JEE Challenger
More from Laws of Motion
PhyMedAdv 2014 • Paper 2

Matching Friction Expression on Inclined Plane for Two Contact Blocks

Two blocks with masses m1=1 kgm_1 = 1\text{ kg} and m2=2 kgm_2 = 2\text{ kg} are placed together on an inclined plane with inclination angle θ\theta, as shown in the diagram. List I lists different values of θ\theta. The friction coefficient between block m1m_1 and the plane is zero, whereas both static and dynamic friction coefficients between block m2m_2 and the plane equal μ=0.3\mu = 0.3. List II contains expressions representing the friction force acting on block m2m_2. Match each inclination angle from List I with the corresponding friction expression in List II, and select the correct choice. Let gg represent the acceleration due to gravity.

[Useful information: tan⁡(5.5∘)≈0.1;tan⁡(11.5∘)≈0.2;tan⁡(16.5∘)≈0.3\tan(5.5^\circ) \approx 0.1; \tan(11.5^\circ) \approx 0.2; \tan(16.5^\circ) \approx 0.3]

List IList IIP. θ=5∘1. m2gsin⁡θQ. θ=10∘2. (m1+m2)gsin⁡θR. θ=15∘3. μm2gcos⁡θS. θ=20∘4. μ(m1+m2)gcos⁡θ\begin{array}{ll} \text{List I} & \text{List II} \\ \text{P. } \theta = 5^\circ & \text{1. } m_2 g \sin\theta \\ \text{Q. } \theta = 10^\circ & \text{2. } (m_1 + m_2)g \sin\theta \\ \text{R. } \theta = 15^\circ & \text{3. } \mu m_2 g \cos\theta \\ \text{S. } \theta = 20^\circ & \text{4. } \mu (m_1 + m_2)g \cos\theta \end{array}
Question Diagram 1

Options

A

P-1, Q-1, R-1, S-3

B

P-2, Q-2, R-2, S-3

C

P-2, Q-2, R-2, S-4

D

P-2, Q-2, R-3, S-3

Correct

Topics & Concepts

Laws of MotionFrictionNewton's Laws

Step-by-Step Solution

To determine the friction force acting on block m2m_2 for different angles of inclination θ\theta, let us analyze the force equilibrium of the two-block system placed on the inclined plane.

1. Analysis of the System and Critical Angle Calculation

Block m1m_1 (mass m1=1 kgm_1 = 1\text{ kg}) is frictionless (μ1=0\mu_1 = 0), so it tends to slide down the incline under its own component of weight, pushing block m2m_2 (mass m2=2 kgm_2 = 2\text{ kg}) from behind. Block m2m_2 experiences a friction force from the inclined surface, where the coefficient of friction is μ=0.3\mu = 0.3.

The total downward gravitational force parallel to the incline for both blocks is: Fdown=(m1+m2)gsin⁡θF_{\text{down}} = (m_1 + m_2)g \sin\theta

The normal force on block m2m_2 from the inclined plane is: N2=m2gcos⁡θN_2 = m_2 g \cos\theta

Thus, the maximum static friction force that can act on block m2m_2 is: fmax=μN2=μm2gcos⁡θf_{\text{max}} = \mu N_2 = \mu m_2 g \cos\theta

For the two blocks to remain in static equilibrium (no slipping): Fdown≤fmaxF_{\text{down}} \le f_{\text{max}} (m1+m2)gsin⁡θ≤μm2gcos⁡θ(m_1 + m_2)g \sin\theta \le \mu m_2 g \cos\theta

Dividing both sides by (m1+m2)gcos⁡θ(m_1 + m_2)g \cos\theta, we get the condition for no slipping: tan⁡θ≤μm2m1+m2\tan\theta \le \frac{\mu m_2}{m_1 + m_2}

Substitute the given numerical values m1=1 kgm_1 = 1\text{ kg}, m2=2 kgm_2 = 2\text{ kg}, and μ=0.3\mu = 0.3: tan⁡θc=0.3×21+2=0.63=0.2\tan\theta_c = \frac{0.3 \times 2}{1 + 2} = \frac{0.6}{3} = 0.2

From the useful information provided, tan⁡(11.5∘)≈0.2\tan(11.5^\circ) \approx 0.2. Therefore, the critical angle of inclination is: θc≈11.5∘\theta_c \approx 11.5^\circ

2. Evaluation for Different Inclination Angles

  • Case P: θ=5∘\theta = 5^\circ Since θ=5∘<θc≈11.5∘\theta = 5^\circ < \theta_c \approx 11.5^\circ, the system is in static equilibrium and does not slip. The static friction force balances the total force acting down the incline: f=(m1+m2)gsin⁡θf = (m_1 + m_2)g \sin\theta This corresponds to List II - 2.

  • Case Q: θ=10∘\theta = 10^\circ Since θ=10∘<θc≈11.5∘\theta = 10^\circ < \theta_c \approx 11.5^\circ, the system remains in static equilibrium. The static friction force is: f=(m1+m2)gsin⁡θf = (m_1 + m_2)g \sin\theta This corresponds to List II - 2.

  • Case R: θ=15∘\theta = 15^\circ Since θ=15∘>θc≈11.5∘\theta = 15^\circ > \theta_c \approx 11.5^\circ, the blocks overcome static friction and slide down the incline. The friction force is kinetic friction: f=μm2gcos⁡θf = \mu m_2 g \cos\theta This corresponds to List II - 3.

  • Case S: θ=20∘\theta = 20^\circ Since θ=20∘>θc≈11.5∘\theta = 20^\circ > \theta_c \approx 11.5^\circ, the blocks are sliding down the incline. The friction force is kinetic friction: f=μm2gcos⁡θf = \mu m_2 g \cos\theta This corresponds to List II - 3.

Conclusion

Matching List I with List II:

  • P →\rightarrow 2
  • Q →\rightarrow 2
  • R →\rightarrow 3
  • S →\rightarrow 3

This matches the combination given in Option D.

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