The coefficient of static friction between Teflon and scrambled eggs is about ${}^{\mathbf{0}\mathbf{.}\mathbf{04}}$.What is the smallest angle from the horizontalthat will cause the eggs to slide across the bottom of a Teflon-coated skillet?

Coefficient of static friction,${}^{{\mu }_s=0.04}$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagram, The smallest angle from the horizontal that will cause the eggs to slide across the bottom of a Teflon-coated skillet can be found.

Formula:

$F_{net}=\sum _{}ma$

where,Fis the net force,mis mass andais an acceleration.

Free body Diagram of egg over the skillet:

By using Newton’s 2nd law along vertical direction (along y),

$\sum _{}{F}_y=m{a}_y$

Since, block is not moving upward,${}^{a_y=0}$

$$\begin{gathered} N-F_g\cos \left(\theta \right)=0 \\ N=F_g\cos \left(\theta \right) \end{gathered}$$

Relation between Static frictional force and Normal force is,

$$\begin{gathered} f_s={\mu }_{s}N \\ ={\mu }_s{F}_g\cos \left(\theta \right) \end{gathered}$$ (i)

Now, applying Newton’s 2nd along the x direction,

$\sum _{}{F}_x=m{a}_x$

Since, block is not moving,${}^{a_x=0}$

$$\begin{gathered} f_s-F_g\sin \left(\theta \right)=0 \\ {f}_s=\sin \left(\theta \right) \end{gathered}$$

By using equation (i) in the above equation,

role="math" localid="1654077871119" $$\begin{gathered} {\mu }_s{F}_g\cos \left(\theta \right)=F_g\sin \left(\theta \right) \\ {\mu }_s\cos \left(\theta \right)=\sin \left(\theta \right) \\ \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}={\mu }_s \\ \tan \left(\theta \right)={\mu }_s \\ \theta ={\tan }^{-1}\left({\mu }_s\right) \\ ={\tan }^{-1}\left(0.04\right) \\ =2° \end{gathered}$$

Hence, the smallest angle from the horizontalthat will cause the eggs to slide across the bottom of a Teflon-coated skillet is$2°$