A tangential force equal to the weight of 500kg is applied to the opposite face of a cube of aluminium of side 10cm has its one face fixed. Find the shearing strain produced. Rigidity modulus of aluminium is `2.5 xx 10^(10 ) Nm^(-2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Force
The weight of the object is given as 500 kg. To find the force (F) in Newtons, we use the formula:
\[ F = m \cdot g \]
where \( m \) is the mass (500 kg) and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

\[ F = 500 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 4905 \, \text{N} \]

### Step 2: Calculate the Area of the Cube Face
The side length of the cube (L) is given as 10 cm, which we convert to meters:
\[ L = 10 \, \text{cm} = 0.1 \, \text{m} \]
The area (A) of one face of the cube is given by:
\[ A = L^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Calculate the Shear Stress
Shear stress (\( \tau \)) is defined as the force per unit area:
\[ \tau = \frac{F}{A} \]
Substituting the values we have:
\[ \tau = \frac{4905 \, \text{N}}{0.01 \, \text{m}^2} = 490500 \, \text{N/m}^2 \]

### Step 4: Use the Shear Modulus to Find Shear Strain
The relationship between shear stress (\( \tau \)), shear modulus (G), and shear strain (\( \gamma \)) is given by:
\[ \tau = G \cdot \gamma \]
We can rearrange this to solve for shear strain:
\[ \gamma = \frac{\tau}{G} \]
Substituting the values of shear stress and shear modulus:
\[ G = 2.5 \times 10^{10} \, \text{N/m}^2 \]
\[ \gamma = \frac{490500 \, \text{N/m}^2}{2.5 \times 10^{10} \, \text{N/m}^2} = 1.962 \times 10^{-5} \]

### Final Answer
The shearing strain produced is approximately:
\[ \gamma \approx 1.96 \times 10^{-5} \]