A tow truck is pulling a car out of a ditch by means of a steel cable that is 10.2 m long and has a radius of 0.50 cm. When the car just begins to move, the tension in the cable is 870 N. How much has the cable stretched?
COnstant for Steel cable Y=2.0e11
Please mention the units when there is a mix of units. (10.2 m, 0.5 cm, 2.0e11 Pascal, etc.)
Also, note that cables do not have a solid cross section. Lack of better information, we will assume the section to be solid. The actual deformation will be higher than calculated.
Modulus of elasticity, E = 2.0e11 pascals
tension, T=870 N
length, L=10.2 m
radius of cable, r = 0.005 m.
Area of cable, A = πr²
deformation, ΔL
=(stress/Mod.Elasticity)*L
=(T/A / E )*L m
To calculate the amount of stretch in the steel cable, we need to use Hooke's Law, which states that the amount of stretch or deformation in a material is directly proportional to the force applied to it.
Hooke's Law can be expressed as:
ΔL = (F * L) / (Y * A)
Where:
- ΔL is the change in length of the cable
- F is the force applied to the cable (tension)
- L is the original length of the cable
- Y is the Young's modulus (constant for the steel cable)
- A is the cross-sectional area of the cable
In this case, we are given:
- F = 870 N (tension in the cable)
- L = 10.2 m (original length of the cable)
- A = π * r^2 (cross-sectional area of the cable, where r is the radius)
First, we need to calculate the cross-sectional area of the cable:
A = π * (0.005 m)^2
A = 0.00007854 m^2
Now, we can substitute the known values into Hooke's Law to find the change in length:
ΔL = (870 N * 10.2 m) / (2.0e11 N/m^2 * 0.00007854 m^2)
ΔL ≈ 0.558 x 10^-4 m
Therefore, the steel cable has stretched by approximately 0.558 x 10^-4 meters.