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 = 3.51·108 N/m2.
The applied weight is
M g = 675.2 N
The maximum allowable strain is
deltaL/L = 0.0101/51.5 = 1.96*10^-4
The maximum allowable stress is
Ynylon*(strain) = 6.88*10^4 N/m^2
That equals 675.2 N divided by the minimum allowable area. Solve for that area and the string diameter.
To find the minimum diameter of the nylon string, we need to use Hooke's Law which states that the elongation of an elastic material is directly proportional to the applied force.
In this case, Hooke's Law can be written as: F = k * x
Where F is the force applied to stretch the string, k is the spring constant, and x is the elongation of the string. In this case, F is calculated as the weight of the load (68.9 kg) multiplied by the acceleration due to gravity (9.8 m/s^2), which gives us a force of 674.62 N.
To find the spring constant (k), we can use the formula: k = Y * A / L
Where Y is the Young's modulus of the material (given as 3.51 * 10^8 N/m^2), A is the cross-sectional area of the string, and L is the length of the string.
We need to rearrange this formula to solve for A: A = k * L / Y
Since we want to find the minimum diameter of the string, we can assume it is a circular cross-section. The cross-sectional area of a circle is given by the formula A = π * r^2, where r is the radius.
Since we want the minimum diameter, we can assume the string is a cylinder, and the radius (r) is half the diameter (d/2).
Substituting the formula for A into the equation above, we have: π * (d/2)^2 = k * L / Y
Rearranging to solve for d, we get: d = √(4 * k * L / (π * Y))
Plugging in the given values, we have: d = √(4 * 674.62 * 51.5 / (π * 3.51 * 10^8))
Evaluating this expression gives us the minimum diameter of the nylon string, which is approximately 0.022 meters or 2.2 cm.