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.
(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.
The angles were calculated using trigonometry, specifically, the inverse of tan (arctan).
The orange angle was found by:
\text{tan}^{-1}\left(\dfrac{300}{400}\right)=36.87^0
We can now write our equations of equilibrium. We will assume forces going \rightarrow^+ to be positive and \uparrow+ to be positive.
F_{BC}\text{cos}\,(45^0)\,+\,F_{BD}\text{cos}\,(36.87^0)\,-\,240\,=\,0 (eq.1)
+\uparrow \sum \text{F}_\text{y}\,=\,0
F_{BC}\text{sin}\,(45^0)\,-\,F_{BD}\text{sin}\,(36.87^0)\,=\,0 (eq.2)
Let us solve for F_{BC} and F_{BD}.
Isolate for F_{BC} in eq.2.
F_{BC}\,=\,\dfrac{F_{BD}\text{sin}\,(36.87^0)}{\text{sin}\,(45^0)} (eq.3)
Substitute this value into eq.1.
\dfrac{F_{BD}\text{sin}\,(36.87^0)}{\text{sin}\,(45^0)}\text{cos}\,(45^0)\,+\,F_{BD}\text{cos}\,(36.87^0)\,-\,240\,=\,0
(solve for F_{BD})
F_{BD}\,=\,171.4 N
Substitute the value of F_{BD} we just found into eq.3 to find F_{BC}
F_{BC}\,=\,145.4 N
how did you come up with the 171.4N as the Fbd?
So you just solve the equation for F_BD. It’s a single equation with 1 variable so it’s doable ?