Lecture Notes on Gravity, Weight, and Normal Force

Lecture Overview
  • Course: PHYS 225 - Fundamentals of Physics: Mechanics

  • Instructor: Prof. Meng (Stephanie) Shen

  • Institution: California State University, Fullerton

Learning Goals
  • Understand the concept of Weight

  • Comprehend the Force of Gravity in general

  • Learn about Normal Force

  • Analyze Free Body Diagrams (FBD)

Key Concepts in Gravity and Forces
1. Weight
  • Weight is defined as the force of gravity acting on an object near the Earth’s surface.

  • Mathematically expressed as:
    W=F<em>g=ma</em>y=gjextwheregextisapproximately9.8extm/s2W = F<em>g = m a</em>y = -g \boldsymbol{j} ext{ where } g ext{ is approximately } 9.8 ext{ m/s}^2

2. Free Fall in a Vacuum
  • A feather and a penny dropped from the same height will land at the same time in a vacuum due to the absence of air resistance.

3. Newton’s Law of Gravity
  • An attractive force described by:
    F<em>g=Gracm</em>1m2r2|F<em>g| = G rac{m</em>1 m_2}{r^2}

  • G = Gravitational constant

  • The direction of force is always towards the other object.

4. Earth-Moon Gravity Interactions
  • Both Earth and Moon exert equal gravitational forces on one another, thus:

    • Answer: c) They pull on each other equally.

5. Force of Gravity Example: Earth
  • Parameters:

    • Earth radius (R_E) = 6371 km

    • Earth mass (M_E) = 5.972imes1024extkg5.972 imes 10^{24} ext{ kg}

    • Gravitational constant (G) = 6.674imes1011extNm2extkg26.674 imes 10^{-11} ext{ N m}^2 ext{ kg}^{-2}

  • The acceleration due to gravity on Earth's surface can be calculated as:
    a<br>ightarrow(6.674imes1011)(5.972imes1024)/(6.371imes106)2<br>ightarrowaextisapproximately9.8extm/s2a <br>ightarrow (6.674 imes 10^{-11})(5.972 imes 10^{24}) / (6.371 imes 10^6)^2 <br>ightarrow a ext{ is approximately } 9.8 ext{ m/s}^2

6. Relationship Between Gravity and Weight
  • For an object of mass m:
    F<em>g=GracmM</em>ERE2|F<em>g| = G rac{m M</em>E}{R_E^2}

  • Force of gravity or weight can also be expressed as:
    Fg=W=mgjF_g = W = -m g \boldsymbol{j}

  • Where gextisequivalentto9.8extm/s2g ext{ is equivalent to } 9.8 ext{ m/s}^2 on Earth.

7. Normal Force
  • The normal force is the support force exerted by a surface that supports the weight of an object resting on it.

  • Direction: Always perpendicular to the surface.

  • Magnitude: Adjusts to be exactly equal to the weight force when in equilibrium.

8. Free Body Diagrams (FBD)
  • FBDs provide a visual representation of all forces acting on an object.

  • Each force is shown as a vector to indicate direction.

  • Forces not aligned with coordinate axes should be decomposed along these axes.

Steps to Draw a Free Body Diagram (FBD)
  1. Draw the coordinate system (x, y, z axes).

  2. Indicate all forces acting on the object or system.

  3. Decompose any non-axial forces as necessary.

Example Problem
  • Scenario: A block of mass 1 kg sliding with external forces F1 = 2N and F2 = 8N along a frictionless surface.

  • Use FBD to solve for acceleration and normal force acting on the block.

Conclusion
  • This lecture emphasized understanding gravity, how to analyze forces in free fall, normal forces, and using Free Body Diagrams for problem-solving in mechanics.

Key Equations and Concepts
  1. Weight
    Weight (bW) is defined as the force of gravity acting on an object near the Earth’s surface.
    Mathematically expressed as:
    W=F<em>g=ma</em>y=gj where g is approximately 9.8 m/s2W = F<em>g = m a</em>y = -g \boldsymbol{j} \text{ where } g \text{ is approximately } 9.8 \text{ m/s}^2

  2. Free Fall in a Vacuum
    A feather and a penny dropped from the same height will land at the same time in a vacuum due to the absence of air resistance.

  3. Newton’s Law of Gravity
    An attractive force described by:
    F<em>g=Gm</em>1m2r2|F<em>g| = G \frac{m</em>1 m_2}{r^2}

    • G = Gravitational constant

    • The direction of force is always towards the other object.

  4. Force of Gravity Example: Earth
    Parameters:

    • Earth radius (R_E) = 6371 km

    • Earth mass (M_E) = 5.972×1024 kg5.972 \times 10^{24} \text{ kg}

    • Gravitational constant (G) = 6.674×1011 N m2 kg26.674 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}

    • The acceleration due to gravity on Earth's surface can be calculated as:
      a(6.674×1011)(5.972×1024)(6.371×106)2a is approximately 9.8 m/s2a \rightarrow \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{(6.371 \times 10^6)^2} \rightarrow a \text{ is approximately } 9.8 \text{ m/s}^2

  5. Relationship Between Gravity and Weight
    For an object of mass m:
    F<em>g=GmM</em>ER<em>E2|F<em>g| = G \frac{m M</em>E}{R<em>E^2} Force of gravity or weight can also be expressed as: F</em>g=W=mgjF</em>g = W = -m g \boldsymbol{j}
    Where g is equivalent to 9.8 m/s2g \text{ is equivalent to } 9.8 \text{ m/s}^2 on Earth.

  6. Normal Force
    The normal force is the support force exerted by a surface that supports the weight of an object resting on it.

    • Direction: Always perpendicular to the surface.

    • Magnitude: Adjusts to be exactly equal to the weight force when in equilibrium.

  7. Free Body Diagrams (FBD)
    FBDs provide a visual representation of all forces acting on an object.

    • Each force is shown as a vector to indicate direction.

    • Forces not aligned with coordinate axes should be decomposed along these axes.