Find the minimum diameter of a 51.5-m-long nylon string that will stretch no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end. Assume that Ynylon = 3.51·10^8 N/m2

physics - bobpursley, Saturday, October 29, 2011 at 10:32am
Hookes Law:
elongation= force/Y * length/area
.00101=68.9*9.8/3.51*10^8 * 51.5/area
solve for area, then solve for diameter
I have solved the area and after that I have solved for diameter d=sqrt4*A/3.14 with this formula and I've founded the diameter 0.11 ! But the result is not correct! Please can you help me :(

The allowed elongation is 0.0101 m, not 0.00101

yes I know. I solved with the value of elongation 0.0101 m. and I've found diameter 0.11 ! But it's wrong :( I tried many times and i always found 0.11 diameter.. I don't know where is the problem..

To find the minimum diameter of the nylon string, we need to use Hooke's Law, which relates the force applied to a spring-like material to the resulting elongation. Hooke's Law equation is:

elongation = force / (Young's modulus * length / area)

In this case, the elongation (strain) is 1.01 cm = 0.0101 m, the force is 68.9 kg * 9.8 m/s^2 (convert from kg to N), the length is 51.5 m, and the Young's modulus for nylon is 3.51 * 10^8 N/m^2.

Let's solve the equation for the area of the nylon string:

0.0101 m = (68.9 kg * 9.8 m/s^2) / (3.51 * 10^8 N/m^2) * (51.5 m / area)

Multiply both sides by the area:

0.0101 m * area = (68.9 kg * 9.8 m/s^2) / (3.51 * 10^8 N/m^2) * 51.5 m

Divide both sides by 0.0101 m:

area = [(68.9 kg * 9.8 m/s^2) / (3.51 * 10^8 N/m^2) * 51.5 m] / 0.0101 m

Now we can solve for the area of the nylon string using a calculator. After finding the area, we can use the formula for the diameter of a circle:

diameter = 2 * sqrt(area / π)

Let's plug in the value of the area into this formula, and calculate the diameter.

diameter = 2 * sqrt(area / π)

Finally, we get the minimum diameter of the nylon string.