Math Problem Statement

A 17.6 kg sled is pulled with a 25.7 N force at a 22.4° angle across the ground where the coefficient of friction (μk) is equal to 0.135. What is the force of friction on the sled?

Solution

To determine the force of friction (FfF_f) on the sled, we can use the following steps:

Given Data:

  • Mass of sled, m=17.6kgm = 17.6 \, \text{kg}
  • Applied force, F=25.7NF = 25.7 \, \text{N}
  • Angle of applied force, θ=22.4\theta = 22.4^\circ
  • Coefficient of kinetic friction, μk=0.135\mu_k = 0.135

Solution

  1. Calculate the Normal Force (FNF_N): Since the force is applied at an angle, the vertical component of the applied force will reduce the normal force. The normal force is given by: FN=mgFsin(θ)F_N = mg - F \sin(\theta) where g=9.81m/s2g = 9.81 \, \text{m/s}^2 (acceleration due to gravity).

  2. Calculate the Force of Friction (FfF_f): The force of friction is given by: Ff=μkFNF_f = \mu_k \cdot F_N

Step-by-Step Calculation:

Step 1: Calculate FNF_N

FN=(17.6kg)(9.81m/s2)(25.7N)sin(22.4)F_N = (17.6 \, \text{kg})(9.81 \, \text{m/s}^2) - (25.7 \, \text{N}) \sin(22.4^\circ)

  1. Calculate mgmg: mg=17.6×9.81=172.656Nmg = 17.6 \times 9.81 = 172.656 \, \text{N}
  2. Calculate Fsin(θ)F \sin(\theta): Fsin(22.4)=25.7×sin(22.4)9.78NF \sin(22.4^\circ) = 25.7 \times \sin(22.4^\circ) \approx 9.78 \, \text{N}
  3. Substitute back to find FNF_N: FN=172.6569.78=162.876NF_N = 172.656 - 9.78 = 162.876 \, \text{N}

Step 2: Calculate FfF_f

Ff=μkFN=0.135×162.87621.99NF_f = \mu_k \cdot F_N = 0.135 \times 162.876 \approx 21.99 \, \text{N}

Answer:

The force of friction on the sled is approximately: Ff=21.99NF_f = -21.99 \, \text{N} (negative because friction opposes the direction of motion).

Would you like more detailed steps or have any questions?

Follow-up Questions:

  1. How would the friction force change if the angle of applied force increased?
  2. What would happen if the coefficient of friction was zero?
  3. How does the mass of the sled affect the normal force?
  4. What is the role of the vertical component of the applied force in this scenario?
  5. How would the friction force change if the sled was on an incline?

Tip:

Always break forces into components when they’re applied at an angle, as it helps in calculating the normal force and friction correctly.

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Math Problem Analysis

Mathematical Concepts

Physics
Forces
Friction
Trigonometry

Formulas

Normal force formula: FN = mg - F * sin(θ)
Friction force formula: Ff = μk * FN

Theorems

Newton's Second Law

Suitable Grade Level

Grades 10-12