What forces must be applied to the ends of the wire in order to stretch it by 1 mm. wire stiffness 100kN \ m?

Given:

k = 100 kN / m = 10 ^ 5 N / m – coefficient of spring stiffness;

dx = 1 millimeter = 0.001 meter – spring tension.

It is required to determine the forces F1 and F2 (Newton) that must be applied to the ends of the spring in order to stretch it by the dx value.

First, let’s find the value of the resultant force required to stretch the spring:

F = dx * k = 0.001 * 10 ^ 5 = 100 Newtons.

Since, according to the condition of the problem, the forces are applied to the ends of the spring and are equal, then:

F1 + F2 = F;

2 * F1 = F;

F1 = F2 = F / 2 = 100/2 = 50 Newtons.

Answer: forces equal to 50 Newtons must be applied to the ends of the spring.