112k views
1 vote
A copper wire is 1.0 meters long and it's diameter is 1.0 millimeters if the wire hangs vertically how much weight must be added to its free end in order to stretch it 3.0 millimeters?

2 Answers

1 vote

Final answer:

To stretch the given copper wire 3.0 mm, a weight of approximately 3.467 Newtons, or a mass of about 0.353 kilograms, must be added to its free end.

Step-by-step explanation:

Calculating the Weight Needed to Stretch a Copper Wire:

To determine the weight needed to stretch the copper wire 3.0 mm, we need to apply the formula for the elastic deformation of materials, which relates the force applied, the cross-sectional area, Young's modulus, the initial length, and the elongation of the material:

F = (Y × A × ΔL) / L0

Where:

  • F is the force (weight) needed,
  • Y is Young's modulus for copper,
  • A is the cross-sectional area of the wire,
  • ΔL is the change in length (elongation), and
  • L0 is the original length of the wire.

Young's modulus for copper (Y) is typically about 1.17 × 1011 N/m2. The cross-sectional area (A) can be calculated using the diameter (d):

A = π(d/2)2

Here, d = 1.0 mm = 1.0 × 10-3 meters. Therefore:

A = π(1.0 × 10-3m / 2)2 = 7.854 × 10-7 m2

With the elongation (ΔL) of 3.0 mm = 3.0 × 10-3 meters and the original length (L0) of 1.0 m, the force required to stretch the wire 3.0 mm can be calculated:

F = (1.17 × 1011 N/m2 × 7.854 × 10-7 m2 × 3.0 × 10-3 m) / 1.0 m = 3.467 Newtons

To find the weight in more familiar units, remember that weight (W) is given by W = mg, where g is the acceleration due to gravity (9.81 m/s2). Thus:

W ≈ 3.467 N * (1 kg · m/s2/N) * (1/9.81 m/s2) = 0.353 kg

This value represents the mass of the weight. To get the force (or weight in newtons), we simply use F = mg:

Weight needed = 0.353 kg * 9.81 m/s2 ≈ 3.467 Newtons

User Zardosht
by
8.4k points
5 votes

Answer:

The weight required to stretch a wire can be calculated using the formula:

W = (π/4) x d^2 x L x S

where:

d = diameter of the wire

L = length of the wire

S = stress applied to the wire

π = pi (3.14)

In this case, the wire has a diameter of 1.0 millimeters and a length of 1.0 meters. We want to stretch it by 3.0 millimeters, which means the stress applied to the wire is:

S = ΔL / L = 3.0 / 1000 = 0.003

Now we can plug in the values and calculate the weight required:

W = (π/4) x (1.0)^2 x 1.0 x 0.003

W = 0.0002355 kg or 0.2355 grams

Therefore, we need to add 0.2355 grams of weight to the free end of the wire to stretch it by 3.0 millimeters.

User Kieren
by
8.0k points