A chandelier of mass 250 kg is hung vertically from a ceiling using a 1.25 m long brass rod of square cross-section of sides 0.4 cm each. How much will the rod stretch when the chandelier is hung? By what factor is the minimum breaking load greater than the weight of the chandelier?

You asked that already, using a different name, rk. It has been ansqwered.

Thanks for answering my problem.But I did not ask it using different name.I used the same name "sand"

To find the amount by which the rod will stretch when the chandelier is hung, and by what factor the minimum breaking load is greater than the weight of the chandelier, we need to consider the concepts of stress, strain, and Young's modulus.

1. Finding the rod's stretch:
To calculate the stretch in the rod, we need to determine the stress and strain. Stress (σ) is the force applied per unit cross-sectional area, while strain (ε) is the ratio of deformation to the original length.

The chandelier's weight is given by:
Weight = mass × acceleration due to gravity
Weight = 250 kg × 9.8 m/s^2
Weight = 2450 N

Next, we need to find the cross-sectional area of the rod. Since it has a square cross-section with sides measuring 0.4 cm each, the area (A) is given by:
A = side length × side length
A = 0.004 m × 0.004 m
A = 1.6 × 10^-5 m^2

Now, we can calculate the stress:
Stress (σ) = Weight / Cross-sectional area
Stress (σ) = 2450 N / 1.6 × 10^-5 m^2
Stress (σ) ≈ 1.53125 × 10^8 Pa (Pascals)

Next, we need to find the Young's modulus (E) of brass. For brass, the Young's modulus is typically around 101 × 10^9 Pa.

Finally, we can calculate the strain:
Strain (ε) = Stress / Young's modulus
Strain (ε) ≈ 1.53125 × 10^8 Pa / 101 × 10^9 Pa
Strain (ε) ≈ 0.001516 (~ 1.52 × 10^-3)

The stretch in the rod can be found by multiplying the original length with the strain:
Stretch = Original length × Strain
Stretch = 1.25 m × 1.52 × 10^-3
Stretch ≈ 1.9 mm (millimeters)

Therefore, the rod will stretch by approximately 1.9 mm when the chandelier is hung.

2. Finding the factor by which the minimum breaking load is greater than the weight of the chandelier:
The minimum breaking load is the load at which the rod will break. It is greater than the weight of the chandelier. To find this factor, we need to calculate the breaking load and then compare it to the weight of the chandelier.

Using the stress equation, we can rearrange it to solve for the breaking load (L):
L = Stress × Cross-sectional area

Substituting the values:
L = 1.53125 × 10^8 Pa × 1.6 × 10^-5 m^2
L ≈ 2450 N (same as the weight of the chandelier)

To find the factor by which the breaking load is greater than the weight of the chandelier:
Factor = Breaking load / Weight
Factor = 2450 N / 2450 N
Factor = 1

Therefore, the factor by which the minimum breaking load is greater than the weight of the chandelier is 1.