10/28 tut

Introduction

• Reminder about deadline for regrade requests.
• Students must check solutions before filing a request.

Review of Pulleys and Force Analysis

• Pulley Demonstration
  • Discussed previous class where a demonstration with a single pulley was shown.
  • Rope wrapping around the pulley affects the exerted force.
  • The exerted force is not equal to the object's weight; it is a fraction.
• Calculation of Tension
  • If the rope wraps around n times, the force exerted is given by:
  • $T = \frac{W}{n}$,
  • Where W is the weight of the object.
  • Example with Weight:
  • Object's weight = 60 pounds.
  • Wrapped around 3 times, tension is $T = \frac{60}{3} = 20 ext{ pounds}$.
  • Wrapped around 4 times, tension is $T = \frac{60}{4} = 15 ext{ pounds}$.
  • Assumes constant speed; if accelerating, more force is required.

Applications of Pulleys

• Real-world example of a car winch.
  • Car uses a winch wrapped around a tree to pull itself out.
  • Modifies the force in question; when wrapped around a tree, the force division changes to half instead of a quarter.

Problem-Solving with Inclines

• Combining two concepts: Forces on an incline with a pulley and determining acceleration direction.
• Summative question discussed regarding acceleration direction:
  • Two objects at different angles connected via a rope.
• Need to determine whether the system moves left, right, or remains stationary. Prompted discussion among students regarding their reasoning.

Weight Components on Inclines

• Weight Decomposition:
  • Objects on an incline have their weight decomposed into:
  • Perpendicular Component (Normal)
  • Parallel Component (Pulling Down the Incline)
  • Mathematical representation:
  • $W_{parallel} = mg \sin(\theta)$
  • Example of weight calculations:
    • For mass 1 (10 kg, angle 23°):
      $W_{1 parallel} = 10 kg \times 9.8 m/s² \times \sin(23°) \approx 38 ext{ Newtons}$.
    • For mass 2 (8 kg, angle 40°):
      $W_{2 parallel} = 8 kg \times 9.8 m/s² \times \sin(40°) \approx 50 ext{ Newtons}$.
• The direction of movement is determined by comparing both components; the net force dictates the overall direction.

Calculating Acceleration

• Discussion on Atwood's Machine for vertical force application:
  • Set up an equation to find the net acceleration when two weights are involved.
• Total mass is needed for acceleration calculation:
  • $F<em>{net} = m</em>{total} \cdot a$,
  • Where net force is the difference of the weights pulling in the system.
• Calculated example:
  • Difference in Forces (50N - 38N) leads to:
  • $a = \frac{F<em>{net}}{m</em>{total}} = \frac{12 ext{ N}}{18 ext{ kg}} = 4.67 m/s²$.

Determining Tension in the System

• Internal Forces:
  • Tension is an internal force; when analyzing the whole system, it cancels out.
• To find tension:
  • Consider either block separately and set up forces:
  • $T - W<em>{parallel1} = m</em>{1} a$ or
  • $T - W<em>{parallel2} = m</em>{2} a$.
• Both should yield the same tension value:
  • Expected tension calculated to be 45N.

Summary of Key Concepts

• Tension: A pull force transmitted via ropes/pulleys; must consider friction and masslessness of ropes.
• Pulleys: Frictionless and massless assumptions lead to uniform tension across the system.

Transition to Chapter 6

• Introduces one-dimensional circular motion.
• New forces encountered – centripetal forces critical for objects moving in circular paths.
• Example: Blood centrifuge as an application of centripetal force.

Circular Motion and Centripetal Force

• Concepts underpinning acceleration:
  • Acceleration: $a = \frac{\Delta v}{\Delta t}$; requires consistent forces acting on the object circling.
  • Centripetal force defined as force directing towards the circle's center.
• Centripetal equation:
  • $F_{c} = \frac{m v^2}{r}$, where $v$ is linear speed, $r$ is radius of the circular path.

Misconceptions About Centrifugal Force

• Centrifugal Force: Often seen as a force pushing objects outward but is a byproduct of inertia against centripetal force.
• While in circular motion, an object's inertia attempts to maintain a straight path unless compelled by an inward force (centripetal).

Practical Examples and Applications

• Discussion on friction in car movement:
  • Static friction allows for changes in direction in circular motion; transitioning to kinetic friction when limits are exceeded.
• Free-body diagram considerations when analyzing objects in circular motion demonstrate varied forces at play (e.g., static friction, normal force, etc.).
• An example with cars on a racetrack highlights how forces maintain a circular path.

Conclusion

• Speed limits on roads based on safe turning radii determined by friction coefficients.
• Emphasize understanding the difference between centripetal and centrifugal forces to avoid misinterpretations in physics applications.