Ch 05: Friction, Inclines, SystemsWorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Example #1: Kinetic Friction Problems

Transcript

Hey guys so I want to go over a few motion problems with friction. So we're going to combine the idea friction with the three or four equations of motion or kinematic equations that we covered earlier. Now all of these I'm gonna have simply a box or crates moving on the surface so it's going to be sliding on the surface. So we're really going up against kinetic friction on all these examples. Let's check it out. All right. So on this first one here I have a 10 kilogram box on a horizontal surface and it is initially at rest let me draw that. 10 Kilogram block. It is initially at rest on the horizontal surface when you pull on it with a force of 80. In the x axis so it looks like this. It moves. Now since I'm telling that it moves you don't have to check if the force is enough to overcome kinetic static friction. You just know that it won't move. Therefore. It will have a kinetic friction opposing it. So it wasn't moving. And then it starts moving in this direction which means that there was an acceleration in this direction and I'm going to call this the direction of positive. If meu points five. Find a speed after you pull on it for six seconds. OK. So I want to know Vfinal after six seconds.

So this is a motion problem because I'm looking for. Final velocity given time. And you might notice you have a lot of motion variables here. Right. For example the initial velocity is zero fine velocity is what I'm looking for the time between these two. Is six seconds. Now there are five variables remember that and the other two here that are missing are a and delta x. Now to solve this problem. Remember also I need to know three or five variables if you look around. I only know. Two. Now. This is what we're looking for so that's not going to be it. My third variable is going to be either a or Delta x. The only way I can get out is using one of the motion equations themselves. So I'm kind of stuck. I can't get delta x yes what I can get though outside of motion outside of kinematics is a because I can find a is if the final forces using F equals ma so instead of kinematics I can use dynamics . So that's what we're going to do. Delta x will be ignored variable but I will be able to find a third variable to be able to solve this and to find the way. To find a. I'm going to use F equals ma, the sum of our forces equals ma, I'm going along the x axis so I can write that because I mean the x axis. So you didn't have to put the axis because there's nothing happened in the wire. So the only acceleration you do have is in the x. So that's kind of optional. OK. What are the forces I have 80 in the opposite direction. I have F in the positive direction and I have. Friction kinetic in the negative direction.

The mass is 10 and I'm looking for the acceleration. OK. So friction kinetic is meu kinetic normal. I only have one meu which means this is both kinetic and static. That's the one I will use. Point five normal but I have to figure out what normal is. Now in this case I have a block. That's been pulled down by an mg. The mass is 10. If I use gravity is 10 I get that the mass. Of the mg is simply one hundred. And since there are no other forces in the y axis I can say that normal must equal mg so that they cancel. So normal is 100 as well. Okay That's what goes here 100 from normal. And this gives me a friction of 50. So friction's 50 so I can now plug this and this is 50 here, 80 positive because it's the right, 50 negative because it's to the left, 10a. And if I solve for this I get that the acceleration is three meters per second square. I got a positive make sense and point to the right that's positive. It moves to the right. That's positive. So acceleration is three I now know three out of five variables. And I can just pick a motion equation one of these three guys here to solve it three or four guys remember some professors only want you to use the top three alright. Delta x is my ignored variable the only equation that doesn't have Delta x is the first one. So going to be able to use that first equation to solve this. Vfinal equals Vinitial plus at, Vfinal equals Vinitial zero acceleration is three and T is six. So this part's pretty straightforward. And this is just 18 meters per second. And that's the end of this one. I want you to try a practice one which is very similar to this and hopefully you get.

Practice: A 15 kg block moving on a frictionless horizontal surface with a constant 40 m/s enters a long, rough patch. If the coefficient of friction between the box and the patch is 0.7, calculate the block’s total stopping distance.

Practice: A 1,000-kg car leaves a skid mark of 80 m while coming to a stop. If the coefficient of friction between the car and the road is 0.7, find the car’s initial velocity before braking.

Example #2: Kinetic Friction Problems

Transcript

All right so here I'm launching a block a horizontally on the surface so basically I'm giving this block an initial velocity of ten. And the block covers a distance of twenty meters in a time delta T of three seconds now here might be tempted to think that this implies that the final velocity zero but it actually isn't OK and if you look around I have three variables if you want you can actually calculate what the final velocity is and you'll be able to see that it's not zero so just to be explicit I'm going to put here does does not mean that the final velocity is zero ok, the fifth variable that I'm missing here is acceleration the question asking for the coefficient of friction between the block and a surface now the coefficient of friction is going to be kinetic because the block's moving so that's the only one I could find out of this problem so how would I go about finding this well you might think where do where is the you know where can I find of coefficient of friction what equation is a show up right obviously I have to make it show up in the equations I can solve for it well the question of friction only shows up is here, friction is meu normal, so I have to find friction so I can find meu k and how can I find friction a problem like this, well friction is a force so I can use F equals ma right now you gave this block an initial velocity this way. But once you did that there's no other forces pushing forward right and there's friction so friction is going to be to the left and that's going to be the only the only force in the X. axis will be friction kinetic I have mg down here and five kilograms times I only use gravity is ten so this is fifty and I have normal up here, fifty as well. OK So for me to find friction I can simply write since I don't have meu I'm not going to be able to figure it out this way right I'm going to have to write F equals ma.

Since I don't know these two guys the only way to get out of here is to write F equals ma, the only force is friction. It's negative because we're going to say they're going to the right is positive and this equals ma five a so I can expand friction into meu k normal since what I want is actually not friction I want meu k and that equals five a, now I know normal but I don't know way and I have to have both of these guys to be able to find meu but I can't find a using kinematics or motion equations because I know three out of five variables, so here my ignored variable is the final velocity so that tells me that I can use the third equation to find acceleration so I will find acceleration and I'm going to use the third equation delta X. equals Vot plus a half of a T. squared. Delta axis twenty the initial velocity is ten times three.

And acceleration is what we're looking for here. Ok So this is thirty I'm going to move this over here twenty minus thirty then this is three square for this nine divided by two four and a half. And if I move things around I get the acceleration is negative two point two two meters per second squared makes make sense and have a negative acceleration. Now I'm ready to go and plug this back in here OK, meu k is normal which is fifty times five times acceleration which is negative two point two two. There's a negative here that I almost forgot right. I noticed that because meu was about to be negative if I didn't have the negative there so those two negatives will cancel and I can just move over the fifty so I have five two point two two divided by fifty and the answers point two two two. That's my meu, remember meu is unit less so that's your final answer here all right that's it for this one, let me know if you have any questions.

Example #3: 2D Push / Pull with Friction

Transcript

Hey guys. So I want to show you some problems of two dimensional forces with friction. So we're going to be either pulling or pushing on the block and there will be friction as well. So let's check it out. So remember first of all the forces at an angle must be decomposed into the x and y components. Also realize that force is applied in the y axis if you have a forces that are applying on the Y-axis that will affect your friction and the reason is force in the y axis will affect normal and normal will affect friction. OK let's check it out. So here a 10 kilogram block is initially at rest. So initial velocity is zero on a flat surface right there the coefficient of friction between the block and the surface are point five point six. You pull the up with 100 Newtons so I'm going to draw like this. F equals 100 at positive 37. Right. Calculate all the forces on the block and then calculate the blocks ax and ay. So the first thing you will do here is decompose this F, Fx and Fy. Fx is F cosine of theta so it's 100 cosine of 37 if you play this in the calculator you get 80, Fy is 100 sine of 37. And this gives you a sixty. Ok. What other forces do I have. I have an mg pulling down. mg will be. 10. I'm going to use gravity 10 as well so mg equals 100. Do I have normal force here. I do. Because even though I'm pulling the box up with 60 that's not strong enough to lift it from the floor. OK so I can write here that since I don't want to have to justify it I could just calculate the normal here but I can write since. Fy is less than mg. There is a normal force. Ok And also not there's only a normal force. But we know now that's the acceleration in the y axis must be zero because box isn't getting lifted. It also doesn't break through the ground. Right. Normal here would be 40. And that's because remember top cancels with bottom. So if you want to write that you can write normal plus Fy. Those are the two forces going up have to equal mg that's the force going down. And then you can use this to calculate normal to be 40. Ok. Normal's forty. There's a coefficient of friction so this is a rough surface and there's a normal. Remember friction is meu normal. So I have a meu and I have a normal soof friction. And if I'm pulling to the right friction will pose then that tends to move the box to the right. It will pose it to the left here with friction. Now we have to figure out whether this friction is enough whether this force here is enough for me to move. So one way to do that because I can just calculate friction static to see if this box will move at all friction static is meu static normal meu static is the bigger of the two here so it's point six times normal which is 40. This is 24. Ok. Let me calculate kinetic real quick, kinetic is point five 40. So it's 20. All right to determine if it moves I have to see is this number greater than this number in the race. So again here I can write since Fx is greater than static friction max. Remember this is my static fiction Max. Right. This means that there will be an acceleration in the x axis the acceleration in the x axis is not zero. This object will move it also means I'm going up against kinetic friction instead. OK. So this will be kinetic friction because I'm pulling hard enough. So it will actually be a 20 over here. Right. And I want to know the acceleration in the x and y, I already know that the acceleration in y would be zero because the forces are canceling their. To find the acceleration of the x axis I'm gonna write F equals ma, sum of all forces in the x equals mAx and Ax is what I want. The forces in the x axis are these two green forces over here. So it's Fx and friction. So if you want you can write positive Fx plus negative friction kinetic. Mass is 10 and Ax is what I want, this is 80 positive to the right. And this is 20 to the left. So if you do this you've got the acceleration that is six meters per second squared. Notice how. The fact that pull at an angle gave me an Fy. That's the big point here. This Fy change my normal if it wasn't for this Fy normal would have been 100. And this changing normal affected the change in both friction static and kinetic. OK. So that's it for this one. I want to try the next question which is very similar except that we are pushing down the object instead of pulling it. So let's give this a shot.

Practice: A 5 kg block is initially at rest on a flat surface. The block-surface coefficients of friction are 0.4 and 0.5. If you push on the block with F = 130 N directed at –53°, calculate its acceleration (select a = 0 if the block does NOT move).

Example #4: 2D Push / Pull with Friction

Transcript

Hey I so hope you try this one and you got it right ten kilograms walk here it's moving with an initial velocity of twenty when you get to the bottom of a long smooth inclined plane that makes an angle of fifty three with the horizontal, I put all the information in the picture so over here my initial velocity is twenty and I want to know how far up the plane will go how far up the plane remember this is now my x axis so I want to know what is my Delta x. Before it switches position or switches directions if it's switching directions because it's going to the top stopping and coming back so the idea is I want to know the velocity until it reaches zero OK. Smooth means no friction along basically just tells you that you don't have to worry about the fact this block might sort of go all the way to the plane and get sort of projectile motion out of here OK it will do the plane's long enough you don't have to worry about that not that you would but that's why they say that to be very precise OK so I'm looking for I'm looking for Delta x. I hope you realize that the way to start this is by writing your motion equation or at least listing of variables because the delta x is a motion variable, Vinitial, Vfinal, a, delta x, delta t.

Initial velocity is twenty final velocity is zero, delta x is what I'm looking for if you look here I know two things but I'm missing a third variable so it's going to be either a or t, well inclined plane problems are almost always forced problems so the idea is that I'm going to be getting my a from F equals ma, sum of all forces along the x axis on the inclined plane equals ma ok, what are the forces, let me draw a very quick free body diagram here here's my x axis, this plane is block is this way mg now mgx is always going to be down never up this way always down this way because gravity is pulling it down and mgx is always into the plane, I'm sorry mgy always into the plane and I have a normal this way, since I'm moving off the plane I'm going to say that that's the positive direction of X. so I'm moving with the positive a positive velocity and if this is the positive direction then this means that this guy here is a negative force and that should make sense it's slowing it down so we're going to call it negative so the only force acting here is actually a negative mgx.

Remember mgx is just mg sine of theta So this means that the acceleration is just negative g sine of theta and I want to show you how this is basically the same thing as it was actually when we did this going down this way right if you do this going down you get a g sine of theta instead of negative g sine of theta ok, let's plug this in and you get a negative eight meters per second squared now that I have eight I can go here and I can find a negative eight right there. I can find the answer OK My ignored variable is delta T so I can use the second equation Vfinal square equals Vinitial squared plus 2a delta x the delta x is what I'm looking for final velocity is zero the initial velocity is twenty the acceleration is negative eight, delta x this is four hundred I'm going to move it over to the left so this is negative four hundred, this is negative sixteen so I move it over here and then I end up with Delta x be twenty five meters. So this thing is going to go twenty five meters up the plane stop and come back down, that's it for this one hope you got it let me know if you guys have any questions.