shear modulus, then a little force is required to deform that object. Mathematically it is represented as:
$\eta = \dfrac{{\left( {F/A} \right)}}{{\left( {\Delta l/l} \right)}}$
Where $\eta =$ shear modulus
$F =$ the force applied in a parallel direction
$A =$ area of cross-section
$l =$ length
$\Delta l =$ change in length
The force is applied parallel to the narrow side of the slab having dimensions:
Length,
$\,l = 50cm$
$\Rightarrow l = \dfrac{{50}}{{100}}m\,\left( {\because \,1cm = \dfrac{1}{{100}}m} \right)$
$\Rightarrow l = 0.5m$
Thickness,
$t = 10cm$
$\Rightarrow t = \dfrac{{10}}{{100}}m$
$\Rightarrow t = 0.1m$
Area of the narrow side,
$A = l \times t$
$\Rightarrow A = 0.5m \times 0.1m$
$\Rightarrow A = 0.05{m^2}$
The shearing force,
$F = 9 \times {10^4}N$
Shear modulus of lead, $\eta$ $= 5.6 \times {10^{ - 9}}\,N{m^{ - 2}}$
Let the displacement caused by the tangential shearing force be $\Delta l$ . The formula of shear modulus is:
$\eta = \dfrac{{\left( {F/A} \right)}}{{\left( {\Delta l/l} \right)}}$
$\Rightarrow \Delta l = \dfrac{{Fl}}{{\eta A}}$
$\Rightarrow \Delta l = \dfrac{{9 \times {{10}^4} \times 0.5}}{{5.6 \times {{10}^9} \times 0.05}}$
$\Rightarrow \Delta l = \dfrac{{4.5 \times {{10}^4}}}{{0.28 \times {{10}^9}}}$
$\Rightarrow \Delta l = 16.07 \times {10^{ - 5}}m$
$\Rightarrow \Delta l = 16.07 \times {10^{ - 5}} \times 1000\,mm\,\left( {\because 1m = 1000mm} \right)$
$\Rightarrow \Delta l = 16.07 \times {10^{ - 2}}\,mm$
$\Rightarrow \Delta l \approx 0.16\,mm$
Therefore, option A is the correct answer.

Note:
Fluids also experience the shear stress when it flows in between solid boundaries. Fluids have different levels and each level flows at a different speed due to shear stress. This is also the reason for erosion on river beds. Real-life examples of shear stress are cutting vegetables, chewing food, painting, etc.