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
.0101=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 :(
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 :(
physics - drwls, Sunday, October 30, 2011 at 7:34am
The allowed elongation is 0.0101 m, not 0.00101
physics - maria, Sunday, October 30, 2011 at 7:39am
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 can use Hooke's Law, which states that the elongation of an elastic material is directly proportional to the force applied to it.
First, let's rearrange the formula for elongation to solve for the area:
elongation = force/Y * length/area
Given values:
elongation = 1.01 cm = 0.0101 m
force = 68.9 kg * 9.8 m/s^2 = 674.62 N
length = 51.5 m
Ynylon = 3.51 * 10^8 N/m^2
Plugging in the values, we get:
0.0101 = (674.62)/(3.51 * 10^8) * (51.5)/area
Simplifying the equation, we have:
area = (674.62 * 51.5)/(0.0101 * 3.51 * 10^8)
area ≈ 9.95 * 10^-4 m^2
To find the diameter, let's rearrange the formula for the area of a circle:
area = π * (diameter/2)^2
Plugging in the value for the area, we have:
9.95 * 10^-4 = π * (diameter/2)^2
Solving for the diameter, we get:
(diameter/2)^2 = (9.95 * 10^-4)/π
diameter/2 ≈ sqrt((9.95 * 10^-4)/π)
diameter ≈ 2 * sqrt((9.95 * 10^-4)/π)
Using a calculator, the approximate value of the diameter is:
diameter ≈ 0.0631 meters or 6.31 cm
So, the minimum diameter of the nylon string that will stretch no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end is approximately 6.31 cm.