A tow truck is pulling a car out of a ditch by means of a steel cable that is 9.3 m long and has a radius of 0.50 cm. When the car just begins to move, the tension in the cable is 1070 N. How much has the cable stretched?

Answer in m

To calculate the amount the cable has stretched, we can use Hooke's law, which states that the change in length (ΔL) of an elastic object is directly proportional to the force (F) applied to it and inversely proportional to its elastic constant (k).

The formula for Hooke's law is: ΔL = (F * L) / (k * A)

Where:
ΔL = change in length
F = force applied
L = original length of the object
k = elastic constant of the material
A = cross-sectional area of the object

In this case, the force applied is the tension in the cable (1070 N), the original length is the length of the cable (9.3 m), and the material of the cable is steel. The elastic constant for steel is approximately 2.0 x 10^11 N/m^2.

To calculate the cross-sectional area of the cable, we can use the formula for the area of a circle: A = π * r^2
where r is the radius of the cable.

Let's plug in the values into the formula:

ΔL = (1070 * 9.3) / (2.0 * 10^11 * π * (0.005)^2)

ΔL = (9941) / (3.14 * 10^-9)

ΔL ≈ 9.99 x 10^-6 meters

Therefore, the cable has stretched by approximately 9.99 x 10^-6 meters.

To find the amount of stretch in the cable, we need to calculate the increase in length when the tension is applied.

The increase in length of the cable can be determined using Hooke's Law, which states that the extension of a spring or elastic material is directly proportional to the applied force.

Hooke's Law can be expressed as:

F = k * ΔL

Where:
F is the applied force (tension in the cable)
k is the spring constant (a measure of the cable's stiffness)
ΔL is the change in length of the cable

In this case, the tension in the cable is given as 1070 N, and we need to find ΔL.

To find the spring constant (k), we can use the formula for the stiffness of a cable:

k = (π * E * r^2) / L

Where:
E is the Young's modulus (a measure of the cable's elasticity)
r is the radius of the cable
L is the original length of the cable

The radius of the cable is given as 0.50 cm, which is equivalent to 0.005 m.
The original length of the cable is given as 9.3 m.

Now we need to find the Young's modulus (E) for the steel cable. Assuming the steel used in the cable has a typical Young's modulus of about 200 GPa (200 × 10^9 N/m^2), we can substitute the given values into the stiffness formula to find k.

k = (π * 200 × 10^9 N/m^2 * (0.005 m)^2) / 9.3 m

Next, we can rearrange Hooke's Law to solve for ΔL:

ΔL = F / k

Substituting the given values:

ΔL = 1070 N / k

Now we can calculate ΔL by substituting the value we obtained for k:

ΔL = 1070 N / ( π * 200 × 10^9 N/m^2 * (0.005 m)^2 / 9.3 m)

Evaluating this expression will give us the amount of stretch in the cable.

Multiply the length L of the cable by the strain, (deltaL)/L

You can get the strain from Hooke's law and Young's modulus of steel, E.

Young's Modulus for steel is
E = 2.0 x10^11 N/m2

Stress = Strain/E

Stress = 1070 N/(cable area)
Cable area = pi*r^2 = 0.7854 cm^2
= 7.854*10^-5 m^2
Stress = 1.362*10^7 N/m^2
Strain = 6.8*10^-5

Cable Length increase = 6.3*10^-4 m