Question

In a large hall, a pendulum is to be made by suspending a 40-kg ball from the end of a steel wire 15 m long. Tensile strength of steel is 0.48 x 10^11 Pa, and Y is 200 x 10^9 Pa.

a) What cross-sectional area should the wire have if the applied stress in it is to be only 10% of its breaking stress? [8.1 x 10^-6 m²]
b) How far will the ball stretch? [3.6 mm]

Answer 1

The cross-sectional area of the wire should be approximately 6.47 × 10^-6 m². The elongation of the wire is 4.5 mm.

Step-by-step explanation:

To calculate the cross-sectional area of the wire, we can use the formula for stress: stress = force / area. Rearranging the formula, we have area = force / stress. The breaking stress of the steel wire is given as 6.00 × 10^6 N/m² and the applied stress is to be 10% of the breaking stress, so the applied stress is 0.10 * 6.00 × 10^6 N/m² = 6.00 × 10^5 N/m².

Substituting the values into the formula, we get area = 40 kg * 9.8 m/s² / (6.00 × 10^5 N/m²) = 6.47 × 10^-6 m².

Therefore, the cross-sectional area of the wire should be approximately 6.47 × 10^-6 m².

To calculate the elongation of the wire, we can use Hooke's law: stress = Young's modulus * strain. Rearranging the formula, we have strain = stress / Young's modulus. Substituting the values, we get strain = 6.00 × 10^5 N/m² / (200 × 10^9 N/m²) = 3.00 × 10^-6.

Finally, the elongation of the wire can be calculated using the formula: elongation = original length * strain. Substituting the values, we get elongation = 15 m * 3.00 × 10^-6 = 0.045 m = 4.5 mm.