# An object with a mass of 4 kg is lying still on a surface and is compressing a horizontal spring by 1/4 m. If the spring's constant is 6 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Jan 7, 2017

${\mu}_{s} \approx 0.04$

#### Explanation:

We will solve this first by setting up a force diagram and writing out sum of forces statements.

Where $\vec{n}$ is the normal force, ${\vec{F}}_{g}$ is the force of gravity, ${\vec{F}}_{s p}$ is the force of the spring, and ${\vec{f}}_{s}$ is the force of static friction. I will define rightward as positive.

Statements of the net force:

$\sum {F}_{x} = {\vec{F}}_{s p} - {\vec{f}}_{s} = m {\vec{a}}_{x}$

$\sum {F}_{y} = \vec{n} - {\vec{F}}_{g} = m {\vec{a}}_{y}$

The object is not moving, and experiences no net acceleration in either $x$- or $y$-direction.

$\sum {F}_{x} = {\vec{F}}_{s p} - {\vec{f}}_{s} = \cancel{m {\vec{a}}_{x}} = \vec{0}$

$\sum {F}_{y} = \vec{n} - {\vec{F}}_{g} = \cancel{m {\vec{a}}_{y}} = \vec{0}$

We can rearrange these statements to find:

${\vec{F}}_{s p} = {\vec{f}}_{s}$

$\vec{n} = {\vec{F}}_{g}$

Where ${\vec{F}}_{g} = m g$

The equation for the frictional force is $\vec{f} = \mu \vec{n}$. Therefore, we have:

${\vec{F}}_{s p} = \mu m g$

The equation for the spring force is given by Hooke's law as ${\vec{F}}_{s p} = k \left(\Delta s\right)$, where $k$ is the spring constant and $\Delta s$ is the displacement from equilibrium (how far the spring is stretched or compressed).

$k \left(\Delta s\right) = \mu m g$

Solving for $\mu$, the coefficient of friction:

$\mu = \frac{k \left(\Delta s\right)}{m g}$

Using our known values, we can calculuate $\mu$:

$\mu = \frac{\left(6 \frac{N}{m}\right) \left(\frac{1}{4} m\right)}{4 k g \cdot 9.8 \frac{m}{s} ^ 2}$

$\mu = 0.038 \ldots \approx 0.04$

Therefore, the minimum coefficient of static friction to keep the object still is $\approx 0.04$.