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 $9.8 \frac{m}{s^2}$. Denoted as $g$.
• Acts on all masses equally.

Force of Friction

• Represented as: $F_{friction} = \mu N$
  • $\mu$: Coefficient of friction (depends on surfaces).
    • Large for cement.
    • Small for ice.
  • $N$: Normal force.

Normal Force

• Rule of thumb: $N = mg$ (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 $g$, resulting in a force of $mg$.

Example: Person on a Hill

• Scenario: Person standing on a frictionless hill with a $30$-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

• $N - mg\sin(60^\circ) = 0$
• No acceleration in the y-direction.
• $N = mg\sin(60^\circ)$
• Demonstrates normal force being less than the force of gravity.

Sum of Forces in X-Direction

• No friction.
• $F_x = -mg\cos(60^\circ)$
• $F<em>x$ must equal mass times acceleration: $F</em>x = ma_x$
• $-mg\cos(60^\circ) = ma_x$

Acceleration Calculation

• Masses cancel out.
• $a_x = -g\cos(60^\circ)$
• $a_x = -9.8 \frac{m}{s^2} \times \cos(60^\circ) = -4.9 \frac{m}{s^2}$
• Interpretation: The person slides downwards with an acceleration of $-4.9 \frac{m}{s^2}$.
• 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 $10 N$ (approximately $2$ pounds).
• Skater's mass: $50 kg$
• Coefficient of friction: $\mu = 0.01$
• Goal: Calculate the skater's horizontal acceleration.

Free Body Diagram

• Forces involved:
  • Gravity (downwards).
  • Normal force (upwards).
  • Applied force (horizontal, $10 N$).
  • Friction (opposite to motion).

Y-Direction Analysis

• Summation of forces: $N - mg = 0$
• $N = mg = 50 kg \times 9.8 \frac{m}{s^2} \approx 500 N$

X-Direction Analysis

• Summation of forces: $F<em>{applied} - F</em>{friction} = ma_x$
• $F_{friction} = \mu N = 0.01 \times 500 N = 5 N$

Acceleration Calculation

• $a<em>x = \frac{F</em>{applied} - F_{friction}}{m} = \frac{10 N - 5 N}{50 kg} = 0.1 \frac{m}{s^2}$
• Interpretation: The skater accelerates at $0.1 \frac{m}{s^2}$.

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).