Question

A block of mass m is hanged at an angle $\beta$ with line of greatest slope on an inclined of angle $\alpha$ as shown. Minimum value of $\mu$ for no sliping is, T is tension at minimum $\mu$.

• A. $\mu_{min} = \frac{\sin \alpha \sin \beta}{\cos \alpha}$
• B. $\mu_{min} = \frac{\sin \alpha \cos \beta}{\cos \alpha}$
• C. T = mg sin $\alpha$ cos$\beta$
• D. T=mg cos$\alpha$ sin $\beta$

Answer 1

Answer: (A), (C)

A, C

Answer 2

To solve this problem, we need to analyze the forces acting on the block and apply the conditions for equilibrium (no slipping).

1. Set up a Coordinate System: Let's define the axes:

• x-axis: Along the line of greatest slope, pointing downwards.
• y-axis: Perpendicular to the inclined plane, pointing outwards.
• z-axis: Perpendicular to the line of greatest slope, in the plane of the incline (horizontal on the incline).

2. Identify and Resolve Forces:

• Gravitational Force (mg):

  • Component perpendicular to the incline: mg cos α (along -y direction).
  • Component parallel to the incline (down the slope): mg sin α (along +x direction).

• Normal Force (N):

  • Acts perpendicular to the incline: N (along +y direction).

• Tension (T): The tension T is applied at an angle β with the line of greatest slope.

  • Component along the line of greatest slope (up the slope): T cos β (along -x direction).
  • Component perpendicular to the line of greatest slope (sideways): T sin β (along +z direction).

• Friction Force (f): The friction force will oppose the tendency of motion. For no slipping, the static friction force must be less than or equal to μN. On the verge of slipping, f = μN.

3. Apply Equilibrium Conditions:

• Equilibrium Perpendicular to the Incline (y-axis): Since there is no acceleration perpendicular to the inclined plane: ΣF_y = 0 N - mg cos α = 0 N = mg cos α

• Forces in the Plane of the Incline (x-z plane): The forces tending to cause motion are:

  • Along x-axis: F_x = mg sin α - T cos β (down the incline)
  • Along z-axis: F_z = T sin β (sideways)

  For the block to be on the verge of slipping, the magnitude of the resultant force tending to cause motion must be balanced by the maximum static friction force μN. The magnitude of the resultant force F_resultant is: F_resultant = \sqrt{F_x^2 + F_z^2} F_resultant = \sqrt{(mg \sin \alpha - T \cos \beta)^2 + (T \sin \beta)^2}

  For no slipping, F_resultant \le \mu N. For the minimum value of μ (i.e., on the verge of slipping): F_resultant = \mu_{min} N Substituting N = mg cos α: \mu_{min} = \frac{\sqrt{(mg \sin \alpha - T \cos \beta)^2 + (T \sin \beta)^2}}{mg \cos \alpha}

4. Minimize μ with respect to T:

To find the minimum μ, we need to minimize the numerator with respect to T. Let's analyze the term inside the square root: K = (mg \sin \alpha - T \cos \beta)^2 + (T \sin \beta)^2 Expand the expression: K = (mg \sin \alpha)^2 - 2 T (mg \sin \alpha) \cos \beta + T^2 \cos^2 \beta + T^2 \sin^2 \beta K = (mg \sin \alpha)^2 - 2 T (mg \sin \alpha) \cos \beta + T^2 (\cos^2 \beta + \sin^2 \beta) Since cos^2 β + sin^2 β = 1: K = T^2 - (2 mg \sin \alpha \cos \beta) T + (mg \sin \alpha)^2

This is a quadratic equation in T of the form aT^2 + bT + c, where a=1, b = -2 mg sin α cos β, and c = (mg sin α)^2. The minimum value of a quadratic aT^2 + bT + c (with a > 0) occurs at T = -b / (2a). So, the tension T at which μ is minimum is: T = \frac{-(-2 mg \sin \alpha \cos \beta)}{2 \times 1} T = mg \sin \alpha \cos \beta

This matches option (C).

5. Calculate μ_{min}:

Now substitute the value of T back into the expression for K: K_{min} = (mg \sin \alpha \cos \beta)^2 - (2 mg \sin \alpha \cos \beta) (mg \sin \alpha \cos \beta) + (mg \sin \alpha)^2 K_{min} = (mg \sin \alpha)^2 \cos^2 \beta - 2 (mg \sin \alpha)^2 \cos^2 \beta + (mg \sin \alpha)^2 K_{min} = (mg \sin \alpha)^2 (1 - \cos^2 \beta) K_{min} = (mg \sin \alpha)^2 \sin^2 \beta

Now, substitute K_{min} into the expression for μ_{min}: \mu_{min} = \frac{\sqrt{K_{min}}}{mg \cos \alpha} \mu_{min} = \frac{\sqrt{(mg \sin \alpha)^2 \sin^2 \beta}}{mg \cos \alpha} \mu_{min} = \frac{mg \sin \alpha \sin \beta}{mg \cos \alpha} (Assuming sin β >= 0 for the relevant range of β) \mu_{min} = \frac{\sin \alpha \sin \beta}{\cos \alpha}

This matches option (A).

Both options (A) and (C) are correct based on the derivation.