# Determine the stretch in each of the two springs

Determine the stretch in each of the two springs required to hold the 20-kg crate in the equilibrium position shown. Each spring has an unstretched length of 2 m and a stiffness of k = 300 N/m.

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

#### Solution:

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We will first write each force in the cables/springs in Cartesian vector form. To do so, we need to figure out the position vectors for each force.

Notice in the diagram that springs OA and OB lie in the x and y axis, respectively. Therefore, we only need to find a position vector for cable OC.

The position vector for OC:

$r_{OC}\,=\,\left\{6i+4j+12k\right\}$ m

A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Now, we calculate the magnitude:

magnitude of $r_{OC}\,=\,\sqrt{(6)^2+(4)^2+(12)^2}\,=\,14$ m

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

Now, we write the unit vector for the position vector.

$u_{AB}\,=\,\left(\dfrac{6}{14}i+\dfrac{4}{14}j+\dfrac{12}{14}k\right)$

The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was $r\,=\,ai+bj+ck$, then unit vector, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$

We can now express the force in the cable OC in Cartesian form:

$F_{OC}\,=\,\left\{0.428F_{OC}i+0.286F_{OC}j+0.857F_{OC}k\right\}$

Now the other forces:

$F_{OA}\,=\,\left\{0i-F_{OA}j+ok\right\}$

$F_{OB}\,=\,\left\{-F_{OB}i+0j+ok\right\}$

$F\,=\,\left\{0i+0j-(20)(9.81)k\right\}\,=\,\left\{0i+0j-196.2k\right\}$

(Force F, is the force exerted on the ring O by the crate, which has a mass of 20 kg. It is applied directly downwards, meaning it only has a z-component)

Now that we have written all forces in Cartesian vector form, we can write our equation of equilibrium. All forces added together must equal zero because the system is in equilibrium.

$\sum \text{F}\,=\,0$

$F_{OC}+F_{OA}+F_{OB}+F\,=\,0$

As all the forces added together equals zero, then each component (x,y, z-components) added individually must also equal zero.

x-components:

$0.429F_{OC}-F_{OB}\,=\,0$

y-components:

$0.286F_{OC}-F_{OA}\,=\,0$

z-components:

$0.857F_{OC}-196.2\,=\,0$

Solving the three equations gives us:

$F_{OC}\,=\,229$ N

$F_{OB}\,=\,98$ N

$F_{OA}\,=\,65.5$ N

We can now find the stretch of the springs using the following equation (Hooke’s Law):

$F=ks$

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

Starting with the force in spring OA:

$F=ks$
$65.5\,=\,300s$

(solve for s)

$s\,=\,0.218$ m

Now, the force in spring OB:

$F=ks$
$98\,=\,300s$

(solve for s)

$s\,=\,0.327$ m