To determine the mass that will lead to failure of the material, we will calculate the maximum load (force) that the steel hook can withstand before reaching its ultimate tensile strength (UTS), and then convert this force into mass.
-
Determine the cross-sectional area (A) of the hook:
The diameter \( d \) of the hook is given as 20 mm. Therefore, the radius \( r \) is: \[ r = \frac{d}{2} = \frac{20 , \text{mm}}{2} = 10 , \text{mm} = 0.01 , \text{m} \]
The cross-sectional area \( A \) of the circular hook can be calculated using the formula: \[ A = \pi r^2 = \pi (0.01 , \text{m})^2 \approx 3.14 \times 10^{-4} , \text{m}^2 \]
-
Calculate the maximum load (F) using the ultimate tensile strength (σ):
The ultimate tensile strength (UTS) is given as 700 MPa, which can be converted to Pascals: \[ \sigma = 700 , \text{MPa} = 700 \times 10^6 , \text{Pa} \]
The maximum load \( F \) that the material can withstand is given by: \[ F = \sigma \cdot A \] Substituting the known values: \[ F = 700 \times 10^6 , \text{Pa} \times 3.14 \times 10^{-4} , \text{m}^2 \] \[ F \approx 219,900 , \text{N} \]
-
Convert the force (F) to mass (m):
To find the mass that corresponds to this load, we use the relation between force and mass. From Newton's second law: \[ F = m \cdot g \] where \( g , (\text{acceleration due to gravity}) \) is approximately \( 9.81 , \text{m/s}^2 \).
Rearranging for mass \( m \): \[ m = \frac{F}{g} = \frac{219,900 , \text{N}}{9.81 , \text{m/s}^2} \approx 22,394 , \text{kg} \]
-
Final result:
Therefore, the mass that will result in the failure of the steel hook is approximately 22,394 kg.