Physics Problems: Friction and Inclines

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Suppose we take a simple object and put it on an incline plane and let it go. What's going to happen? Simple application of Newton's second law. And you just draw the force diagram, and the answer is it accelerates down the ramp. And the acceleration is given by the formula g sine theta, where theta is the angle of the ramp. Let's just put it on. The formula, which is easily derived, is clearly not working. The acceleration appears to be zero. That's because that formula was derived in my mind assuming no friction. Always assuming no friction when I'm trying to think about problems in a simple way. In the real world, friction is clearly an issue that you have to think about. And can we figure out what's the story when we include friction? The answer is sure. It's just an application of Newton's second law.
First of all let's just watch what happens. If I tilt this thing up to a slightly higher angle, it's still sticking, and a higher angle, still sticking, still sticking, still sticking, and all of a sudden at a certain critical angel, finally it breaks free, and it starts to accelerate down the ramp. And it would be nice to be able to work out the details. For instance, what is that critical angle? Can we figure it out from first principles?
All you've got to do is apply Newton's second law. It's the usual story. You can solve, basically, any problem. Let's draw the force diagram and think about what's going on. Here's the force diagram. I've got a ramp. It's at some angle theta. There's a normal vector. Remember the normal vector is perpendicular to the surface, whether the surface is frictionless or frictionful, I don't know if that's a word. The normal vector is perpendicular to the surface. And the weight of the object, that's mg, is straight down. And in an ideal world, these would be the only force vectors. And if you add these two vectors to find the net force, you can sort of see with your eye that they add up to something that points down the hill. And so the net force is along the hill, and the object should accelerate at any angle.
If there's friction, and it's stuck, what do you have to do? You just add a friction vector. And which way does it point? Well it's static friction. Friction always is parallel to the surface. And how big is it? Well I can draw an arrow, but I really have to compute how big it is using Newton's second law. If it's sticking, the net acceleration is zero, and so this arrow will adjust itself so that the sum of all three arrows is equal to zero. No net force, it sits there stuck. As I increase theta, the length of this arrow will increase, because the steeper the angle, the more the object wants to slide down the hill, and the more friction will have to be present, static friction, to hold it up. Static friction works that way. The harder you push on something, as long as it's still sticking, the harder static friction pushes back.
So what would be the critical angle? That would be the angle at which static friction has gotten as large as it can possibly be. And if you increase the angle just an iota beyond that, then static friction can't get any bigger than it's maximum, and it'll break free. Remember the formula for the maximum static friction. The maximum that static friction can ever be is given by a number, mu[s], which depends on the coefficient of friction between say wood and aluminum, in this case. It's a number that you just have to look up or know or measure, times N, the normal force. It's not times the weight. It's times the normal force, the contact force, between the surface and the object.
So knowing the maximum force is going to allow us to compute the maximum angle. So what do we have to do? We're going to use Newton's second law. And it's the usual story. On an incline plane, we'd really like to work in components, because we're going to have to look at the x and y components separately. I don't think you want to pick x is horizontal and y is vertical. That's a usual convention in flat problems, but when you have an incline plane, it's typically much better to pick x along the direction of the incline, and y to be perpendicular to that. And the reason is that we know the object is going to move in the direction of the incline. And so that way, all the motion is going to be in this direction, the x direction. It won't rise up off the surface, or sink down into the surface. There won't be any acceleration in the y direction. So Newton's law in the y direction will be particularly simple.
Look at the picture. Newton's law in the y direction will involve n, which is pure y, and a component of the w vector. Friction is irrelevant. Whether or not there is friction doesn't affect the formula for the normal vector. So let's do the y direction first, because then we'll know the normal vector. And then we'll know what friction is. And then we can move into the x component. So all I need is this piece, the y projection or component of the weight vector.
So let's look at that picture blown up a little bit. Here's the weight vector pointing straight down. I've labeled it by its magnitude, W. It's perpendicular to horizontal. And this angle theta is the angle of the plane. And then I project this vector has a W[x] projection in this direction, labeled by its length, and the W[y] labeled by its length. And the arrows tell me that this down arrow has got a positive x component, and a negative y component. W[y] is pointing in the opposite direction from what we are calling plus y.
It's just geometry. This picture is a little confusing. It's worth drawing it yourself and thinking about it. This is the angle theta that's given. This one has not been given, and I have to do a little geometry. The way I usually think of it is, let's see, this triangle is a right triangle, so this is 90 minus theta. And then this together is 90 degrees. So that leaves this one is theta. But just think about it yourself and convince yourself that this angle is theta in this little triangle, now we're all set. W is mg. And so I can figure out W[y], it's just mg cos theta. And W[x], opposite, mg sin theta.
So let's go ahead and write Newton's second law in the y direction. Sum of the forces in the y direction should equal to m times a in the y direction. And a in the y direction, of course, is zero, because it's not moving in the y direction. What have we got in the y direction? Look at our force diagram. We've got plus N, and then we've got the y component of W, which we just worked out. It's negative mg cosine of theta. So there's our formula that we wanted for N, , and warning it's not mg, normal force doesn't have to be mg. You've got to work it out in the problem. And that's what we're going to need for the motion in the x direction. So let's look at motion in the x direction. Sum of the forces in the x direction is equal ma[x]. If there was no friction, this formula would tell me the acceleration in the x direction. But there is friction. We're imagining slowly ramping up theta, and we're looking for the critical angle. It's the last angle before it breaks free. So if it hasn't broken free, it's still just sticking. A[x] is still zero. So as long as we're sticking, the right-hand side is going to be zero. What's the sum of the forces in the x direction? Well I've got a force of friction, and it's in the negative x direction. So that's going to be negative f[s]. And then I've got the x component of the weight, which is just mg sine theta. And that's it. There's no x component of the normal vector. And now what's f[s]? If theta is an arbitrary theta, this would be the equation I would use to solve for static friction. But if we're at the critical angle, that's the maximum theta, let me add f[s] to both sides, f[s] is mg sine theta. This will be true at any angle. At the critical angle, that means the maximum f[s]. As theta gets bigger and bigger, you can see that this formula is getting bigger and bigger. The maximum value of f[s], when this is critical, would be mu[s] times N. That's the formula. And N, we've already worked out is mg cosine of theta. And since we're at the critical angle, it's cos theta critical.
So here's the formula we were after, which connects theta and friction. And notice something interesting. This is an equal sign. N is on both sides. It cancels. Kind of curious, it doesn't matter how heavy the cart is, the angle at which it starts to break free is independent of the mass, even independent of g. I can do this little demo on the moon, and the critical angle where it breaks free would be the same. I could divide both sides by cosine theta critical to get a nifty little formula. Sine over cosine is tangent. Tangent theta critical is equal to mu[s]. That's a nice little formula. It tells me, for instance, that if I want to measure static friction coefficients, I can just set up this little demo, tilt it, and when I get to the critical angle, that will immediately tell me what's the coefficient. If I already know the coefficient of static friction, then I can use this formula to predict the critical angle.
What would happen if I tipped it just a little bit past the critical angle? Well we've pretty much done all of the work that we need to do. The force diagram is going to look basically the same. There's a subtle difference. If we've tipped theta past critical, it's starting to slide down the ramp. So its motion is to the right. The friction is no longer static friction. Now it's kinetic friction. What's the direction? Is it like this or like this? Well if it's broken free, and it's sliding down the ramp, kinetic friction always opposes motion, so you have to know what the direction of motion is before you can draw this arrow. And kinetic friction is easier than static friction. We know exactly what this frictional force is. The kinetic friction force in magnitude is just mu[K] times N. And the N story is the same as before. It's mg cosine theta. And so we know what this arrow is. So now I can write down the x components of Newton's law. I would get minus mu[K] times N, mg cos theta, that's the frictional force, plus mg sine theta. That's the downward component in the x direction actually. That's the x component of the weight. And that is going to equal to ma[x]. And so here's my formula to solve for the acceleration. And indeed, there will be an acceleration. As soon as it breaks free, it starts to accelerate down the ramp.
Adding real life friction to Newton's second law problems is really a fairly straightforward matter. There's nothing fundamentally different about adding friction compared to any other forces that you put into force diagrams. You draw the diagram. One thing about friction is you have to think a little bit carefully about the direction of the friction. It's going to oppose motion if you're moving. And it's going to adjust itself if you're static so that you'll stay static. There's no formula for static friction. There's only a formula for the maximum static friction. And once you have got all of that stuff set up, you can solve real life two or even three-dimensional dynamics problems including friction.
Dynamics
The Forces of Friction
Problems of Friction and Inclines Page [1 of 2]