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.