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

The Correct Option is A

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:

\[ \frac{1}{4} \, m. \]

Answer 2

The problem asks for the maximum height at which a block can be placed on a parabolic surface without slipping, given the equation of the surface and the coefficient of static friction.

Concept Used:

1. Condition for Static Equilibrium on an Inclined Plane: A block placed on an inclined plane at an angle \( \theta \) with the horizontal will remain at rest without slipping if the angle of inclination is less than or equal to the angle of repose, \( \theta_s \). The condition is given by:

\[ \tan\theta \le \mu_s \]

where \( \mu_s \) is the coefficient of static friction. The maximum angle at which the block can be placed without slipping is when \( \tan\theta = \mu_s \).

2. Slope of a Curve: For a curve given by an equation \( y = f(x) \), the slope of the tangent line at any point \( (x, y) \) is given by its derivative, \( \frac{dy}{dx} \). This slope is also equal to the tangent of the angle of inclination \( \theta \) of the tangent line with the horizontal axis.

\[ \frac{dy}{dx} = \tan\theta \]

Step-by-Step Solution:

Step 1: Identify the given information.

The equation of the vertical cross-section of the surface is:

\[ y = \frac{x^2}{4} \]

The coefficient of friction is \( \mu = 0.5 \).

Step 2: State the condition for the block to be at the point of slipping.

For the block to be at the maximum height without slipping, it must be on the verge of sliding down. At this point, the angle of inclination \( \theta \) of the surface is equal to the angle of repose. The condition is:

\[ \tan\theta = \mu \]

Step 3: Find the slope of the tangent to the curve.

The slope of the tangent at any point \( (x, y) \) on the curve is the derivative of \( y \) with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^2}{4} \right) = \frac{2x}{4} = \frac{x}{2} \]

Step 4: Relate the slope to the angle of inclination and the friction condition.

The slope of the tangent is equal to \( \tan\theta \). Therefore:

\[ \tan\theta = \frac{x}{2} \]

At the point of maximum height, we substitute \( \tan\theta = \mu \):

\[ \frac{x}{2} = \mu \]

Step 5: Solve for the x-coordinate of the position of maximum height.

Substitute the given value \( \mu = 0.5 \):

\[ \frac{x}{2} = 0.5 \] \[ x = 1 \]

Final Computation & Result:

Step 6: Calculate the maximum height (\(h\)) by finding the y-coordinate corresponding to \( x = 1 \).

The height \( h \) is the value of \( y \) at this x-coordinate.

\[ h = y = \frac{x^2}{4} \]

Substitute \( x = 1 \):

\[ h = \frac{(1)^2}{4} = \frac{1}{4} = 0.25 \, \text{m} \]

The maximum height above the ground at which the block can be placed without slipping is 0.25 m.