Ch 05: Friction, Inclines, SystemsWorksheetSee all chapters
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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
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Ch 38: Quantum Mechanics

Concept #1: Forces in Connected Objects

Transcript

Hey guys. So in this video I want to show you how to solve problems where we have multiple objects a system of objects that is connected to each other by a rope. And usually we're going to have the rope going through some sort of pulley. And how do we do this let's check it out. So a system of objects is just a collection of objects that move together. And we're going to have some pullies as you can see down here. Right. And the first thing that you need to know is that objects that are connected will have the same velocity and the same acceleration. They move together so they have to have the same velocity and the same acceleration. Ok. So here for these objects instead of calling the velocity of the first object V one and the velocity of second object V two since they're said they're the same. I you just call it V. Right. Another way to think about this or another were another way that they can work this and the problem is the velocity of a system. So if I ask what is the velocity of the first object is the same things as asking what is the velocity the second object or what is the velocity the whole thing. They're all the same thing. Same thing with the acceleration a one and a two are the same. So instead of having two variables a1 and a2 we're just going to call it a. So I have one variable instead and it's easier ok. There are four steps to do this. The first three are sort of the main one of the fourth one that will depend on whether you get asked more information we will get there. But there's a lot of text here right. And you might think you have to remember remember a lot of stuff in but I think once you do this you'll see it's pretty straightforward. Basically what you're doing is just two things writing F equals ma. And then you're drawing a free body diagram and then you're putting it all together which is three things right which is how we always solve force problems free body diagram, F equals ma. But here if you have two objects you just do that twice and then you put it together. So for each object we're going to draw free body diagram. And then we're going to calculate all the forces along the direction of motion along the direction of motion, let me talk about that real quick and I'll use this example here as an example so if you have a pully and two blocks which eve block is the heavier one the system will move in the direction of the heavier one. Right. So here I have a 6 and here I have four. So I know that the system will go sort of the pulley will spin this way. This thing will go down and this thing will go up. OK. We're going to say that that direction is the direction of positive. So the acceleration is this way. So this is the direction of positive. Ok knows how far one object the directional power that is down for you or for the other one is up. And it might seem weird at first but realize that these are really the same direction so it's going like this. Right. So positive is this way. OK. They kind of you can kind of imagine that they sort of connect here and the positive direction is this way. OK. And so. We need to calculate all the forces along the direction of motion. In other words what are the forces along this direction here. We're gonna calculate that. All right. So we'll that in a bit. So and after we do that we're going to compare the forces to determine the direction of acceleration, we already did this. This is heavier than this one. So it's going to go the other way. And you choose the direction of positive to follow acceleration. It's about the acceleration not the velocity. And we already did that as well. OK then we're going to write to sum of all forces equals ma along the direction of motion other words for this block this way. And for this block in this direction.

So both of these cases we're going to write F equals ma for the Y-axis and then we're going to add all the equations to solve for a. Let me show you which is much easier to see. So the first step is we're going to draw a diagram for both of these guys. I have a six kilogram here. And it has two forces acting on it and has its own mg which is just mass six gravity when rounded at 10. And this is a 60 and it has a tension here T. Now you could think of it as there is a tension here T one that touches the m1 and there's a tension here T-2 that touches m2 but because it's the same rope. Right. I can say that these tensions are the same. Now I want to talk a little bit more about that at the end because that's not exactly right but it's good enough for now. So if the same rope will have the same tension. So instead of calling this T1 and T2 since they are same I can just call it T. So istead of two variables I have one variable. All right. So I have this here this is the free body diagram here and I'm gonna writa F equals ma for this block. Sum of all forces ma remember for the six the direction of positive direction was going down. So I have that way which means that the mg is positive because going down and tension is negative. So when I write this I'm going to have mg minus tension equals ma and it's 60 minus T equals 6a because mass is 6.

And I can I'm kind of stuck here there's nothing else I can do with this equation because I have one equation and two unknowns so I can't solve this. But that's OK because you have two objects and you just have to do this to both objects. When you do that. You get two equations and that's going on that enable you to solve this problem. So let me draw a free body diagram for this guy. So the four kilogram is here. It also has an mg pulling it down. And it also has a tension. The difference is if you look at the mg at the second block here is the acceleration for this block is up. Which means the tension will be positive and the mg will be negative. The mass is four so the mg is 40. Ok and I can write sum of all forces equals ma, the forces are T minus equals the mass is four so 4a. And notice how this equation cannot be solved either, two unknowns came for each one of those two equations but because these two unknowns are the same in both equations. If I combine the equations I can do this I can solve it. Ok. The long way to do this is you can solve for T here. Right. Move everything to the other side so T is by itself and then you plug the T in here. But that's too long there's a better way. And there is a better way that's always going to work which is you're going to add these equations together.

It says you then add all the equations to solve for a. And you can add equations you may not remember how to do this but Ill show you real quick. So 16 minus T equals 6a That's the first equation here. And put a little box and then I'm gonna add it to this one so, plus this if you don't remember you can add equations. And this will work 100 percent of the time. The way you write equations is you add everything to the left side and that equals everything to the right of the equal side. So if you add everything to the left I have 60 minus T plus T. Notice how a minus T here and a plus T here will cancel, this is going to happen 100 percent of the time your tension will always cancel and what you left is 60 minus 40. Which is 20 right on the right side you have six a plus four a so that's 10 a. This is 20 equals 10a So a equals 2, 2meters per second square. All right. So look at those steps here. It says Draw a free body diagram calculate all the forces with that that right we calculate all the forces compared to forces to determine the direction of a. Well this was a 60 and this was a 40. So this says what I call the winning force. So things will move in the direction of 40 and we determine that this isn't. And so this is the direction of the acceleration ok. And then. You're right F equals ma, you end up with these two equations here. And then you put it all together. All right. That's how you do it. We're going to a few more of these. The fourth step here says to find other values and the other value is most of the time is tension right. Two big things you gonna ask in these questions is what is the acceleration and what is the tension. Then all you have to do is plug your acceleration into one of the original equations. So now that I know that a equals two all I have to do is plug them in here and I can find T I can plug in here or I can plug it in here just pick which everyone is easiest. I like this one because the T is already positive so I don't have to move the T around. Ok so let's do that. So T minus 40 equals 4a. So T equals 4a, a is 2. And then this negative 40 goes the right side becomes positive 40. So you get a 48 Newtons. So tension is 48 Newtons. All right. So to find any other value you just have to plug it in there. So no matter what these problems we're always going to want to find the acceleration first and then the tensions. Or should you hear I'm asking for acceleration tension. The way to solve is you always find a first and then T, some questions we're just asking for T. Right. That could happen but you need to know that you have to find acceleration first. There's one last point to make right here. OK and I told you that I was going to come back to this for massless pulleys, the tension is the same on both sides of the string and rope. Towards the T-1 equals T-2, that is only the case because this pully is massless. But when you first see these kinds of problems the pulley will be massless and there's no reference to the mass of the pulley. And so you don't have to worry about the problems where the pull is not massless. You might see them later but not yet. OK. So for massless pulleys as long as it's the same rope it's the same tension. So that's how you do these problems. I have a practice problem here. Very similar but now I have three objects so you can end up with three equations. You have to put them all together. Let's give this a shot.

Practice: If the table below is frictionless, find the tension on each of the two cables (find the system’s acceleration first).

Example #1: More Connected Objects

Transcript

Hey guys so we're going to solve a few more problems where we have multiple objects connected to each other. They form a system and we're going to be running through pulleys. The most important thing to remember is that we're going to have one equation for each object we have and then we're going to put all these equations together. Right so let's do this one here. Here it says if the block if the table block coefficient of friction and lowers the coefficient of friction between the table and the block are point four, point five we want to know what is the acceleration of the system and what is the tension on the rope. OK so what's the acceleration. This whole thing. It's a system it's going to move together. The system can move two ways it can move this way or this way and it's always going to be or it's always going to move in the direction that the pulley can spin the spin can. The pully can spin this way or this way. So those are the two possibilities. I have two coefficients of friction here you should remember that the small one when you have two is the kinetic and the larger one is the static. All right. So ww want to write equations but before we can write our F equals ma equation we have to know which way this is going to move and to figure out which way it moves we're going to look at all the forces on all these blocks along the direction of motion right. So the direction of motion is this. So for this guy it's up or down. And for this guy left or right. So I'm going to look at mg here and tension here. And then there's a tension here. And because there's friction, friction is going to be this way. And the way that I can tell that friction is to the left is that if there was no friction the system would for sure be going this way. So friction would have to be holding the system back by pulling this way. So often the way to figure out the direction of friction is to try to figure out which way things would move without friction. All right. So I have ma, Tension, tension and friction. Now there's two options here, either the system moves this way or it doesn't move at all and it wouldn't move at all if the mg was not strong enough to overcome friction. And in this particular problem you have to check first if it moves. It doesn't tell you that moves it could very well be that it doesn't move. And then the acceleration would be zero to check if it moves. I have to check if mg is greater than the maximum static friction. So let's check that out, mg is 20 and friction static which has a maximum friction is meu static normal. And we're talking about this block here friction on this block to find normal There are no forces on the y axis other than normal and mg. So I can say that normal equals mg. They have to kids a lot and this is a 30 and rounding that using gravity 10 to make life a little bit easier here.

So meu is point five, normal is 30 so friction is 15 Newtons. Right. So the maximum friction is 15. The actual friction is the maximum friction that you have to overcome to move. And notice that mg is in fact greater than that. OK. So I can say since mg is greater than the static friction where the maximum static friction. We know that the object will move or accelerate. We know that the acceleration will not be zero. The acceleration would not be zero which means we can start moving and that also means if you're actually going to be going up against kinetic friction. So these are the three conclusions we're going to move. There's no there is an acceleration and you're going up against kinetic friction. So this 15 here was just used to make sure that you move and you do and because you move the friction is actually a type kinetic and friction kinetic is meu kinetic normal, meu kinetic is point 4 normal is 30 and this is 12 newtons. Static friction only gets used to make sure that the test to see if you're moving. And if you don't move the acceleration is zero. If you do move this is the actual friction that you have going up against which you will always be less. Right. So I got 20 on one end and I have 12 on the other end. OK. So now I can I could draw diagrams for these objects independently and then write equations I'm going to write equations because we don't have to draw the diagram Obviously if your professor wants you to then you would. But we can just kind of look at the picture here and since I'm drawing everything in the picture see what I'm going to get. We know we have acceleration the has to be this way because this is a greater force. So the system will accelerate this way so I'm gonna say that this is the direction of positive and this is the direction of positive. And real quick look at the forces mg is going down and for this object the direction of positive was down so mg positive but tension is negative here tension's positive because it's going in the direction of positive for that object. Right. And then here negative. OK. Remember what I pointed out last time that these are you're always going to have an even number of tensions if you have a positive tension somewhere, there is going to be that same tension is going to show up as a negative somewhere else because they're supposed to cancel So let's now write the equation to some of our forces on a two kilogram I'm going to call this block two equals m2a and sum of all forces on block 3 is m3a. The accelerations are the same. So we just call that a the forces on block 2 right here are mg going down is a positive which is a 20, so I'm gonna put positive 20 and tension up. So it's a negative there. The mass is 2a, notice I have two unknowns here, T and a so I can't do anything. But that's fine because you already know what's going to happen here. You know you don't need to have two equations. So we're not worried about the fact that you can't solve this right away. Force on three I have T positive and friction negative, T positive friction is a 12 so it's plus the negative 12 equals 3a and if you look here I'm also missing a T, also missing a.

But it's OK because I have two equations missing the same two variables and I'm going to do what we always do with these problems which is combine and add up these two equations to get the answer. OK. Notice how in doing this we didn't worry about normal or mg because they're not in the direction of motion right. They only worry about the force doing this way and then the force that are going this way. Let's combine these two equations here. I have. I'm going to go ahead and put this equation right here. So here's the first one. And I'm gonna put the second one here. And it's going to be T minus 12 equals 3a. So notice how the T is canceled they always will cancel. And this is 20 minus 18, I' sorry 12. The answer will be 18 equals 5a. So a will be this is 18 then divide by 5 and 18 divided by 5. I'm sorry it's 8 divide by 5 g's 18 divide by 5 and the answer is 1.6 meters per second squared. Boom that's. Right. And then for part B the tension on the rope you know how to do this you just have to plug this number back into either one of the original equations. So for part B to find tension I'll have to just pick any of these equations. I'm going to pick this the second equation here just because tensions are positive. All I got to do is move the 12 over so T minus 12 equals 3a. So T equals three. The acceleration is one point six and then the 12 comes to the other side as a positive. And if you do this you get that the tension is sixteen point eight Newton's. So this question is very similar to what we've done in previously the only difference is that we had friction and we had to actually check to see friction if the system is moving or not. And once we found out that it was moving we were going to use kinetic friction. OK other than that it's the same process right. Equations put them together find a then find your T. All right. I've got a practice problem here. It has no friction but it is at an inclined plane. And let's try to solve that one.

Practice: If the table below is frictionless, calculate the system’s acceleration. Find the rope’s tension.

Example #2: More Connected Objects

Transcript

All right so in this example we're going to combine two blocks that are connected to each other a system of two blocks on an incline with friction right it's going to get a little tricky but I think you can handle it let's do it so we know that the coefficient of friction is point two because I'm only given one coefficient of friction I know that's both static and kinetic coefficient of friction friction OK So that's just my meu equals point two and I don't have to distinguish between them which is good makes a little bit easier and I don't know the magnitude and direction of the systems acceleration so same old stuff I'm going to know what is the acceleration I got two objects I have to write F equals ma twice combine those equations before I can do that I have to compare forces to determine the direction of motion and I have mg which were going around as to one twenty. Twelve times nine point eight or twelve times tenish tensions up this way there is a tension same tension pulling this guy in that direction and then the other force here is an mgx, I can calculate mgx before I do that. When you talk about friction you might be tempted to place to pick a direction for friction but you don't actually know which way friction is going because you don't know which way the system is moving so you need to know the direction of a before you can write F equals ma but you also have to know which force is greater so that you know which way friction will go friction will pose the intended direction so if the system without friction is going to the right and systems to the left so you have to first analyze the system without friction so I'm going to right here you we gonna find thdirection of acceleration without friction what direction will the system move if there was no friction and then you can know that friction is opposite to that right so the way to do this is I have to compare mg with mgx. OK if I calculate mgx, mgx is going to be mg sine of fifty three names in the correct position their position we want it at and this gives me this gives me one forty four. Ok this gives me one forty four. But here one twenty, one foutry four so it's pretty clear that the system without friction will move this way now that means that friction must be going opposite to the intended direction of motion the reason I say intended direction of motion is that a friction is strong enough the system will actually not move at all OK so I now know that it's going to move this way without friction which means frictions opposing it and I have to check if it's actually moving, in case you find the action of a without friction and then you're going to know that friction is opposite to. The direction of A that you found. OK And then you have to test or check your system moves. OK So luckily here I only have one type of friction Otherwise one type of coefficient of friction otherwise you would be checking against the static friction right but here it's the same one so I can just write let's write here that friction is meu normal in this case we're talking about static friction but the meu static is just point two and normal on the eighteenth kilograms is mgy Right normal mgy. OK so normal equals mgy which is mg cosine of theta which is mg cosine of 53 and if you put it all together you get 108. OK 108 is the number that's going to go right here OK I'm going to put this in sort of a big parentheses here friction is meu which is point two, point two 108 and if you do this this number becomes twenty one point six, twenty one point six friction is twenty one point six So let's look at all the forces that are happening here I'm going to draw sort of a new diagram. This is the eighteen kilograms it's been just because we've got a lot of stuff going on here already it's been pulled this way by by a one forty four it's been pulled this way by a friction of twenty one point six. And then on the other side you have the twelve being pulled this way by a one twenty so the first thing you need to look at are these three forces remember the tension this is holding this thing together doesn't really do anything right so.

The one forty four is pulling down the plane and the one twenty in the twenty one point six are pulling the eighteen block the 18 kilogram block up the plane so the question here is the greater force of all of these and not to overcome this plus this in the answer's yes because if you combine these two forces the net force that's pulling up adds up to one forty one point sixty or one forty one point six right it's a combination of these two forces and you make them a different color. If you combine this and this you get this and one forty one is still greater OK So I gonna write here that you have to check. If the sytem meu I'm going to say here that I have the answer to that in this case the direction of acceleration was down the plane which means the acceleration will be up the plane is a system moving we gonna say since mgx was greater than mgx this is mg for the eighteen hundred kilogram this is mg for the twelve kilogram plus the friction on the eighteen kilogram right so the force pulling this way was greater than all the other forces and this system will fact move in this direction OK So this is a direction of positive this is a direction of positive this way the direction of positive is this way right cool so you got to go through all that.To make sure that it's moving OK.All right now let's write some equations sum of our forces on the eighteen equals 18a, sum of all forces on the twelve equals 12a. How many forces in the eighteen along the direction of motion remember along the direction of motion is this direction here and there are three forces there's mgx down and then tension and normal I'm sorry tension and friction up so I'm gonna write, this is positive and the other two are negative I'm gonna write that there's a positive mgx, a negative friction. Negative tension equals 18a and I know these numbers mgx is one forty four friction is twenty one point six tension we don't know but that's OK So that's our first equation the second equation we can combine is a little bit further. Let's just keep going here, so for the second equation I have only T which is a positive and mg which is a negative OK. So I have T plus negative mg equals 12a I have T minus one twenty equals 12a OK Now when we put this all together we're going to combine these equations.OK And this equation is going to come here.We put this all together we have T. minus one twenty equals 12a notice that the T.'s cancel and everything else will combine so I have one forty five for minus twenty one point six minus one twenty equals twelve plus eighteen that is 30a. All right and so we put this all together. I don't have it here but this is a two point four equals 30a so a equals two point four divided by thirty, let me do that real quick.

That's point zero eight. Zero point zero eight meters per second squared All right.Again the annoying part was to check whether or not the system moves at all and then once you do that it's pretty straightforward you just have to write the two equations and put them all together now on a quick note you couldn't solve this question will be easier if you just assume that the system moves right you could have assumed the system moves and you could have the sooner the system moves that say in this direction and then you could have worked out the problem with all the signs in the right way and then if the system didn't move or moved in the opposite direction you would get a negative sign and then if you got a negative sign it means you have to basically start over right it does't mean you just go in the other direction in easier to start over and try everything again go in the opposite direction OK a negative sign will just meant that it doesn't actually work that's because if it's the other way the friction you know now plays a different role and it could have stopped the whole thing from moving all together so you can assume a direction of motion calculate the number if you get a positive number you're good and if you get a negative number you're basically gambles in trying to make it easier and failed and then you have to start over so I think it's generally better than problems like this unless it's obvious by these masses been very very different from each other here the words right this thing barely moved at all but unless it's obvious that the masses are different enough that the system will move in a particular way when you're told that the system moves in a particular way then I would take my time to find the direction of motion so that you can solve the symbols right at seven if you have any questions.

Practice: If m1 = 5 kg, m2 = 3 kg, Θ1 = 37°, and Θ2 = 53°, find the magnitude and direction of the system’s acceleration.

Consider the figure given below. Mass M 1 = 15 kg is free-hanging, supported by the tension in the rope. The mass M2 = 5 kg is at rest on the surface, under the influence of static friction with a coefficient μs = 0.4, which is balancing the tension in the rope pulling it to the right. In order for the rope to connect the masses, it passes over a pulley with mass m = 100 g supported by a rope anchored to the ceiling. In order for this system to be in equilibrium, what must the tension in the rope supporting the pulley be? Assume that the rope holding the pulley is attached to the center of the pulley, allowing it to freely change its angle θ, and that both ropes are massless.
In the figure below, a cart is pushed with a force F. If the coefficient of static friction between the cart and the box is 0.45, what is the minimum value of F in order for the box to not slide down the front of the cart?
In the figure below, the two boxes are initially at rest when a crane lifts the pulley with a force F. What is the motion of each box if the force is equal to 110 N?
A box, mass m, hangs from the pulley system shown to the right. What is the tension in the rope that runs over the pulleys? (Assume the rope and pulleys are massless and frictionless.) A) mg B) mg/2 C) 2mg D) mg/4 E) mg/6
A 5 kg box rests on top of a 10 kg box. The friction coefficients between the 10 kg box and the floor are μs1 = 0.4 and μk1 = 0.3, and the friction coefficients between the 10 kg box and the 5 kg box are μs2 = 0.35 and μk2 = 0.25. If a horizontal force of 50 N is applied on the 10 kg box, which of the following statements is true A) There is a static friction force on the 10 kg box, but no friction force on the 5 kg box B) There is a static friction force on the 10 kg box and a static friction force on the 5 kg box C) There is a kinetic friction force on the 10 kg box and a static friction force on the 5 kg box D) There is a kinetic friction force on the 10 kg box and a kinetic friction force on the 5 kg box
A 5 kg box rests on top of a 10 kg box. The friction coefficients between the 10 kg box and the floor are μs1 = 0.4 and μk1 = 0.3, and the friction coefficients between the 10 kg box and the 5 kg box are μs2 = 0.35 and μk2 = 0.25. A horizontal force, F, is applied on the 10 kg box. What is the minimum value of F required to start moving the boxes? A) 29.4 N B) 39.2 N C) 44.1 N D) 58.8 N 
A 10.0 kg block on a table is connected by a string to a 63-kg mass, which is hanging over the edge of the table. Assuming that frictional forces may be neglected, what is the magnitude of acceleration of the 10.0 kg block when the other block is released? A) 9.0 m/s2 B) 7.5 m/s2 C) 8.1 m/s2 D) 8.5 m/s2
A force F pulls three blocks with masses m 1, m2, and m3 along a surface as shown the the picture below. The coefficient of kinetic friction between the blocks and the surface is given by μ, the tension in the strings between the blocks are T1 and T2, and the acceleration of the blocks is a. The acceleration of gravity is g. What is a correct equation for the tension T2 between the blocks with masses m2 and m3? 1. T2 = m2 (μg − a) − T1 2. T2 = T1 + m2 (μg + a) 3. T2 = T1 − m2 (μg − a) 4. T2 = T1 + m2 (μg) 5. T2 = F + T1 − m2 (μg + a) 6. T2 = T1 − m2 (μg + a) 7. T2 = T1 + m2 (μa) 8. T2 = T1 − m2 (μg) 9. T2 = T1 − m2 (a − μg) 10. T2 = T1 − μm2g + F − m 2a 
A flatbed truck is carrying a 20.0-kg crate along a level road. The coefficient of static friction between the crate and the bed is 0.400. What is the maximum acceleration that the truck can have if the crate is to stay in place?[A] 3.92 m/s2[B] 8.00 m/s2[C] 31.2 m/s2[D] 78.5 m/s2[E] 196 m/s2
Box A, 15 kg, rests on a frictionless horizontal surface connected to box  B, 5.0 kg, by a string as shown in the diagram. What is the magnitude of box B’s acceleration?A) 0.75 m/s2B) 9.8 m/s2C) 2.5 m/s2D) 1.5 m/s2E) 3.3 m/s2
A 5 kg box rests on top of a 10 kg box. The friction coefficients between the 10 kg box and the floor are μs1 = 0.4 and μk1 = 0.3, and the friction coefficients between the 10 kg box and the 5 kg box are μs2 = 0.35 and μk2 = 0.25. If a horizontal force of 50 N is applied on the 10 kg box, draw a free body diagram for each box, with the appropriate values of each force.
In the figure below, two masses, each 150 kg, are NOT at rest and are attached to each other by thin unstretchable cable. Find the acceleration of this system of masses and cable. The angle of the ramp is as shown in the figure. (Hint: Draw a free-body diagram for each mass) Find the magnitude and direction of the acceleration of the system if the coefficient of friction between block and ramp is 0.
In the figure below, two masses, each 150 kg, are NOT at rest and are attached to each other by thin unstretchable cable. Find the acceleration of this system of masses and cable. The angle of the ramp is as shown in the figure. (Hint; Draw a free body diagram for each mass) Find the magnitude and direction if the coefficient of friction is 0.100.