Physics Exam Review: Forces, Friction, and Circular Motion

Contact Information for Help

James’ Contact Info

  • Email: jhowar02@uoguelph.ca

  • Drop-in Help Hours:

    • Location: MacN 304 (third floor of the building with the Campus Bookstore)

    • Time: Mondays 12:30 PM - 2:30 PM

  • Additional Information: Students can also email James to make an appointment for in-person or Zoom meetings.

Week 4 Overview

  • Focus Areas: Centripetal Force, Friction, Newton’s 3rd Law

  • Overview: This week covers concepts that may be light on new information, emphasizing practice problems and study guides.

Centripetal Force and Uniform Circular Motion

  • Definition: An object moves at a constant speed vv in a circle of radius rr, experiencing centripetal acceleration aca_c.

  • Formula for Centripetal Acceleration:

    • ac=v2ra_c = \frac{v^2}{r} toward the center of the circle.

  • Resultant Force: There is a resultant force acting to produce centripetal acceleration, denoted as:

    • ΣF<em>c=ma</em>c=mv2r\Sigma F<em>c = ma</em>c = \frac{mv^2}{r}.

  • Definition of Centripetal Force (FcF_c):

    • Centripetal force arises from various forces (e.g., gravity, tension, friction, normal force) acting on the object toward the circle's center.

Important Clarification on Forces

  • External Forces: Forces like gravitational force (F<em>GF<em>G), tension, and normal force (F</em>NF</em>N) exist independently, and it’s conventional to label the sum of these forces acting towards the center as the centripetal force (FcF_c).

  • Nature of Centripetal Force:

    • Key Point: Centripetal force is not a distinct force but rather a resultant force from already known forces (gravity, normal force, tension, friction).

Circular Motion Examples

1. Example 1: Tension in String

  • Scenario: A child swings a piece of wood (mass = 0.47 kg) tied to a string in a horizontal circle of radius 83 cm at a constant speed of 3.83 m/s, with the string making an angle of 29° above the horizontal.

  • Forces Breakdown:

    • Horizontal: ΣF<em>x=T</em>x=mac\Sigma F<em>x = T</em>x = ma_c

  • Radial Tension Calculation:

    • Tcos⁡θ=mv2rT \cos \theta = \frac{mv^2}{r}

  • Solve for T:

    • T=mv2rcos⁡θT = \frac{mv^2}{r \cos \theta}

  • Substitute values:

    • T=0.47 kg×(3.83 m/s)20.83 mcos⁡(29°)T = \frac{0.47 \text{ kg} \times (3.83 \text{ m/s})^2}{0.83 \text{ m} \cos(29°)}

  • Resulting in a tension value T≈9.5 NT \approx 9.5 \text{ N}.

2. Example 2: Normal Force on a Car

  • Scenario: A car with a mass of 1.0×103 kg1.0 \times 10^3 \text{ kg} traveling at 21 m/s21 \text{ m/s} crosses over a hill with a radius of curvature 1.0×102 m1.0 \times 10^2 \text{ m}.

  • Determining Forces (part a):

    • ΣF<em>y=F</em>g−F<em>N=F</em>c\Sigma F<em>y = F</em>g - F<em>N = F</em>c

    • F<em>N=F</em>g−FcF<em>N = F</em>g - F_c

    • FN=mg−mv2rF_N = mg - \frac{mv^2}{r}

  • Substitute values:

    • FN=(1000 kg)(9.8 m/s2)−(1000 kg)(21 m/s)2(100 m)F_N = (1000 \text{ kg})(9.8 \text{ m/s}^2) - \frac{(1000 \text{ kg})(21 \text{ m/s})^2}{(100 \text{ m})}

    • Resulting in FN≈5.4×103 NF_N \approx 5.4 \times 10^3 \text{ N}.

  • Determining Velocity (part b):

    • Using condition when normal force is zero (FN=0F_N = 0):

    • 0=mg−mv2r0 = mg - \frac{mv^2}{r}

    • This leads to v2=grv^2 = gr, so v=grv = \sqrt{gr}

  • Substitute values to find speed:

    • v=(9.8 m/s2)(100 m)≈31 m/sv = \sqrt{(9.8 \text{ m/s}^2)(100 \text{ m})} \approx 31 \text{ m/s}

3. Example 3: Tension on a Ball

  • Scenario: A ball is swung on a rope through circular motion.

  • Question: Where is the tension the greatest?

    • a) Top

    • b) Bottom

    • c) Left-hand side

    • d) Right-hand side

    • e) Tension is the same everywhere

  • Answer: Tension is greatest at the bottom, due to maximum centripetal force needed to keep the ball in circular motion.

Dynamics: Friction Concepts

  • Types of Friction:

    • Static Friction (FsF_s): Acts against movement of objects that are at rest.

    • Kinetic Friction (FkF_k): Acts against objects that are already moving.

  • Dependency: Frictional force depends on the mass of the object and the material type in contact with it.

Defining Static and Kinetic Friction

  • Static Friction Formula:

    • F<em>sF<em>s is dependent on normal force F</em>NF</em>N:

    • F<em>s≤μ</em>sFNF<em>s \le \mu</em>s F_N

  • Kinetic Friction Formula:

    • F<em>k=μ</em>kFNF<em>k = \mu</em>k F_N

Coefficients of Friction

  • Materials Examples (Approximate Values):

    • Steel on Steel (dry):

    • Coefficient of Static Friction (μs\mu_s) = 0.74

    • Coefficient of Kinetic Friction (μk\mu_k) = 0.57

    • Waxed wood on wet snow:

    • μ<em>s\mu<em>s = 0.14, μ</em>k\mu</em>k = 0.10

    • Ice on ice:

    • μ<em>s\mu<em>s = 0.1, μ</em>k\mu</em>k = 0.03

    • Synovial fluid in human joints:

    • μ<em>s\mu<em>s = 0.01, μ</em>k\mu</em>k = 0.003

  • Key Finding: Static friction is typically greater than kinetic friction, highlighting that it takes more force to start an object moving than to maintain its motion.

Calculating Maximum Static Frictional Force

  • Formula:

    • F<em>s,max=μ</em>sFNF<em>{s,max} = \mu</em>s F_N

  • It indicates the maximum static friction that can act on an object by a given surface, suggesting the minimum force needed to initiate movement.

Note on Normal Force (FNF_N)

  • Observation: The normal force F<em>NF<em>N is often equal and opposite to gravitational force F</em>gF</em>g or Fgcos⁡(θ)F_g \cos(\theta) in inclined planes.

  • However, it can vary depending on other forces acting on the object, necessitating careful evaluation in specific problems.

Friction Example Problems

1. Problem Statement: Refrigerator on a Floor

  • A person exerts a horizontal force of 512 N on a stationary refrigerator (mass = 117 kg) on a horizontal floor. Coefficients of friction are μ<em>s=0.69\mu<em>s = 0.69 and μ</em>k=0.60\mu</em>k = 0.60.

  • Determine if the fridge moves:

    • Sum forces in y-direction:

    • ΣF<em>y=F</em>N−F<em>g=0⇒F</em>N=mg\Sigma F<em>y = F</em>N - F<em>g = 0 \Rightarrow F</em>N = mg

    • F<em>N=117 kg×9.8 m/s2F<em>N = 117 \text{ kg} \times 9.8 \text{ m/s}^2 results in F</em>N=1143.66 NF</em>N = 1143.66 \text{ N}

    • Calculate maximum static friction:

    • F<em>s,max=μ</em>sF<em>N⇒F</em>s,max=0.69×1143.66 N=790.04 NF<em>{s,max} = \mu</em>s F<em>N \Rightarrow F</em>{s,max} = 0.69 \times 1143.66 \text{ N} = 790.04 \text{ N}

    • Since F{s,max} > F{push} (790.04 N > 512 N), the fridge does not move.

2. Situation with a Static Box

  • Mass of box = 10 kg.

  • Coefficients of friction: μ<em>s=0.1\mu<em>s = 0.1, μ</em>k=0.05\mu</em>k = 0.05.

  • A force of 7 N is applied, but the box doesn’t move.

  • Find the static frictional force:

    • Since static friction equals external force when the object is at rest and the applied force is less than F<em>s,maxF<em>{s,max}: F</em>s=7 NF</em>s = 7 \text{ N} as it balances the applied force.

3. Rope Force Problem: Box with Angled Pull

  • A rope exerts a force of 21 N at an angle of 31° above horizontal on a box (mass = 3.3 kg) at rest. Coefficients between the box and the floor: μ<em>s=0.55\mu<em>s = 0.55 and μ</em>k=0.50\mu</em>k = 0.50.

  • Questions: (a) Does the box move? (b) If it moves, what’s the acceleration?

  • Resolve forces vertically (y-direction):

    • ΣF<em>y=0=F</em>N+F<em>ay−F</em>g\Sigma F<em>y = 0 = F</em>N + F<em>{ay} - F</em>g

    • F<em>N=mg−F</em>ay=(3.3 kg×9.8 m/s2)−(21 Nsin⁡(31°))F<em>N = mg - F</em>{ay} = (3.3 \text{ kg} \times 9.8 \text{ m/s}^2) - (21 \text{ N} \sin(31°)), leading to FN≈21.52 NF_N \approx 21.52 \text{ N}.

  • Calculate maximum static friction:

    • F<em>s,max=0.55×F</em>N=0.55×21.52 N≈11.8 NF<em>{s,max} = 0.55 \times F</em>N = 0.55 \times 21.52 \text{ N} \approx 11.8 \text{ N}.

  • Calculate horizontal component of the exerted force:

    • Fax=21 Ncos⁡(31°)≈18 NF_{ax} = 21 \text{ N} \cos(31°) \approx 18 \text{ N}

  • Since F{ax} > F{s,max} (18 N > 11.8 N), the box moves.

  • Find acceleration (using kinetic friction once it moves):

    • First, calculate kinetic friction: F<em>k=μ</em>kFN=0.50×21.52 N≈10.76 NF<em>k = \mu</em>k F_N = 0.50 \times 21.52 \text{ N} \approx 10.76 \text{ N}

    • Resolve forces horizontally (x-direction) for acceleration:

    • ΣF<em>x=ma</em>x=F<em>ax−F</em>k\Sigma F<em>x = m a</em>x = F<em>{ax} - F</em>k

    • a<em>x=F</em>ax−Fkma<em>x = \frac{F</em>{ax} - F_k}{m}

    • ax=18 N−10.76 N3.3 kg≈2.2 m/s2a_x = \frac{18 \text{ N} - 10.76 \text{ N}}{3.3 \text{ kg}} \approx 2.2 \text{ m/s}^2

Summary of Friction

  1. Static Friction (FsF_s):

    • Acts to prevent movement of stationary objects.

    • Maximum static friction defined as F<em>s,max=μ</em>sFNF<em>{s,max} = \mu</em>s F_N.

  2. Kinetic Friction (FkF_k):

    • Acts on moving objects.

    • Kinetic friction force given by F<em>k=μ</em>kFNF<em>k = \mu</em>k F_N.

  3. General Rule: Problems pertaining to friction often necessitate the use of Newton’s 2nd Law in conjunction with analyzing forces in x and y directions to resolve unknown quantities, as they may be simultaneously affected by static and kinetic friction dynamics.

Newton’s 3rd Law

  • Definition: If object A exerts a force on object B, then B exerts an equal and opposite force on A.

Newton's 3rd Law in Examples

1. Example: Ship and Iceberg Collision

  • Scenario: A ship collides with an iceberg.

  • Question: Which object experiences more force during the collision?

  • Answer: The magnitude of force experienced is the same for both objects due to equal and opposite reactions as per Newton's 3rd Law.

2. Acceleration Question

  • Question: Which object experiences greater acceleration from the collision?

  • Answer: Since the ship and iceberg have different masses, the object with smaller mass (iceberg) experiences greater acceleration (a=F/ma = F/m).

3. Pushing Two Objects

  • Scenario: Two glasses are pushed across a table, exerting a total applied force of 2.50 N.

  • Larger glass mass (m1=1.20 kgm_1 = 1.20 \text{ kg}), accelerating at 1.20 m/s21.20 \text{ m/s}^2.

  • Calculate mass of smaller glass (m2m_2):

    • Using Newton's 2nd Law for the system: ΣF=(m<em>1+m</em>2)a\Sigma F = (m<em>1 + m</em>2)a

    • F<em>a=(m</em>1+m2)aF<em>a = (m</em>1 + m_2)a

    • m<em>1+m</em>2=Faam<em>1 + m</em>2 = \frac{F_a}{a}

    • m<em>2=F</em>aa−m1m<em>2 = \frac{F</em>a}{a} - m_1

  • Substitute values:

    • m2=2.50 N1.20 m/s2−1.20 kgm_2 = \frac{2.50 \text{ N}}{1.20 \text{ m/s}^2} - 1.20 \text{ kg}

    • Resulting in m2≈0.88 kgm_2 \approx 0.88 \text{ kg}.

Example Calculations of Forces in Contact

  • Contact Forces:

    • When analyzing two objects, the force exerted by the first object on the second, and vice versa, can be calculated using dynamic equations accounting for their acceleration.

    • The force accelerating the smaller glass (m<em>2m<em>2) is the contact force from the larger glass (m</em>1m</em>1):

    • F<em>contact=m</em>2aF<em>{contact} = m</em>2 a

    • Plugging values: Fcontact=(0.88 kg)(1.2 m/s2)≈1.06 NF_{contact} = (0.88 \text{ kg})(1.2 \text{ m/s}^2) \approx 1.06 \text{ N}.

    • The net force on the larger glass (m1m_1) is the applied force minus the contact force:

    • F<em>a−F</em>contact=m1aF<em>a - F</em>{contact} = m_1 a

    • This confirms that the total applied force F<em>a=F</em>contact+m1a=1.06 N+(1.20 kg)(1.20 m/s2)=2.50 NF<em>a = F</em>{contact} + m_1 a = 1.06 \text{ N} + (1.20 \text{ kg})(1.20 \text{ m/s}^2) = 2.50 \text{ N}, matching the problem statement.

Inclined Plane with Friction

  • Example of a sled on an 18° incline:

    • Total mass = 71 kg, coefficients: μ<em>s=0.15\mu<em>s = 0.15 and μ</em>k=0.095\mu</em>k = 0.095.

  • Forces on the incline are primarily gravitational and frictional:

    • a) Normal force magnitude:

    • Calculated as: F<em>N=mgcos⁡(18°)=71 kg×9.8 m/s2cos⁡(18°)F<em>N = mg \cos(18°) = 71 \text{ kg} \times 9.8 \text{ m/s}^2 \cos(18°) yielding approximately F</em>N=662 NF</em>N = 662 \text{ N}.

    • b) Acceleration of sled:

    • Determine if it slides:

      • Compare gravitational pull down the incline (F<em>gx=mgsin⁡(18°)F<em>{gx} = mg \sin(18°)) and maximum static friction (f</em>s,max=μ<em>sF</em>Nf</em>{s,max} = \mu<em>s F</em>N):

      • Fgx=71 kg×9.8 m/s2sin⁡(18°)≈214.9 NF_{gx} = 71 \text{ kg} \times 9.8 \text{ m/s}^2 \sin(18°) \approx 214.9 \text{ N}

      • fs,max=0.15×662 N≈99.3 Nf_{s,max} = 0.15 \times 662 \text{ N} \approx 99.3 \text{ N}

      • Finding that F{gx} > f{s,max} (214.9 N > 99.3 N) leads to the conclusion that the sled will slide down.

      • Calculate kinetic friction: f<em>k=μ</em>kFN=0.095×662 N≈62.99 Nf<em>k = \mu</em>k F_N = 0.095 \times 662 \text{ N} \approx 62.99 \text{ N}

      • Apply Newton's 2nd Law along the incline: ΣF<em>x=ma</em>x=F<em>gx−f</em>k\Sigma F<em>x = ma</em>x = F<em>{gx} - f</em>k

      • $$