# An object with a mass of 10 kg is lying still on a surface and is compressing a horizontal spring by 5/6 m. If the spring's constant is 24 (kg)/s^2, what is the minimum value of the surface's coefficient of static friction?

Mar 30, 2017

mu_(s*"min")=0.2041

#### Explanation:

If an object is lying still then that means the acceleration in any direction is $0 {\text{m/s}}^{2}$. Since $\vec{a} = \frac{\Sigma \vec{F}}{m}$ this means the net force in any direction is $0 \text{N}$. That means the force of static friction has to be equal in magnitude and opposite in direction to the force exerted by the spring because the horizontal net force is $0 \text{N}$.

The force exerted by a spring is given by $\vec{{F}_{s}} = \text{-} k \vec{x}$ where $k$ is the spring constant and $x$ is the displacement from equilibrium. Plugging in the values from the problem, $\vec{{F}_{s}} = \text{-"(24"kg/s"^2)(5/6"m")="-"20"N}$, so the force of static friction $\vec{{F}_{f}} = 20 \text{N}$.

The magnitude of the force of static friction is defined as ${F}_{f} = {\mu}_{s} {F}_{N}$, so ${\mu}_{s \cdot \text{min}} = {F}_{f} / {F}_{N}$.

Since the acceleration in the vertical direction is $0 {\text{m/s}}^{2}$, the net force in the vertical direction is $0 \text{N}$, therefore the normal force is equal in magnitude and opposite in direction to the force of gravity. Since $\vec{{F}_{g}} = \text{-} m g$, $\vec{{F}_{N}} = m g$.

mu_(s*"min")=F_f/(mg)=(20"N")/((10"kg")(9.8"m/s"^2))=0.2041#