https://youtu.be/oDt7AmZTAZg
HYS 1101: Lecture Thirteen, Part Seven
This next example I want to do for you doesn’t involve Newton’s Third Law. It is just a single object that we’re focused on, but its motion along an incline plane and an efficient way of doing problems like this is to use a rotated coordinate system. Let me walk you through how that goes and what you need to do.
Our problem states that we need to determine if the block is going to slide along the slope and if it does, what is the acceleration?
It’s a single object. We’re applying the acceleration is equal to the net force divided by the mass. This is the tail to tip sum of all the forces, leads to the same direction as the acceleration. We know that mathematically we handle that by breaking it down into two one dimensional problems.
So far when we’ve done that, we’ve had motion that is either vertical or horizontal and we’ve used the standard coordinate systems of horizontal and vertical to handle it. What’s critical when we take this general resultant vector form of Newton’s second law and break it up into the two one-dimensional problems is that these two directions are simply perpendicular to each other.
They don’t have to be horizontal and vertical; they just have to be perpendicular. That’s what is critical to separating it out one part of the motion from the other part and leading to two one-dimensional problems.
When should you choose axes that are horizontal and vertical compared to at an angle? Let’s go in for this problem and draw the forces that we’re dealing with. Well we know we’ve got the force due to gravity straight down on this block and we can start on the right side, work our way around the block. Nothing touching it, nothing touching it, nothing touching it, then I get to the surface here and I know I have a normal source. A surface can only push and that has to be perpendicular to the surface. That has to be 90 degrees.
I know that I can have roughness and potential friction. I’ll have static friction if there is no motion, but there’s a tendency to motion. If this plane turned into a sheet of ice, yes indeed the block would slide down and to the right, so there has to be a friction force that’s keeping it from doing that accelerating down.
The first part of this question has us assessing whether the block slides or not, so a good approach to that is to calculate the max this friction force can be, given this coefficient of static friction, and then determine if that’s enough to give us an acceleration that’s zero. If it’s not big enough, and there is a net force down the slope to the right, then I have acceleration along the slope to the right.
Okay. I’ve got all my forces, I have a consideration of the direction of acceleration should I have it.
Let me point out to you the relative orientations of these forces. We have, if you view it horizontal and vertical, we have one, two, three, three forces that are at an angle to horizontal and vertical we would have to apply trigonometry to in order to properly execute these two equations. Picture however, if you rotated your head and you considered this as the vertical axis, this is my plus y, and let’s consider this as my plus x.
Can you picture in this rotated coordinate system now I have one, two, three vectors that are already parallel to an axis and I only have one vectors that is at an angle, that I have to do trigonometry to.
That’s the reason we’re going to go with this tilted coordinate system. It’s just that simple. I’m lazy. I don’t like to do trigonometry. I’m going to do it as seldom as I can. So I’m going to use this tilted coordinate system and I’m going to go in and do trigonometry with this vector as being a hypotenuse and I need to find the components of this vector along an axis that’s in this direction parallel to y and in this direction parallel to x. This is going to be my component w sub y and this is going to be my component w sub x.
Okay. To do our trigonometry on a right triangle, we know we have to get an angle inside that triangle. Here’s my right angle. Usually for these tilted problems along a slope, you’re given the angle from the horizontal of this slope like this 30 degrees that I show you.
Whatever angle you’re given, you just have to do some geometry here to try to bootstrap your way to getting an angle inside this right triangle. As an example, I know this is a right angle. This is 90 degrees. This is 90 degrees. Well, if this is 90 and this is 30, I know the angles inside of any triangle, therefore this triangle, have to add up to 180 degrees. This angle then has to be 60.
Well, 60 degrees is part of the way. This has to be 30 degrees then to make this 90. Let me grab my pink pen here and let’s call that 30. That’s what I mean by bootstrap your way there, you just have to consider angles inside of a triangle have to add up to 180. Where are your right angles? Where are your similar angles? Et cetera.
Now I’ve got one. An angle inside the triangle up here is 30. The more of these you do, you’ll soon realize that with a force that’s straight down, and drawing this kind of a right triangle, this angle becomes this tucked up angle inside this triangle here. Okay. Now with that right triangle I suspect that everyone now is practiced on their trig enough to be able to do that trigonometry to solve for wy and wx given that the hypotenuse has a magnitude of m times g.
We would then be done with our trigonometry, applying it only to this one vector rather than having to apply it to three vectors and that’s why we’re going to go with this tilted coordinate system. I have erased here this important note. Now I have a force picture here broken down for this object. I’m going to use the tilted coordinate system. Let’s go about solving the problem now.
I’m going to solve it in two steps. For part a, or step a, I’m going to calculate the maximum value for the static friction. Then whether the object slides or not is going to be determined by comparing that to, or calculating the value of a x. Given this maximum value of the static friction, what would be fnet?
Let’s just work through those numbers and then think back to the consequences. Let’s start out by going after fs max. Well, the static friction force at most can be the coefficient of static friction times the normal force. Remember we had to multiply by the normal force or it impacted how large the friction was because this fn represents the load, so to speak, that the surface would be bearing, so in turn is a measure of how pushed together these two surfaces are.
Okay. Well our coefficient is given. We know that. How big is the normal force? Remember our rule of thumb for any normal force is that we have to look to Newton’s second law and what it tells us the normal force has to be in order for the acceleration to have no component away from the surface in this perpendicular direction.
So right away I’m looking at my drawing and appreciating that I don’t quite like the length that I’ve drawn fn. It’s not consistent with that scenario. With this magnitude of a hypotenuse for this right triangle representing my gravity force, the component of gravity that’s balanced, or should be balanced with the normal force is only w sub y.
I was a little heavy handed with my normal force I can see. It should only be this long. That is, to determine fn, I’m going to solve, starting with my generic equation a sub y is equal to the balance of the y forces divided by the mass.
I’m not accelerating pulling away from the surface, I have no y component to my acceleration perpendicular to the slope, so my forces have to balance. What do I have? In my picture I have a plus fn and a minus w sub y. Okay. Telling me that the normal force equals the y component of my force due to gravity.
Well, let’s go in and do that. We actually didn’t follow through with our trigonometry and get a real number for our components here. Let’s do that off to the left.
Wy is the adjacent side to this angle where my hypotenuse has a magnitude of the mass times g, so wy is going to be m times g times the cosine of 30 degrees. Putting in my values I have a 10 kilogram mass, times 9.8 meters per second squared and the cosine of 30 degrees.
That gives me a force component that is 84.87 Newtons. At least this is the magnitude of it. Doing my trig to just find the length of this right side, that side of the right triangle. Notice over here when I plugged it into my vector equation, I included the right sign to represent the direction.
What’s my horizontal component? How large is this side of the right triangle? That’s the opposite of that angle so I need to use the sine. Mg times the sine of 30 degrees. Plugging in the 10 kilograms and the 9.8 into that equation and I ended up with 49 newtons.
Okay. With those number now, I know that my normal force is 84.87 Newtons. I’ve taken the sign to represent direction into account so what I represent here by just this variable wy is going to be just the magnitude. The sine is taken care of so I just need to plug in the magnitude here. The normal force has a magnitude of 84.87 Newtons.
I now know the magnitude of the friction force. At least the maximum possible value. It’s the coefficient of static friction, which was .2 for our problem. There’s no units to that coefficient, just a number times my 84.87 newtons gives me 17.3 newtons.
Okay. So the frictional force can be less than this. This represents the maximum possible value. Let’s look at the x components now for this scenario. What does our a(x) equation look like? What are the forces that are aligned to the x axis?
Well, my force due to gravity I have replaced with the two components aligned to x and aligned to y so I can look at those. All I need to consider is the x aligned components. I have minus the static friction and I have plus the x component to gravity. So I have plus the x component of gravity minus the static friction.
Well the x component to gravity is 49 newtons. The maximum friction force I can have is 17.3 newtons, so if I put in that maximum value, I can see that that friction force is definitely less than the part of gravity that’s pulling or accelerating that mass down the slope. It’s not going to stay put.
Had this turned out to be less than the static friction, then my static friction would not need to be the maximum value to balance and give me a(x) of zero. Let me just write that this is not zero so the box slides.
Visually said, another way looking at our picture here, if I had drawn the vector that represents the maximum static friction along the slope, if this represents the component of gravity along the slope, it’s larger than this maximum possible friction and so the friction can’t hold it. The block is going to break free and start moving down the slope.
What’s its acceleration down the slope though? Is it equal to what this number would calculate out to be? The answer to that is no. This calculation tells us simply whether the box is going to break free or not, but it’s a strange thing about friction that when it’s not moving, it’s coefficient will be different than when the surfaces are sliding with respect to each other.
As soon as that block or object breaks free, I have overcome the maximum static friction value. Now I have real sliding, it’s kinetic, and this is the coefficient that I need to use to figure out the value of the frictional force.
So here I’ve copied that picture for you again and I’ve written that out explicitly. Now that we know it will slide down the hill, so we’ve answered the first part of the question, we have to change the static friction into a kinetic frictional force.
I’m going to go over here to my free body diagram and I’m going to make that force smaller and I’m going to call it f sub k for kinetic friction and I’m going to remind myself that the value of that coefficient was .1. It’s going to be smaller because the frictional force is the coefficient times the normal force. The normal force won’t change once the block is sliding. The normal force is always the value of w sub y, but the coefficient now is half as big as it was for the static friction.
Now I can write that it’s 0.1 times the normal force which was w sub y. It’s 84.87 newtons. In fact I could write that the kinetic frictional force is the same as the static divided by 2 because the static coefficient was twice as big. So that static coefficient divided by 2 is .1. Then that maximum value of the static frictional force was 17.3 newtons. That’s the numerator and I just have to divide by 2. That gives me 8.65 newtons.
Okay. We’ve evaluated what the frictional force is as it’s sliding. The last part of the question is what is the acceleration down the slope? That means we are solving for the acceleration component along the slope which we’ve defined to be our x direction. I just need to add up all of the components along the slope, the forces, and divide by m. I know what those are now.
It’s positive wx minus the kinetic friction divided by m. Wx again was plus 49 newtons, my kinetic friction is minus 8.65. Again the sign captures the negative direction of that force. I have to divide the whole thing by the 10 kilograms, the mass. I end up with 4.04 meters per second squared. That will be my acceleration down the slope.