Chapter 5 Lecture: Normal Forces: Examples

Principles of the Normal Force

• Variable Nature of the Normal Force: The normal force ($n$) exerted by a supporting surface is not a constant value, nor is it inherently equal to the weight ($w$) of the resting object.

• Condition for Equality with Weight: The magnitude of the normal force equals the magnitude of the object's weight ($n = w$) if and only if:

  • The gravitational weight is the sole vertical force acting on the object.

  • The vertical acceleration of the object is zero ($a_y = 0\,\text{m/s}^2$).

• Response to Applied Forces: As external vertical forces applied to an object increase or decrease, the supporting surface deforms (bends) accordingly to adjust the normal force until either static equilibrium is maintained or the surface reaches its structural threshold and breaks.

Systematic Procedure for Applying Newton's Second Law

• Step 1: Choose a Coordinate System:

  • Establish vertical and horizontal axes.

  • Explicitly define positive and negative directions (e.g., standard convention sets upward as the positive y-direction, $+y$, and downward as the negative y-direction, $-y$).

• Step 2: Identify All Forces:

  • Determine all contact forces (e.g., applied push/pull forces, surface normal forces) and non-contact forces (e.g., gravitational force/weight) acting directly on the target mass.

• Step 3: Construct a Free Body Diagram (FBD):

  • Model the object as a point mass $m$

  • Draw vector arrows originating from the mass for each identified force.

  • Ensure vector directions correspond to physical force orientation.

  • Scale vector lengths proportionally to reflect relative force magnitudes.

• Step 4: Formulate and Solve Newton's Second Law Equations:

  • Sum forces along each dimensional axis independently using correct directional signs: $\sum F_y = m a_y$

  • Substitute known quantities and solve algebraically for unknown variables.

Scenario 1: Object Subjected to Downward Applied Force

• Problem Setup:

  • Object mass: $m = 2.50\,\text{kg}$

  • Applied downward force: $F = 15.0\,\text{N}$

  • State of motion: Resting statically on a horizontal surface ($a_y = 0\,\text{m/s}^2$

• Force Identification and Estimation:

  • Applied Force: $F = 15.0\,\text{N}$ directed downward (negative y-direction).

  • Weight ($w$): Earth's gravitational pull directed downward (negative y-direction).

    • Rough estimation using $g \approx 10\,\text{m/s}^2$: $w \approx 2.50\,\text{kg} \times 10\,\text{m/s}^2 = 25\,\text{N}$

    • Exact calculation using $g = 9.8\,\text{m/s}^2$: $w = m g = 2.50\,\text{kg} \times 9.8\,\text{m/s}^2 = 24.5\,\text{N}$

  • Normal Force ($n$): Exerted upward by the table on the bottom of the block (positive y-direction).

• Free Body Diagram Construction:

  • Downward vector $F$ ($15.0\,\text{N}$) is drawn slightly larger than half the length of downward vector $w$ ($24.5\,\text{N}$).

  • Upward vector $n$ must balance the combined effect of both downward vectors. Using the tip-to-tail vector addition method, the length of $n$ must equal the sum of the lengths of $F$ and $w$

• Mathematical Derivation:

  • Apply Newton's Second Law along the y-axis:         $\sum F_y = n - F - w = m a_y$

  • Substitute $a_y = 0\,\text{m/s}^2$:         $n - F - w = 0$         $n = F + w = F + m g$

  • Substitute numerical values:         $n = 15.0\,\text{N} + (2.50\,\text{kg} \times 9.8\,\text{m/s}^2)$         $n = 15.0\,\text{N} + 24.5\,\text{N} = 39.5\,\text{N}$

• Physical Interpretation: Pushing down on the block compresses the surface further, causing the table to exert an increased normal force of $39.5\,\text{N}$ to prevent vertical acceleration.

Scenario 2: Object Subjected to Upward Applied Force

• Problem Setup:

  • Object mass: $m = 2.50\,\text{kg}$

  • Applied upward force: $F = 15.0\,\text{N}$

  • State of motion: Resting statically on a horizontal surface ($a_y = 0\,\text{m/s}^2$

• Force Identification:

  • Applied Force: $F = 15.0\,\text{N}$ directed upward (positive y-direction).

  • Weight ($w$): $w = m g = 24.5\,\text{N}$ directed downward (negative y-direction).

  • Normal Force ($n$): Exerted upward by the table on the block (positive y-direction).

• Free Body Diagram Construction:

  • The total upward forces must cancel the total downward force to yield zero net force.

  • If $n$ were drawn equal to weight ($24.5\,\text{N}$), the total upward force ($15.0\,\text{N} + 24.5\,\text{N} = 39.5\,\text{N}$) would exceed the downward weight ($24.5\,\text{N}$), causing an unobserved upward acceleration.

  • Correct visual scaling: The normal force vector $n$ must be reduced in length such that the vector addition of $n$ and $F$ exactly matches the length of the weight vector $w$

• Mathematical Derivation:

  • Apply Newton's Second Law along the y-axis:         $\sum F_y = n + F - w = m a_y$

  • Substitute $a_y = 0\,\text{m/s}^2$:         $n + F - w = 0$         $n = w - F = m g - F$

  • Substitute numerical values:         $n = 24.5\,\text{N} - 15.0\,\text{N} = 9.5\,\text{N}$

• Physical Interpretation: Pulling upward partially relieves the load carried by the surface. The table only needs to supply a normal force of $9.5\,\text{N}$ to maintain static equilibrium.

Force Scale Benchmarks and Physical Analogies

• Intuitive Force Quantities:

  • $1\,\text{N}$ of force is approximately equivalent to:

    • The force required to depress an elevator button.

    • The downward weight force exerted by a standard $100\,\text{g}$ apple.

  • $25\,\text{N}$ of force is roughly equivalent to the combined weight of a bag containing 25 apples.

• Distributed Force Analogy:

  • In a classic demonstration where 10 individuals place two fingers beneath a lying person, each individual contributes a small upward force vector.

  • The sum of these individual upward forces balances the total weight of the person ($\sum F_{\text{up}} = w$

  • As external upward support increases, the required force contribution from any single supporting contact surface decreases proportionally.