Question

A block of mass m is placed on a surface having vertical cross section given by $y=\frac{x^2}{4}$. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

• A. $\frac{1}{4} m$
• B. $\frac{1}{2} m$
• C. $\frac{1}{6} m$
• D. $\frac{1}{3} m$

Answer 1

Answer: (A)

$\frac{1}{4} m$

Answer 2

The block is subject to gravitational force and frictional force as it is placed on an inclined surface described by the equation $y = \frac{x}{2}$.

Identify the Forces: The weight of the block $W = mg = 1 \times 10 = 10 \, N$ acts vertically downwards.
The normal force $N$ acts perpendicular to the surface.

Determine the Angle of Incline: From the equation $y = \frac{x}{2}$, we can find the slope:

$\text{slope} = \frac{dy}{dx} = \frac{1}{2} \implies \tan(\theta) = \frac{1}{2}.$

Therefore, the angle $\theta$ can be calculated as:

$\theta = \tan^{-1} \left( \frac{1}{2} \right).$

Apply the Conditions for No Slipping: The maximum frictional force $F_f$ can be expressed as:

$F_f = \mu N = 0.5N,$

where $\mu$ is the coefficient of friction.

Using the Equilibrium of Forces: The component of the weight acting down the incline is:

$W_{\text{parallel}} = mg \sin(\theta).$

The component of the weight acting perpendicular to the incline is:

$W_{\text{perpendicular}} = mg \cos(\theta).$

Therefore, $N = W_{\text{perpendicular}} = mg \cos(\theta) = 10 \cos(\theta)$.

Setting Up the Equation: For the block to not slip, the maximum frictional force must balance the parallel component of the weight:

$F_f \geq W_{\text{parallel}} \implies 0.5N \geq mg \sin(\theta).$

Substituting Values:

$0.5 \times 10 \cos(\theta) \geq 10 \sin(\theta) \implies 5 \cos(\theta) \geq 10 \sin(\theta).$

Rearranging and Solving for Height: Using the relation $h = y$ at the maximum height, where:

$h = \frac{x}{2}.$

Substitute for $h$:

$5 \times \frac{\sqrt{1}}{\sqrt{1 + \left( \frac{1}{2} \right)^2}} \geq 10 \times \frac{1}{2} \cdot \frac{1}{\sqrt{1 + \left( \frac{1}{2} \right)^2}}.$

Final Calculation: This yields:

$5 \geq 10 \times \frac{1}{2} \implies h = \frac{1}{4} \, m.$

Thus, the maximum height above the ground at which the block can be placed without slipping is:

$\frac{1}{4} \, m.$