Question

An inclined plane is bent in such a way that the vertical cross-section is given by $y = x^2 / 4$. If the upper surface is rough with $\mu = 0.5$, the maximum height in cm at which a stationary block will not slip downward is _________ cm.

Hint:

For a curved surface, the condition for slipping is governed by the local tangent. The height is found by identifying where the tangent equals the coefficient of static friction.

Answer 1

Correct Answer: 25

Step 1: A block stops slipping when the slope $\tan \theta \leq \mu$.
Step 2: The slope $\tan \theta = \frac{dy}{dx}$. Given $y = \frac{x^2}{4}$.
Step 3: $\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$.
Step 4: Set $\frac{x}{2} = 0.5 \Rightarrow x = 1$ m.
Step 5: Height $y = \frac{(1)^2}{4} = 0.25$ m.
Step 6: $0.25$ m $= 25$ cm.