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

#### Solution:

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

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

The orange angle was found by:

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

Let us write our equations of equilibrium.

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}.

(simplify)

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:

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

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

#### Final Answers:

Stretch of spring AB = 0.47 m

Stretch of spring AC = 0.79 m

Thank you so much

I have a question similar to this and i couldn’t seem to get the answer. This helps a bunch

You’re very welcome!