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Step-by-Step Solution
Step 1: Identify the Relevant Physical Quantities
• Side of the square slab, l = 60\,\text{cm} = 0.60\,\text{m} .
• Thickness of the slab, t = 15\,\text{cm} = 0.15\,\text{m} .
• Shear modulus (denoted here by Y in the provided solution), Y = 25 \times 10^{9}\,\mathrm{N/m^2} .
• Applied shearing force, F = 18.0 \times 10^{4}\,\mathrm{N} .
• Area of cross section on which force is applied, A : since the force is applied on a narrow face measuring 60\,\text{cm} \times 15\,\text{cm},
A = (0.60\,\text{m}) \times (0.15\,\text{m}) = 0.09\,\text{m}^2.
Step 2: Write the Relation for Shear Deformation
When a shearing force is applied, the shear modulus Y (often denoted as G ) relates stress and strain as:
Y = \dfrac{\text{Shear Stress}}{\text{Shear Strain}}
= \dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}
\quad \Longrightarrow \quad
\Delta l = \dfrac{F \times l}{Y \times A}.
Here, \Delta l is the horizontal displacement (or shear) of the upper edge, l is the dimension over which the shear occurs (here, 0.60\,\text{m} ), F is the shearing force, and A is the area over which this force acts.
Step 3: Substitute the Known Values
Using
\Delta l = \dfrac{F \times l}{Y \times A},
we substitute:
F = 18 \times 10^{4}\,\mathrm{N}, \quad
l = 60 \times 10^{-2}\,\mathrm{m}, \quad
Y = 25 \times 10^{9}\,\mathrm{N/m^2}, \quad
A = (60 \times 15 \times 10^{-4})\,\mathrm{m^2} = 0.09\,\mathrm{m^2}.
Step 4: Calculate the Numerical Value
\Delta l
= \dfrac{(18 \times 10^{4}) \times (60 \times 10^{-2})}{
(25 \times 10^{9}) \times (60 \times 15 \times 10^{-4})
}.
First, let us preserve the original figure and formula steps given:
Y = \frac{Fl}{A \Delta l}
\Delta l = \frac{Fl}{YA}
= \frac{18 \times 10^4 \times 60 \times 10^{-2}}{25 \times 10^9 \times 60 \times 15 \times 10^{-4}}
Carrying out the calculation:
Numerator: 18 \times 10^{4} = 1.8 \times 10^{5}, \quad 60 \times 10^{-2} = 0.60.
Thus, numerator = 1.8 \times 10^{5} \times 0.60 = 1.08 \times 10^{5}.
Denominator: 25 \times 10^{9} \times (60 \times 15 \times 10^{-4}).
Inside parentheses: 60 \times 15 = 900, and 900 \times 10^{-4} = 0.09,
so denominator = 25 \times 10^{9} \times 0.09 = 2.25 \times 10^{9}.
Hence, \Delta l = \dfrac{1.08 \times 10^{5}}{2.25 \times 10^{9}} = 0.48 \times 10^{-4}\,\mathrm{m} = 48 \times 10^{-6}\,\mathrm{m}.
Step 5: Express the Displacement in Micrometers
Since 1\,\mu\text{m} = 10^{-6}\,\text{m},
\Delta l = 48 \times 10^{-6}\,\mathrm{m} = 48\,\mu\text{m}.
Answer
The displacement of the upper edge of the slab is 48\,\mu\text{m}.
