Determine the stretch in springs AC and AB

Determine the stretch in springs AC and AB for equilibrium of the 2-kg block. The springs are shown in the equilibrium position.

Determine the stretch in springs AC and AB

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


Show me the final answers↓

Let us first draw the free body diagram at ring A.

Determine the stretch in springs AC and AB

The angles were calculated using the inverse of tan (arctan).


The brown angle was found by:


The orange angle was found by:



Let us write our equations of equilibrium.

\rightarrow ^+\sum \text{F}_\text{x}\,=\,0

F_{AB}\text{cos}\,(36.87^0)\,-\,F_{AC}\text{cos}\,(45^0)\,=\,0 (eq.1)


+\uparrow \sum \text{F}_\text{y}\,=\,0

F_{AB}\text{sin}\,(36.87^0)\,+\,F_{AC}\text{sin}\,(45^0)\,-\,19.62\,=\,0 (eq.2)
Show me the free body diagram


Let us now solve for F_{AB} and F_{AC}.


Isolate for F_{AB} in eq.1.



F_{AB}\,=\,0.884F_{AC} (eq.3)


Substitute this value into eq.2.


(Simplify and solve for F_{AC})

F_{AC}\,=\,15.85 N


Substitute this value into eq.3 to find F_{AB}


F_{AB}\,=\,14.0 N


Now that we have the force in each spring, we can use Hook’s Law to figure out the stretch.

Hook’s Law states:


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

For spring AB:



s\,=\,0.47 m


For spring AC:



s\,=\,0.79 m


Final Answers:

Stretch of spring AB = 0.47 m

Stretch of spring AC = 0.79 m


This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-18.

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