Math Problem Statement

A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.7 and µk = 0.59, respectively. The acceleration of gravity is 9.8 m/s 2 . 31 kg µ 31◦ What is the frictional force acting on the 31 kg mass? Answer in units of N.

Solution

To find the frictional force acting on the 31 kg mass, we first need to determine whether the block is moving or at rest. This will be based on the static friction.

Step 1: Calculate the gravitational force components

The gravitational force FgF_g acting on the block is:

Fg=mg=31kg9.8m/s2=303.8NF_g = m \cdot g = 31 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 303.8 \, \text{N}

The component of this force parallel to the incline (FF_{\parallel}) is given by:

F=Fgsin(θ)=303.8sin(31)F_{\parallel} = F_g \cdot \sin(\theta) = 303.8 \cdot \sin(31^\circ)

F303.80.515156.45NF_{\parallel} \approx 303.8 \cdot 0.515 \approx 156.45 \, \text{N}

The component of the gravitational force perpendicular to the incline (FF_{\perp}) is:

F=Fgcos(θ)=303.8cos(31)F_{\perp} = F_g \cdot \cos(\theta) = 303.8 \cdot \cos(31^\circ)

F303.80.857260.3NF_{\perp} \approx 303.8 \cdot 0.857 \approx 260.3 \, \text{N}

Step 2: Maximum static friction force

The maximum static friction force fsmaxf_s^{\text{max}} is given by:

fsmax=μsF=0.7260.3182.21Nf_s^{\text{max}} = \mu_s \cdot F_{\perp} = 0.7 \cdot 260.3 \approx 182.21 \, \text{N}

Step 3: Compare the forces

The frictional force needs to balance the parallel component of the gravitational force (FF_{\parallel}) to keep the block at rest. Since the maximum static friction force (182.21 N) is greater than FF_{\parallel} (156.45 N), the block does not move. Therefore, the static friction force is equal to the parallel component of gravity:

fs=F=156.45Nf_s = F_{\parallel} = 156.45 \, \text{N}

Thus, the frictional force acting on the 31 kg block is 156.45 N.

Would you like further details or have any questions?

Here are five related questions to expand on this topic:

  1. How does the angle of the incline affect the frictional force?
  2. What happens if the angle is increased beyond a certain point?
  3. How would the frictional force change if the mass of the block increased?
  4. What would the frictional force be if the block started moving?
  5. How does kinetic friction differ from static friction in this context?

Tip: Remember that static friction adjusts up to its maximum value to prevent motion, while kinetic friction has a constant value once motion starts.

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Math Problem Analysis

Mathematical Concepts

Physics
Static Friction
Inclined Plane
Trigonometry

Formulas

Gravitational force: F_g = m * g
Parallel component of gravitational force: F_parallel = F_g * sin(θ)
Perpendicular component of gravitational force: F_perpendicular = F_g * cos(θ)
Maximum static friction: f_s_max = µ_s * F_perpendicular

Theorems

Newton's Laws of Motion

Suitable Grade Level

Grades 11-12