# The unstretched length of spring AB is 3 m 2

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. 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 determine the stretched length of spring AB. We can use Pythagorean theorem to figure this out.

Length of spring = $\sqrt{4^2+3^2}$

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.

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)

$F\,=\,(30)(2)$

$F\,=\,60$ N

Let us now draw our free body diagram. 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:

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

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)

$m\,=\,\dfrac{W}{g}$
$m\,=\,\dfrac{84}{9.81}$

$m\,=\,8.56$ kg

mass = 8.56 kg

## 2 thoughts on “The unstretched length of spring AB is 3 m”

• Azma

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