Frictional Force Study Guide

Frictional Force Study Guide

1. Problem Statement involving Little Suzy and the Baby Carriage

• Situation: Little Suzy pushes a 6.5 kg baby carriage across a floor.
• Distance: 25.0 meters.
• Force Applied: 25.5 N at an angle of 40° to the horizontal.
• Coefficient of Friction: μ = 0.15.

Calculations

• Weight of the Baby Carriage (Weight):

  • Formula: $F_G = m imes g$
  • Where:
  • $m = 6.5 ext{ kg}$ (mass of the baby carriage)
  • $g = 9.81 ext{ m/s}^2$ (acceleration due to gravity)
  • Calculation:
    $F_G = 6.5 imes 9.81 = 63.67 ext{ N}$

• Vertical Component of the Applied Force (F_V):

  • Formula: $F_V = F imes ext{sin}( heta)$
  • Calculation:
    $F_V = 25.5 imes ext{sin}(40°) = 25.5 imes 0.6428 = 16.36 ext{ N}$

• Normal Force (F_N):

  • Formula:
    $F<em>N = F</em>G + F_V$
  • Calculation:
    $F_N = 63.67 + 16.36 = 80.03 ext{ N}$

• Frictional Force (F_f):

  • Formula:
    $F<em>f = ext{μ} imes F</em>N$
  • Calculation:
    $F_f = 0.15 imes 80.03 = 12.00 ext{ N}$

• Net Force (F_P):

  • Formula:
    $F<em>P = F - F</em>f$
  • Calculation:
    $F_P = 25.5 ext{ N} - 12.00 ext{ N} = 13.50 ext{ N}$

• Acceleration (a):

  • Using Newton's Second Law:
    $F_P = m imes a$
  • Calculation:
    $a = \frac{F_P}{m} = \frac{13.50}{6.5} = 2.08 ext{ m/s}^2$

• Time to Cross the Floor (t):

  • Using the kinematic equation:
    $d = \frac{1}{2} a t^2$
  • Rearrange to solve for time (t):
    $t^2 = \frac{d}{\frac{1}{2} a} = \frac{25.0}{0.5 imes 2.08} = 25.0 / 1.04$
  • Calculation:
    $t ext{ is approximately } ext{4.89 seconds}$

2. Problem Statement Involving a Box Sliding Down an Incline

• Situation: A 36 kg box slides down an incline.
• Incline Angle: 65° to the horizontal.
• Coefficient of Friction: μ = 0.35.
• Length of Incline: 55.0 meters.

Calculations

• Weight of the Box (Weight):

  • Formula: $F_G = m imes g$
  • Calculation:
    $F_G = 36 imes 9.81 = 353.16 ext{ N}$

• Normal Force (F_N):

  • Formula:
    $F<em>N = F</em>G imes ext{cos}( heta)$
  • Calculation:
    $F_N = 353.16 imes ext{cos}(65°) = 353.16 imes 0.4226 = 149.10 ext{ N}$

• Frictional Force (F_f):

  • Formula:
    $F<em>f = μ imes F</em>N$
  • Calculation:
    $F_f = 0.35 imes 149.10 = 52.189 ext{ N}$

• Net Force (F_net) on the Box:

  • Formula:
    $F<em>{ ext{net}} = F</em>G imes ext{sin}( heta) - F_f$
  • Calculation:
    $F_{ ext{net}} = (353.16 imes ext{sin}(65°)) - 52.189 = (353.16 imes 0.9063) - 52.189 = 320.0 - 52.189 = 267.811 ext{ N}$

• Acceleration (a):

  • Formula:
    $F_{ ext{net}} = m imes a$
  • Calculation:
    $a = \frac{F_{ ext{net}}}{m} = \frac{267.811}{36} = 7.44 ext{ m/s}^2$

• Time (t) to Reach the Bottom:

  • Using the formula:
    $d = \frac{1}{2} a t^2$
  • Rearrange to solve for t:
  • Calculation:
    $t^2 = \frac{d}{\frac{1}{2} a} = \frac{55.0}{0.5 imes 7.44} = 55.0 / 3.72$
    $t ext{ is approximately } 3.85 ext{ seconds}$

3. Problem Statement Involving a 10.5 kg Box

• Situation: A force of 150 N is applied to push a 10.5 kg box across a floor with an acceleration.
• Acceleration: 2.68 m/s².

Calculations

• Frictional Force (F_f):

• Net Force (F_N):

  • Given:
    $F_H = 150 ext{ N}$
    $a = 2.68 ext{ m/s}^2$
    $m = 10.5 ext{ kg}$
  • Calculation:
    $F_{net} = m imes a = 10.5 imes 2.68 = 28.14 ext{ N}$
  • Net force in friction (Ff): $F</em>f = F<em>H - F</em>{net} = 150 - 28.14 = 121.86 ext{ N}$

• Coefficient of Friction (μ):

  • Using the formula:
    $μ = \frac{F<em>f}{F</em>N}$
  • Normal Force (F_N):
  • Formula:
    $F_N = m imes g = 10.5 imes 9.81 = 103.4 ext{ N}$
  • Calculation:
  • Plugging into the coefficient formula:
    $μ = \frac{121.86}{103.4} = 1.18$

• Conclusion: The coefficient of friction calculated is approximately 1.184, indicating a very high friction situation - uncommon for typical surfaces.