The spring has a stiffness of k = 800 N/m

The spring has a stiffness of k = 800 N/m and an unstretched length of 200 mm. Determine the force in cables BC and BD when the spring is held in the position shown.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone.
Statics: Mechanics for Engineers.
Singapore: Pearson, 2013.

Solution:

We will first calculate the force of the spring.

From the diagram, we can see that the length of the spring is 500 mm.
We also know that the unstretched length of the spring is 200 mm.
Therefore, the spring is currently stretched by 300 mm (500 – 200 = 300 mm).

The force of the spring can be calculated using Hook’s Law.

F = ks

(Where F is force, k is the stiffness of the spring, and s is the stretch of the spring)

F = (800)(0.3)

(Note that we converted 300 mm to 0.3 m)

F = 240 N

Let us now draw our free body diagram.