The springs AB and BC each have a stiffness of 600 N/m and an unstretched length of 1.5 m. Determine the horizontal force F applied to the cord which is attached to the small pulley B so that the displacement of the pulley from the wall is d = 0.75 m.
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 draw a free body diagram.
Note that the force in spring BA is equal to the tension in spring BC because the force is applied at the center and both springs have the same stiffness.
We can use the pythagorean theorm to figure out how far the spring stretches when d = 0.75 m.
spring stretch = 1.677 m
We can now use hooks law to figure out the tension in each of the springs.
(Where T is force, k is the stiffness of the spring, and s is the length of stretch)
T\,=\,(600)(1.677-1.5)T=106.2 N
Next step is to write our equations of equilibrium. We will assume forces going \rightarrow^+ to be positive and \uparrow+ to be positive.
2(T(\dfrac{0.75}{1.667}))-F=0
(Because the tension in both springs is the same, we can multiple T by 2 and achieve the same answer)
(Substitute the value of T we found and solve for F)
2(106.2)(\dfrac{0.75}{1.667})-F=0F\,=\,95.6N