Animated Solution for Physics - Laws of Motion: A block of mass $m$ is placed on a surface with a vertical cross-section given by $y=\frac{x^3}{6}$. If the coefficient of friction is $0.5$, the maximum height above the ground at which the block can be placed without slipping is

Imagine a block resting on a curved hill. The shape of this hill is given by the mathematical function $y=\frac{x^3}{6}$. We need to find the highest point where the block can sit without sliding down. This problem is a beautiful blend of classical mechanics and differential calculus.

Let's draw the free body diagram of the block at an arbitrary point on the curve. Gravity pulls the block straight down with a force $mg$. The surface pushes back with a normal force $N$, which is perpendicular to the tangent of the curve at that point. Finally, static friction $f$ acts upwards along the tangent to prevent the block from slipping.

At the maximum height, the block is on the verge of slipping. This is the state of limiting equilibrium. Here, the downward pull along the incline, $mg\sin \theta$, is exactly balanced by the maximum static friction, $\mu N$.

Dividing the force equations gives us the classic result for the angle of repose:

Now, how do we connect this physics concept to our mathematical curve? From calculus, we know that the slope of the tangent line, $\tan \theta$, is exactly equal to the derivative of the curve, $\frac{dy}{dx}$.

Let's differentiate our curve equation. The derivative of $\frac{x^3}{6}$ is $\frac{3x^2}{6}$, which simplifies to $\frac{x^2}{2}$. So, our slope $\tan \theta$ is $\frac{x^2}{2}$.

We established earlier that $\tan \theta$ equals $\mu$. So, we equate $\frac{x^2}{2}$ to the given coefficient of friction, $0.5$.

Solving this, we get $x^2=1$, which means $x=\pm 1$. Since we are looking for the height on the positive side of the curve (or due to symmetry, either side gives the same height), we can take the positive $x$-coordinate, $x=1$.

Finally, to find the maximum height $y$, we substitute $x=1$ back into our original curve equation.

And there we have it! The maximum height above the ground at which the block can be placed without slipping is $\frac{1}{6}\text{ m}$.