Physics Exam Review: Forces, Friction, and Circular Motion
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James’ Contact Info
Email: jhowar02@uoguelph.ca
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Time: Mondays 12:30 PM - 2:30 PM
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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 v in a circle of radius r, experiencing centripetal acceleration a_c.
Formula for Centripetal Acceleration:
a_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:
\Sigma Fc = mac = \frac{mv^2}{r}.
Definition of Centripetal Force (F_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 (FG), tension, and normal force (FN) exist independently, and it’s conventional to label the sum of these forces acting towards the center as the centripetal force (F_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: \Sigma Fx = Tx = ma_c
Radial Tension Calculation:
T \cos \theta = \frac{mv^2}{r}
Solve for T:
T = \frac{mv^2}{r \cos \theta}
Substitute values:
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 \approx 9.5 \text{ N}.
2. Example 2: Normal Force on a Car
Scenario: A car with a mass of 1.0 \times 10^3 \text{ kg} traveling at 21 \text{ m/s} crosses over a hill with a radius of curvature 1.0 \times 10^2 \text{ m}.
Determining Forces (part a):
\Sigma Fy = Fg - FN = Fc
FN = Fg - F_c
F_N = mg - \frac{mv^2}{r}
Substitute values:
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 F_N \approx 5.4 \times 10^3 \text{ N}.
Determining Velocity (part b):
Using condition when normal force is zero (F_N = 0):
0 = mg - \frac{mv^2}{r}
This leads to v^2 = gr, so v = \sqrt{gr}
Substitute values to find speed:
v = \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 (F_s): Acts against movement of objects that are at rest.
Kinetic Friction (F_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:
Fs is dependent on normal force FN:
Fs \le \mus F_N
Kinetic Friction Formula:
Fk = \muk F_N
Coefficients of Friction
Materials Examples (Approximate Values):
Steel on Steel (dry):
Coefficient of Static Friction (\mu_s) = 0.74
Coefficient of Kinetic Friction (\mu_k) = 0.57
Waxed wood on wet snow:
\mus = 0.14, \muk = 0.10
Ice on ice:
\mus = 0.1, \muk = 0.03
Synovial fluid in human joints:
\mus = 0.01, \muk = 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{s,max} = \mus 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 (F_N)
Observation: The normal force FN is often equal and opposite to gravitational force Fg or 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 \mus = 0.69 and \muk = 0.60.
Determine if the fridge moves:
Sum forces in y-direction:
\Sigma Fy = FN - Fg = 0 \Rightarrow FN = mg
FN = 117 \text{ kg} \times 9.8 \text{ m/s}^2 results in FN = 1143.66 \text{ N}
Calculate maximum static friction:
F{s,max} = \mus FN \Rightarrow F{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: \mus = 0.1, \muk = 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{s,max}: Fs = 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: \mus = 0.55 and \muk = 0.50.
Questions: (a) Does the box move? (b) If it moves, what’s the acceleration?
Resolve forces vertically (y-direction):
\Sigma Fy = 0 = FN + F{ay} - Fg
FN = mg - F{ay} = (3.3 \text{ kg} \times 9.8 \text{ m/s}^2) - (21 \text{ N} \sin(31°)), leading to F_N \approx 21.52 \text{ N}.
Calculate maximum static friction:
F{s,max} = 0.55 \times FN = 0.55 \times 21.52 \text{ N} \approx 11.8 \text{ N}.
Calculate horizontal component of the exerted force:
F_{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: Fk = \muk F_N = 0.50 \times 21.52 \text{ N} \approx 10.76 \text{ N}
Resolve forces horizontally (x-direction) for acceleration:
\Sigma Fx = m ax = F{ax} - Fk
ax = \frac{F{ax} - F_k}{m}
a_x = \frac{18 \text{ N} - 10.76 \text{ N}}{3.3 \text{ kg}} \approx 2.2 \text{ m/s}^2
Summary of Friction
Static Friction (F_s):
Acts to prevent movement of stationary objects.
Maximum static friction defined as F{s,max} = \mus F_N.
Kinetic Friction (F_k):
Acts on moving objects.
Kinetic friction force given by Fk = \muk F_N.
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/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 (m_1 = 1.20 \text{ kg}), accelerating at 1.20 \text{ m/s}^2.
Calculate mass of smaller glass (m_2):
Using Newton's 2nd Law for the system: \Sigma F = (m1 + m2)a
Fa = (m1 + m_2)a
m1 + m2 = \frac{F_a}{a}
m2 = \frac{Fa}{a} - m_1
Substitute values:
m_2 = \frac{2.50 \text{ N}}{1.20 \text{ m/s}^2} - 1.20 \text{ kg}
Resulting in m_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 (m2) is the contact force from the larger glass (m1):
F{contact} = m2 a
Plugging values: F_{contact} = (0.88 \text{ kg})(1.2 \text{ m/s}^2) \approx 1.06 \text{ N}.
The net force on the larger glass (m_1) is the applied force minus the contact force:
Fa - F{contact} = m_1 a
This confirms that the total applied force Fa = F{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: \mus = 0.15 and \muk = 0.095.
Forces on the incline are primarily gravitational and frictional:
a) Normal force magnitude:
Calculated as: FN = mg \cos(18°) = 71 \text{ kg} \times 9.8 \text{ m/s}^2 \cos(18°) yielding approximately FN = 662 \text{ N}.
b) Acceleration of sled:
Determine if it slides:
Compare gravitational pull down the incline (F{gx} = mg \sin(18°)) and maximum static friction (f{s,max} = \mus FN):
F_{gx} = 71 \text{ kg} \times 9.8 \text{ m/s}^2 \sin(18°) \approx 214.9 \text{ N}
f_{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: fk = \muk F_N = 0.095 \times 662 \text{ N} \approx 62.99 \text{ N}
Apply Newton's 2nd Law along the incline: \Sigma Fx = max = F{gx} - fk
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