The coefficient of static friction between Teflon and scrambled eggs is about \(0.04\). What is the smallest angle from the horizontal that will cause the eggs to slide across the bottom of a Teflon-coated skillet?

The force of static friction can be calculated using the formula \( F_s = \mu_s * F_n \) where \( F_s \) is the force of static friction, \( \mu_s \) is the coefficient of static friction and \( F_n \) is the normal force. However, in this situation, the normal force isn't given directly. When the eggs start to slide, it means that the downward force caused by gravity along the angle equals the force of static friction. As we only need the angle causing this situation, we can ignore the mass of the eggs - it would cancel out on both sides of the equation. So, we can also write \( F_g * sin(\Theta) = \mu_s * F_g * cos(\Theta) \) where \( F_g \) is the force due to gravity, \( \Theta \) is the angle of inclination, and \( sin \) and \( cos \) are trigonometric functions which values depend on the angle \( \Theta \). For the given situation, \( \mu_s \) is 0.04.

Rearranging the formula results in \( tan(\Theta) = \mu_s \) as \( sin(\Theta) / cos(\Theta) = tan(\Theta) \). From the problem, we know that the coefficient of friction \( \mu_s \) is 0.04. So we solve for \( \Theta \) as \( tan^{-1}(0.04) \). The inverse tangent function \( tan^{-1} \) gives us the angle whose tangent is the number inside the parentheses.

The result from step 2 will be in radians. However, the problem asks for the angle in degrees. To convert from radians to degrees, multiply the result by \( 180 / \pi \).