The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.

#### Solution:

Let us first determine theÂ stretched length of spring AB. We can use Pythagorean theorem to figure this out.

Length of spring = 5 m

We know from the question that the unstretched length of the spring is 3 m. That would mean the spring stretched by 2 m (5 m – 3 m = 2m) after the block was hung at ring A.

We can now figure out the force of spring AB using Hook’s Law.

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

F\,=\,(30)(2)F\,=\,60 N

Let us now draw our free body diagram.

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

\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

The next step is to write our equations of equilibrium.

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

60\text{cos}\,(36.87^0)\,-\,F_{AC}\text{cos}\,(45^0)\,=\,0

(solve for F_{AC})

F_{AC}\,=\,67.88 N

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

60\text{sin}\,(36.87^0)\,+\,F_{AC}\text{sin}\,(45^0)\,-\,W\,=\,0

(Substitute the value of F_{AC} we found and solve for W)

W\,=\,84 N

Show me the free body diagram

To figure out the mass, remember that W=mg.

(Where W is weight, *m* is mass, and g is the force of gravity)

m\,=\,\dfrac{84}{9.81}

m\,=\,8.56 kg

#### Final Answer:

mass = 8.56 kg

Hi, I have a question. Why don’t we just calculate the length of AB with square root of 3^2+3^2 ?

Thank you for answering!

So the lengths of the sides are 4m and 3 m. That’s what we have to use to calculate the length of AB. Maybe you misread the dimensions given? Let me know if I am misunderstanding your question.