# 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.

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

#### Solution:

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

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

The brown angle was found by:

$\text{tan}^{-1}\left(\dfrac{3}{3}\right)\,=\,45^0$

The orange angle was found by:

$\text{tan}^{-1}\left(\dfrac{3}{4}\right)\,=\,36.87^0$

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.

$F_{AB}\,=\,\dfrac{F_{AC}\text{cos}\,(45^0)}{\text{cos}\,(36.87^0)}$

(simplify)

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

Substitute this value into eq.2.

$0.884F_{AC}\text{sin}\,(36.87^0)\,+\,F_{AC}\text{sin}\,(45^0)\,-\,19.62\,=\,0$

(Simplify and solve for $F_{AC}$)

$F_{AC}\,=\,15.85$ N

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

$F_{AB}\,=\,(0.884)(15.85)$

$F_{AB}\,=\,14.0$ 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:

$F\,=\,ks$

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

For spring AB:

$s\,=\,\dfrac{F}{k}$

$s\,=\,\dfrac{14}{30}$

$s\,=\,0.47$ m

For spring AC:

$s\,=\,\dfrac{F}{k}$

$s\,=\,\dfrac{15.85}{20}$

$s\,=\,0.79$ m