A 3.8 kg object is supported by an aluminum wire of length 1.9 m and diameter 2.2 mm. How much will the wire stretch?

Young's modulus for Al = Y

= 69*10^9 N/m^2

deltaL/L = stress/Y

Solve for delta L

The stress is M*g/(wire area)

To find the amount the wire stretches, we need to use Hooke's law, which states that the amount a spring or wire stretches or compresses is directly proportional to the force applied to it.

In this case, the weight of the object hanging from the wire is acting as the force that causes the wire to stretch. The weight can be calculated using the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.

Substituting the given values, we find:

W = 3.8 kg * 9.8 m/s^2 ≈ 37.24 N

Next, we need to calculate the effective spring constant of the wire (k). The formula for the effective spring constant of a wire is:

k = (π * d^2 * E) / (4 * L)

Where d is the diameter of the wire, E is the Young's modulus of the material (in this case, aluminum), and L is the length of the wire.

The Young's modulus for aluminum is typically around 70 GPa (gigapascals). However, it's important to note that the units in this equation must be consistent. Since the length of the wire is given in meters, we need to convert GPa to N/m^2:

1 GPa = 1 * 10^9 N/m^2

Therefore, E = 70 * 10^9 N/m^2.

Substituting the given values, we find:

k = (π * (2.2 * 10^-3)^2 * (70 * 10^9 N/m^2)) / (4 * 1.9 m)
≈ 1.04 * 10^6 N/m

Finally, we can use Hooke's law to find the amount the wire stretches (ΔL). The equation for Hooke's law is:

F = k * ΔL

Rearranging the equation, we find:

ΔL = F / k

Substituting the calculated values, we find:

ΔL = 37.24 N / (1.04 * 10^6 N/m)
≈ 3.58 * 10^-5 m

Therefore, the wire will stretch approximately 3.58 * 10^-5 meters.