Newton’s Laws of Motion

To properly account for tensions in a multiple-pulley system:

1. Identify segments where tension is constant:

   • Tension is uniform throughout a massless string/rope between pulleys
   • Tension changes at pulleys where direction changes
   • Each isolated segment between directional changes may have different tension values

2. Analyze motion directions:

   • Components moving in the same direction experience identical accelerations
   • Components moving in perpendicular directions have related but different acceleration components
   • Components connected by a string without sliding maintain equal displacement magnitudes

3. Apply constraint conditions:

   • For inextensible strings, the total string length remains constant
   • For non-sliding contact, the acceleration of connected components must be equal
   • For components at different elevations, vertical and horizontal motions must be analyzed separately

Example: In a system with horizontal and vertical segments, a block moving horizontally and another moving vertically cannot form a single system, even if connected by a string, because their acceleration vectors have different directions.

A pseudo force:

• Is a fictitious force that appears in non-inertial reference frames
• Has magnitude m·a₀ where m is the mass of the object and a₀ is the acceleration of the reference frame
• Acts in the direction opposite to the acceleration of the reference frame
• Is necessary to make Newton's laws work in non-inertial frames

The pendulum in an accelerating car demonstrates this by:

• Appearing to deflect at angle θ where tanθ = a₀/g, despite no actual horizontal force acting on it
• Requiring a pseudo force ma₀ in calculations to explain why the pendulum is in equilibrium at this angle
• Showing that observers in the car must include this pseudo force to correctly analyze the system
• Illustrating that pseudo forces are proportional to the mass of the object, just like gravitational forces

The self-adjusting nature of frictional forces enables coordinated movement through:

1. Automatic equilibration of accelerations:

   • Forward friction f on horse adjusts to match needed propulsion
   • Backward friction f′ on cart adjusts based on rolling resistance

2. Force balancing mechanism:

   • These forces self-adjust such that (F₁-f′)/Mₖ = (f-F₂)/Mₕ
   • This equality ensures the horse and cart maintain the same acceleration

3. Practical implications:

   • Without this self-adjustment, the connection between horse and cart would either break or cause one to drag the other
   • The friction forces respond dynamically to changes in applied muscle force
   • This represents a natural feedback system where forces reach equilibrium values

This coordinated adjustment is why horse and cart move together as a system despite having different masses and experiencing different force components.

The paradox is that if the horse exerts a forward force F₁ on the cart, the cart must exert an equal and opposite force F₂ on the horse (where F₁ = F₂ = F).

Since these forces are equal in magnitude but opposite in direction, their sum is zero. This raises the question: How can the system accelerate forward if the net force appears to be zero?

Resolution:

• The error is in considering only the interaction forces between horse and cart
• Additional forces must be considered, particularly:
  • The force the ground exerts on the horse's hooves (providing forward thrust)
  • Frictional forces between the wheels and ground
• When all forces are properly accounted for, the net force is non-zero, allowing acceleration

This demonstrates the importance of proper system selection when applying Newton's laws.

The tensions in the strings are:

$T₁ = \frac{mg \cos\beta}{\sin(\alpha+\beta)}$

$T₂ = \frac{mg \cos\alpha}{\sin(\alpha+\beta)}$

These expressions are derived by:

• Setting up force equilibrium equations (horizontal and vertical components)
• Horizontal: $T₁\cos\alpha - T₂\cos\beta = 0$
• Vertical: $T₁\sin\alpha + T₂\sin\beta - mg = 0$
• Solving these equations simultaneously for T₁ and T₂

Note: The denominator $\sin(\alpha+\beta)$ shows that as the strings approach the same line ($\alpha+\beta \rightarrow 180°$), the tensions approach infinity.

For the system to remain at rest:

$\sin\theta = \frac{m₂}{m₁}$

Where:

• θ is the angle of the incline
• m₁ is the mass on the incline
• m₂ is the hanging mass

This condition is derived by:

1. Setting T = m₂g (from the hanging mass force balance)
2. For the mass on the incline: T = m₁g·sinθ (component parallel to the incline)
3. Equating these two expressions: m₂g = m₁g·sinθ
4. Therefore: θ = sin⁻¹(m₂/m₁)

The system will remain stationary only at this specific angle, given the mass ratio.

The force magnitude is 625 N.

Solution:

1. Find deceleration using kinematic equation v² = u² + 2ax

   • Initial velocity u = 250 m/s
   • Final velocity v = 0 m/s
   • Distance x = 0.05 m
   • 0 = (250)² + 2a(0.05)
   • a = -625,000 m/s²

2. Apply Newton's Second Law: F = ma

   • m = 10 g = 0.01 kg
   • a = 625,000 m/s²
   • F = 0.01 × 625,000 = 6,250 N

3. Since we're looking for force ON the bullet, F = 625 N

The extremely high deceleration shows how rapidly the bullet's energy is dissipated in a very short distance.

Resolving forces into components is critical because:

1. Vector addition requirement: Forces are vectors and must be added vectorially, not arithmetically

2. Equilibrium conditions: For a body in equilibrium, the vector sum must equal zero, which is easiest to verify by component

3. Non-aligned forces: When forces act at different angles, direct summation is impossible without component analysis

4. Simplified equations: Component analysis converts a complex vector problem into simpler scalar equations

5. Independent equations: Each component direction produces a separate equation, providing enough relations to solve for multiple unknowns

Example process:

• For a mass suspended by strings at angles α and β:
  • Horizontal: T₁cosα - T₂cosβ = 0
  • Vertical: T₁sinα + T₂sinβ - mg = 0
• These independent equations allow solving for both T₁ and T₂

Without component analysis, equilibrium problems with non-collinear forces would be effectively unsolvable.

For a body of mass m on an incline with angle θ:

Parallel component (down the incline): $F_{\parallel} = mg\sin\theta$

Perpendicular component (normal to the incline): $F_{\perp} = mg\cos\theta$

Key relationships:

1. The parallel component increases with incline angle:

   • At θ = 0° (horizontal): F∥ = 0
   • At θ = 90° (vertical): F∥ = mg (maximum)

2. The perpendicular component decreases with incline angle:

   • At θ = 0° (horizontal): F⊥ = mg (maximum)
   • At θ = 90° (vertical): F⊥ = 0

3. The normal force from the incline equals F⊥ for a body in equilibrium

4. For a body to remain at rest when placed on an incline:

   • Either an external force must oppose F∥
   • Or static friction must satisfy: f ≥ F∥

This component relationship is fundamental to analyzing all inclined plane problems.

Determining forces and equations for a suspended mass:

1. Identify all forces:

   • Weight force ($mg$) acting downward
   • Tension forces in each supporting element (strings, ropes, etc.)
   • Any other applied forces

2. Choose coordinate system:

   • Usually align with the geometry of the problem
   • Often horizontal and vertical for suspended masses

3. Apply equilibrium conditions:

   • Sum of all forces in x-direction = 0
   • Sum of all forces in y-direction = 0
   • Sum of all torques = 0 (if rotation is possible)

4. Break vector forces into components:

   • For angled tensions: resolve into x and y components using trigonometric functions
   • Example: A tension T at angle θ has components $T\cos\theta$ (horizontal) and $T\sin\theta$ (vertical)

5. Solve the resulting system of equations to find unknown quantities