Forces and Acceleration

Forces and Acceleration

Introduction

  • Combining previous knowledge of forces to calculate the acceleration of objects due to unbalanced forces.
  • Net acceleration implies movement.

Types of Forces

Force Due to Gravity
  • Creates a nearly constant acceleration of 9.8ms29.8 \frac{m}{s^2}. Denoted as gg.
  • Acts on all masses equally.
Force of Friction
  • Represented as: Ffriction=μNF_{friction} = \mu N
    • μ\mu: Coefficient of friction (depends on surfaces).
      • Large for cement.
      • Small for ice.
    • NN: Normal force.
Normal Force
  • Rule of thumb: N=mgN = mg (mass times acceleration due to gravity), but not always.
    • Can be less on an incline.
    • Can be more if jumping upwards.
Independent Acceleration
  • All forces independently create acceleration.
  • Example: Free fall produces acceleration gg, resulting in a force of mgmg.

Example: Person on a Hill

  • Scenario: Person standing on a frictionless hill with a 3030-degree slope.
  • Goal: Determine the person's acceleration and direction.
Free Body Diagram
  • Forces involved:
    • Gravity (downwards).
    • Normal force (perpendicular to the surface).
Coordinate System
  • Positive y-direction: Normal force direction.
  • Positive x-direction: Up the hill.
  • Reasoning: Acceleration will be either up or down the hill (x-direction).
Sum of Forces in Y-Direction
  • N−mgsin⁡(60∘)=0N - mg\sin(60^\circ) = 0
  • No acceleration in the y-direction.
  • N=mgsin⁡(60∘)N = mg\sin(60^\circ)
  • Demonstrates normal force being less than the force of gravity.
Sum of Forces in X-Direction
  • No friction.
  • Fx=−mgcos⁡(60∘)F_x = -mg\cos(60^\circ)
  • F<em>xF<em>x must equal mass times acceleration: F</em>x=maxF</em>x = ma_x
  • −mgcos⁡(60∘)=max-mg\cos(60^\circ) = ma_x
Acceleration Calculation
  • Masses cancel out.
  • ax=−gcos⁡(60∘)a_x = -g\cos(60^\circ)
  • ax=−9.8ms2×cos⁡(60∘)=−4.9ms2a_x = -9.8 \frac{m}{s^2} \times \cos(60^\circ) = -4.9 \frac{m}{s^2}
  • Interpretation: The person slides downwards with an acceleration of −4.9ms2-4.9 \frac{m}{s^2}.
  • Relevance: Explains why it's difficult to walk on an icy incline.

Example: Ice Skater

  • Scenario: Skater pushes off level ice with a force of 10N10 N (approximately 22 pounds).
  • Skater's mass: 50kg50 kg
  • Coefficient of friction: μ=0.01\mu = 0.01
  • Goal: Calculate the skater's horizontal acceleration.
Free Body Diagram
  • Forces involved:
    • Gravity (downwards).
    • Normal force (upwards).
    • Applied force (horizontal, 10N10 N).
    • Friction (opposite to motion).
Y-Direction Analysis
  • Summation of forces: N−mg=0N - mg = 0
  • N=mg=50kg×9.8ms2≈500NN = mg = 50 kg \times 9.8 \frac{m}{s^2} \approx 500 N
X-Direction Analysis
  • Summation of forces: F<em>applied−F</em>friction=maxF<em>{applied} - F</em>{friction} = ma_x
  • Ffriction=μN=0.01×500N=5NF_{friction} = \mu N = 0.01 \times 500 N = 5 N
Acceleration Calculation
  • a<em>x=F</em>applied−Ffrictionm=10N−5N50kg=0.1ms2a<em>x = \frac{F</em>{applied} - F_{friction}}{m} = \frac{10 N - 5 N}{50 kg} = 0.1 \frac{m}{s^2}
  • Interpretation: The skater accelerates at 0.1ms20.1 \frac{m}{s^2}.
Stopping Distance (Foreshadowing)
  • The skater will eventually stop due to friction.
  • The stopping distance will be focus on the next lecture.

Summary

  • Unbalanced forces cause acceleration in the direction of the dominating force.
  • All forces cause acceleration when acting independently.
  • Specific equations for friction and gravity forces.
  • It is helpful to rotate the x-y axis to simplify problem.
  • Net acceleration leads to movement described by kinematic equations (to be discussed in the next lecture).