Math Problem Statement

(Figure 1) shows a block of mass m resting on a 20∘ slope. The block has coefficients of friction μs = 0.82 and μk = 0.51 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

A block of mass m rests on a 20° incline with static and kinetic friction coefficients μs = 0.82 and μk = 0.51, respectively. It is connected via a massless string and pulley to a 2 kg hanging block. What is the condition for static equilibrium or motion of the system?

Solution

In this problem, we need to analyze the forces acting on both the block of mass mm on the incline and the hanging block of mass 2 kg. The goal is to determine whether the system is in motion or at rest, depending on the friction forces and the conditions of equilibrium or motion.

Step 1: Analyze the forces on the block on the incline

For the block of mass mm on the incline, the forces are:

  1. Weight (gravity): mgmg, acting vertically downward.
  2. Normal force (NN): Perpendicular to the inclined plane.
  3. Frictional force (FfF_f): Opposing the motion, either static or kinetic.
  4. Tension in the string (TT): Acting up the incline.

We can break the weight into components:

  • Parallel to the incline: mgsin(20)mg \sin(20^\circ)
  • Perpendicular to the incline: mgcos(20)mg \cos(20^\circ)

The frictional force is given by Ff=μNF_f = \mu N, where N=mgcos(20)N = mg \cos(20^\circ).

Step 2: Analyze the forces on the hanging block (mass = 2 kg)

For the hanging block:

  • The downward force due to gravity is 2g2g, where g9.8m/s2g \approx 9.8 \, \text{m/s}^2.
  • The tension in the string TT acts upward.

Step 3: Set up equations

  • For the block on the incline, the net force along the incline is: TFfmgsin(20)=maT - F_f - mg \sin(20^\circ) = ma where Ff=μsN=μsmgcos(20)F_f = \mu_s N = \mu_s mg \cos(20^\circ) if the block is at rest.

  • For the hanging block, the net force is: 2gT=2a2g - T = 2a

At the point of impending motion, a=0a = 0 and the tension TT balances the forces on both blocks.

Step 4: Static equilibrium condition (if the system is at rest)

If the system is in static equilibrium (no motion), we compare the maximum static friction force with the weight of the hanging mass.

The tension in the rope is T=2g=2×9.8=19.6NT = 2g = 2 \times 9.8 = 19.6 \, \text{N}.

For the block on the incline to remain stationary: T=μsmgcos(20)+mgsin(20)T = \mu_s mg \cos(20^\circ) + mg \sin(20^\circ) Solve this equation to find the critical mass mm.

Would you like to proceed with solving for mm, or do you have any additional details you'd like to explore?

Here are some related questions:

  1. What happens if the coefficient of friction changes?
  2. How do we solve for mm if the system is accelerating?
  3. What if the angle of the incline were steeper?
  4. How would the tension change if the pulley had mass?
  5. What is the effect of kinetic friction if the block starts moving?

Tip: Always ensure you distinguish between static and kinetic friction when dealing with systems like this!

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Forces on Inclines
Static and Kinetic Friction

Formulas

Net force on incline block: T - F_f - mg sin(θ) = ma
Frictional force: F_f = μN
Normal force: N = mg cos(θ)
Hanging block force: 2g - T = 2a
Tension in static equilibrium: T = 2g = 19.6 N

Theorems

Newton's Second Law of Motion
Friction Laws

Suitable Grade Level

Grade 10-12