Phys1101 - Introductory Physics 1
Phys1101 - Introductory Physics 1
College of Liberal Arts & Sciences

  • Introduction
  • Lecture 01
  • Lecture 02
    • Lecture 2, Part 1: Announcements
    • Lecture 2, Part 2: Units
    • Lecture 2, Part 3: Vector Introduction
    • Lecture 2, Part 4: Adding Vectors Graphically
    • Lecture 2, Part 5: Vector Addition Examples
    • Lecture 2, Part 6: Vector Component Introduction
    • Lecture 2, Part 7: Trigonometry
  • Lecture 03
    • Lecture 3, Part 1: Introduction
    • Lecture 3, Part 2: Where Were We
    • Lecture 3, Part 3: Vector Components in Detail
    • Lecture 3, Part 4: Scalar Component Description
    • Lecture 3, Part 5: Example of Finding Scalar Components
    • Lecture 3, Part 6: Scalar Component Addition
    • Lecture 3, Part 7: Scalar Addition Example
    • Lecture 3, Part 8: Motion Diagrams
  • Lecture 04
    • Lecture 4, Part 1: Introduction
    • Lecture 4, Part 2: Where Were We
    • Lecture 4, Part 3: Location Location Location …
    • Lecture 4, Part 4: How Fast ??? What Direction ???
    • Lecture 4, Part 5: Speeding Up? Slowing Down?
    • Lecture 4, Part 6: What Happens at a Turning Point?
  • Lecture 05
    • Lecture 5, Part 01: Introduction
    • Lecture 5, Part 02: Where Were We
    • Lecture 5, Part 03: Big Picture:  1D Kinematics
    • Lecture 5, Part 04: Kinematic Problem Solving Steps
    • Lecture 5, Part 05: Example 1
    • Lecture 5, Part 06: Example 2
    • Lecture 5, Part 07: Example 3
    • Lecture 5, Part 08: Free Fall
    • Lecture 5, Part 09: Free Fall and Kinematic Equations
    • Lecture 5, Part 10: Example 4
    • Lecture 5, Part 11: Example 5
  • Lecture 06
    • Lecture 6, Part 1: Introduction
    • Lecture 6, Part 2: Where Were We
    • Lecture 6, Part 3: Reading Quiz
    • Lecture 6, Part 4: Graph Basics
    • Lecture 6, Part 5: Practice Makes Perfect…
    • Lecture 6, Part 6: The Tangent Line
  • Lecture 07
    • Lecture 7, Part 1: Introduction
    • Lecture 7, Part 2: Where Were We
    • Lecture 7, Part 3: 2D Motion Diagrams
    • Lecture 7, Part 4: Trajectories
    • Lecture 7, Part 5: Why Work With Components…
    • Lecture 7, Part 6: Key Vectors in 2D
    • Lecture 7, Part 7: Watching 2D Motion
    • Lecture 7, Part 8: Dropping Versus Firing…
  • Lecture 08
    • Lecture 8, Part 1: Introduction
    • Lecture 8, Part 2: Where Were We
    • Lecture 8, Part 3: 2D Kinematic Problems:  The Big Picture
    • Lecture 8, Part 4: 2D Kinematic Problem Solving Steps
    • Lecture 8, Part 5: Example – Part a
    • Lecture 8, Part 6: Example – Part b
    • Lecture 8, Part 7: Your Turn
  • Lecture 09
    • Lecture 9, Part 1: Introduction
    • Lecture 9, Part 2: Where Were We
    • Lecture 9, Part 3: What is Special About Projectile Motion?
    • Lecture 9, Part 4: Example Part a
    • Lecture 9, Part 5: Example Part b
    • Lecture 9, Part 6: Example Part c
    • Lecture 9, Part 7: Your Turn
  • Lecture 10
    • Lecture 10, Part 1: Introduction
    • Lecture 10, Part 2: Where Were We
    • Lecture 10, Part 3: Dynamics:  Why Does Velocity Change?
    • Lecture 10, Part 4: Physical Interpretation of Newton’s Laws
    • Lecture 10, Part 5: What is a Force?
    • Lecture 10, Part 6: Mathematics of Newton’s 2nd Law
  • Lecture 11
    • Lecture 11, Part 1: Introduction
    • Lecture 11, Part 2: Where Were We
    • Lecture 11, Part 3: Free Body Diagram and Vector Nature of Newton’s 2nd Law
    • Lecture 11, Part 4: Common Forces:  Weight
    • Lecture 11, Part 5: Common Forces:  Tension
    • Lecture 11, Part 6: Common Forces:  Normal Force
    • Lecture 11, Part 7: Common Forces:  Friction
    • Lecture 11, Part 8: Problem Solving Steps
    • Lecture 11, Part 9: Example
  • Lecture 12
    • Lecture 12, Part 1: Introduction
    • Lecture 12, Part 2: Where Were We
    • Lecture 12, Part 3: Example 1
    • Lecture 12, Part 4: Example 2
    • Lecture 12, Part 5: Example 3
  • Lecture 13
    • Lecture 13, Part 1: Introduction and Where Were We?
    • Lecture 13, Part 2: Why/When Do We Need Newton’s Third Law?
    • Lecture 13, Part 3: Newton’s 3rd Law
    • Lecture 13, Part 4: Changes To Our Problem-Solving Steps
    • Lecture 13, Part 5: Example 1
    • Lecture 13, Part 6: Ropes and Pulleys
    • Lecture 13, Part 7: Example 2
    • Lecture 13, Part 8: Your Turn
  • Lecture 14
    • Lecture 14, Part 01: Introduction
    • Lecture 14, Part 02: Where Were We ?
    • Lecture 14, Part 03: Uniform Circular Motion:  What You Need To Know
    • Lecture 14, Part 04: Example 1
    • Lecture 14, Part 05: Example 2
    • Lecture 14, Part 06: Example 3
    • Lecture 14, Part 07: Optional Roller Coaster Example
    • Lecture 14, Part 08: Satellite Example
    • Lecture 14, Part 09: The Universal Law of Gravitation
    • Lecture 14, Part 10: Satellite Example Continued
  • Lecture 15
    • Lecture 15, Part 1: Introduction and Where Were We?
    • Lecture 15, Part 2: Energy Conservation:  The Basics
    • Lecture 15, Part 3: How Do You Calculate the Net Work?
    • Lecture 15, Part 4: New Problem Solving Steps
    • Lecture 15, Part 5: Example 1
    • Lecture 15, Part 6: Example 2
    • Lecture 15, Part 7: Last Example
    • Lecture 15, Part 8: Final Quiz Questions…
  • Lecture 16
    • Lecture 16, Part 1: Introduction and Where Were We?
    • Lecture 16, Part 2: Defining Our New “Energy Conservation Starting Equation”
    • Lecture 16, Part 3: Defining Mechanical Energy
    • Lecture 16, Part 4: New Problem Solving Steps
    • Lecture 16, Part 5: First Example
    • Lecture 16, Part 6: Second Example
    • Lecture 16, Part 7: Last Example
    • Lecture 16, Part 8: Redo Example From Last Lecture
  • Lecture 17
    • Lecture 17, Part 1: Lecture
  • Lecture 18
    • Lecture 18, Part 1: Introduction and Where Were We?
    • Lecture 18, Part 2: Momentum Change of a Single Object
    • Lecture 18, Part 3: Conservation of Momentum
  • Lecture 19
    • Lecture 19, Part 1: Introduction and Where Were We?
    • Lecture 19, Part 2: Let’s Start With Another Example
    • Lecture 19, Part 3: Elastic Collisions
    • Lecture 19, Part 4: Remaining Quiz Questions
  • Lecture 20
    • Lecture 20, Part 1: Introduction and Where Were We?
    • Lecture 20, Part 2: Rotational Kinematics:  The Basics
    • Lecture 20, Part 3: Examples
  • Lecture 21
    • Lecture 21, Part 1: Introduction and Where Were We?
    • Lecture 21, Part 2: Describing Motion ALONG the Circular Path…
    • Lecture 21, Part 3: Examples
    • Lecture 21, Part 4: Rolling Motion
  • Lecture 22
    • Lecture 22, Part 1: Introduction and Where Were We?
    • Lecture 22, Part 2: A Net Torque Causes Angular Acceleration
    • Lecture 22, Part 3: Torque Example
    • Lecture 22, Part 4: Equilibrium Example
    • Lecture 22, Part 5: Moment of Inertia
    • Lecture 22, Part 6: Non-Equilibrium Example
    • Lecture 22, Part 7: Another Example
  • Lecture 23
    • Lecture 23, Part 1: Introduction and Where Were We?
    • Lecture 23, Part 2: The Basics of Oscillatory Motion
    • Lecture 23, Part 3: Hooke’s Law
    • Lecture 23, Part 4: Kinematics of Simple Harmonic Motion
    • Lecture 23, Part 5: Example
  • Lecture 24
    • Lecture 24, Part 1: Lecture
  • Lecture 25
    • Lecture 25, Part 1: Introduction
    • Lecture 25, Part 2: The Basics of Wave Motion
    • Lecture 25, Part 3: Motion of a Particle on a Wave
    • Lecture 25, Part 4:  Motion of The Wave Crest
    • Lecture 25, Part 5: Examples
Lecture 14 » Lecture 14, Part 06: Example 3

Lecture 14, Part 06: Example 3

https://youtu.be/DGXZ0j32mcA

PHYS 1101: Lecture Fourteen, Part Six

Hey. Here’s a more complicated example now for us to work out. Sarah puts a box in the trunk of her car. She’s driving along, and then later, she drives around an unbanked curve. That just means flat. The road stays flat as the car goes around, as it goes around a corner. The radius of that curve is 48 meters, about 48 yards, about a half of a football field.

That’s a very gradual corner that she’s taking. The speed of the car as she goes through the curve is 16 meters per second. The box that’s in her trunk, it remains stationary, doesn’t slide in her trunk. We have to determine the minimum coefficient of static friction for the box on the floor of the trunk, for this to be the case.

If her carpet is too smooth, in essence, the car is going to make the corner, but the box won’t. It’ll want to keep going straight. So, it’ll slide in the trunk. But, if the carpet is rough enough, the friction will be enough that the box, like the car, will round the corner and follow this circular motion.

Okay. Step number one is deciding what object to focus on. This is an example where that can be a bit complicated. Always, always focus on the question it’s directly asking you, to focus in on what object you need to consider.

They want the minimum coefficient of static friction for the box on the floor of the trunk. Okay. That’s about a force that is between two objects, the box and the trunk. We got to pick one of those objects. The box is nice to consider because I can imagine the box and cleanly picture what forces are on the box.

If I picture the trunk, well, the trunk’s attached to the car. It’s attached to the road. Less obvious to me that I can get there. I sense I’m going to run into trouble. There’s going to be forces I don’t know about to try to solve it if I focus on the trunk.

My first inclination, hone in on the box. That’s our object. So, there it is. My red box is my object.

Question 19, easy points. A, B, C, or D. You are tempted to think, perhaps, about Sarah, the trunk, the car. You want to focus on the box. Okay. After getting that clearly in your mind, it’s now all about the box. What’s the acceleration for the box? And what are the forces on it?

Question 20: As the box, like the car, rounds this corner, from the side view at this instant, what’s the acceleration?

Question 21. I’m now thinking about the forces on the object, the forces on the box. Which choice looks best to you? I know I’ll have a force of gravity down. I have a normal force up. What else? The normal force, of course, comes from the floor of the trunk that’s in direct contact with the box. A surface always pushes up, but a surface can also be responsible for frictional force. It can keep an object from sliding. What free body diagram captures what has to be going on with the box?

I have to have weight. It’s sitting on the trunk floor. So, I have to have a normal force. The floor is the only thing in contact with the box. So, there can be only one other force. It has to be friction.

Okay. Here’s my list of knowns for this problem. I was given the radius of this corner. This is the radius of the circular motion that this object, the box, is undergoing. This is the speed of this box as it rounds the corner, same as the car. I want the coefficient of static friction. It asked for the minimum coefficient required to keep the box from sliding.

Remember, when we think about a coefficient of friction, ultimately, we’re going to be starting with Newton’s Second Law. So, it’s going to be the frictional force that we’re going to work with. Remember, for static friction, static friction can be less than or equal to a maximum value. That maximum value is the product of the static coefficient times the normal force, the measure of the load that the surface is supporting, how pushed together the two surfaces are.

So, if I want the minimum coefficient of friction, I really want the case where I can consider that the static friction is doing the best that it can. It’s at this maximum possible value to keep the box from sliding. So, I’m going to be focused on using this for my static friction. The variable I’m going to go after is this coefficient. The starting place for the math, Newton’s Second Law.

Let’s do a coordinate system here. Plus y up plus x to the right. I’m inclined to start with the horizontal expression because my acceleration vector, I know, has to point to the center of the corner. That’s horizontal. My only force that is horizontal has to be static friction. I know this because the only thing in contact with the box is a surface, and a surface can only have two possible forces. Surfaces always push up, perpendicular with the normal force. Then, I can have friction.

Friction has to be pointing to the inside of the corner because I have to have a net force in this direction. I know this is the direction of the acceleration for the box. So, there I have it represented.

Now, I begin customizing this equation. Ax… I’ll just rewrite it. Let’s plug in. What do we know about forces for this problem? Well, I have plus the static friction divided by m. There’s only one force here to plug in. I still don’t see the variable yet that I need to solve for. So, I have to keep working the problem. What about a? Do I know a number for a? I don’t, but I do know that I could substitute in v squared over r for a. Let me draw it here off to the side, that I also know that the static friction, I can substitute in the coefficient times the normal force.

How big is the normal force? Let me go back to my problem and have a look. Acceleration is horizontal. Forces up/down have to balance.

My Ay equation tells me what I see, visually, in my force diagram here. The normal force is the only upward force; therefore, it has to completely balance and equal mg. So, for this problem, I know fn is equivalent to mg. So, I make that substitution in that equation. Now, I’m going to take this information back over and go back to my main equation and keep working it.

For fs, into my Newton’s Second Law equation, I now am simplifying more to substitute in that Mu sub s times mg. I have to divide it by the mass of the box again. I notice that the mass drops out. I’m thankful for that because I wasn’t given a mass for the box.

I do know this is the variable I want, that I need to focus on. I have numbers for everything else. I’m set. Just need to do a little algebra. If you rearrange that, you can check my algebra again more carefully. I just need to divide by g on both sides. This coefficient that I’m after is equal to v squared over r times g.

Plug in numbers for that. I have 16 meters per second squared divided by 48 meters divided by 9.8 meters per second squared. Plug that into my calculator, and I ended up with 0.54.

The units cancel. I have 2 meters because it’s squared too in the numerator. That’s going to cancel both of those meters. Then, I have two seconds squared here that are going to cancel with those two seconds squared. These coefficients should always be unit-less and between zero and 1. So, this looks pretty good. That’s checked about as carefully as I can check it.

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