Practice: A 10 kg block moves with 20 m/s when it reaches the bottom of a long, smooth inclined plane that makes 53o with the horizontal. How far up the plane will the block reach before switching directions?

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Intro to Friction | 40 mins | 0 completed | Learn |

Kinetic Friction | 39 mins | 0 completed | Learn |

Static Friction | 15 mins | 0 completed | Learn |

Inclined Planes | 34 mins | 0 completed | Learn |

Inclines with Friction | 73 mins | 0 completed | Learn |

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Concept #1: Intro to Inclined Planes

**Transcript**

Hey guys so in this video I want to introduce you to inclined plane problems which is a very classic type of problem in physics Let's check it out. All right so in inclined plane problems basically whenever you have a block that instead of sitting from this flat surface sits on a surface that's an angle we have to tilt our coordinate system coordinate system just means the x and y axis and we do this because we want to match our acceleration. Let me show you. So typically x `y is like this right y axis here x axis here on the horizontal. And we do this because our objects are usually moving in this direction in the X-axis or in the y axis. Now since we're moving at an angle what I'm going tilt the axis and the reason I do this is so that my x axis over here matches the direction that I'm going to move or the direction of my acceleration. So the red lines here are my old x and y axis. So this is my old y axis and my old x axis. I'm going to be accelerating. This box if there is no friction and whatnot would be accelerating this way. So I want my axes to match that. So I want to be in this green axis over here. So we're going to do is we're going to sort of tilt this so this becomes my new y axis. And this becomes my new x axis, x new y new we no longer use the other ones. All right. Now when you do this when you do this check it out what's the forces here. I have mg down. And I have a normal up remember normal is always perpendicular to the surface. So normal would go this way.

Now here is what happens when you do this notice that normal sits on my positive Y-axis. But mg is somewhere between my negative y axis and then my positive x axis this way by the way we're going to say that the positive x positive y. So normal sits between the x and y when a vector sits between your axes. It's not sitting flat in the x or flat and why it needs to be decomposed so in inclined plane problems we're going to have to decompose mg, mg must be decomposed into mgx and mgy and it's going to look like this. This here is mgx And this here is mgy. So you write that in here this blue arrow of yours is mgx and then this blue arrow here is mgy. OK. So I want to also point out that the x axis is always going to be along the plane so if this is your plane this is your x axis even if you have a really steep plane like let's say 89 degrees right it's almost vertical. This is still the x axis even though it looks a lot more like the y axis. So x axis is always along the plane. All right. So mg has to be decomposed into mgx and mgy. You might remember from Vector decomposition that if I have a force and the angle's here Fx is F cosine of theta. Right. So you might be thinking maybe mgx will be mg cosine of theta and what I want you to do is I want you to remember that in the inclined plane it's the opposite.

So x instead of going with cosine will go with sine so mgx is always mg sine of theta in mgy is mg cosine of theta so it's a little bit different. OK. So for any other problem X go to cosign. But on the inclined plane for mgx and mgy why it's backwards. OK. When we talk about acceleration in these problems which we do a lot. It's always going to refer to the x axis because remember x is along the plane. And that's because the acceleration in the y axis will always be zero. Every single time. OK. If you had acceleration the y axis would mean that this block sort of come off the plane or breaking through the plane. That's weird it's not going to happen. So the acceleration in the y axis will always be zero which means it's in equilibrium in the y axis which means the sum of all forces in the y axis equals zero. This would be the case 100 percent of the time. Now most of the time something like 95 plus percent of the time. OK. There are no extra forces in the y axis. What do you mean by extra forces. You're only going to have normal and mgy. Notice how these are the forces here and the y-axis, normal and mgy. Because those are the only two forces and they cancel then I can say they have the same magnitude normal equals normal equals mgy. Right. Now if you had other forces then that would be the case. For example if I push here with a force of F. Right, now I would be able I would say that normal equals F plus mgy OK. That's not going to happen very often when you get this out of here, few more points. We always want theta to be against the x axis right here. So I'm going to call that theta or theta X just to be super clear that we want to be in the x axis. If it's not in the x axis if you get an angle up here I'm going to call this theta y because it's theta with the vertical between the angle the plane and the vertical. We don't want that we're going to get rid of it and. Find my Theta x, theta x is just 90 minus theta y ok. So we always want the angle to be down there the length angle in the height of a plane are all related. And that's because inclined planes are just basically triangles. So we can use SOHCAHTOA to to go from one to the other. So let me show you this real quick I'm gonna draw a tiny little triangle here, the angle with the x axis length would be this and this would be the height ok. They're all related h equals l sine of theta. This comes from SOHCAHTOA. Now the thing is if you remember y goes with sine, sine has to do with the height of the triangle. That's why h is l sine of theta. You should remember that you're going to see that sometimes. So before I go on one last point here. This is going to happen. About 95 percent of the time as well the angle will be given to you in the right place so you don't have to worry about it that much. This is one last point. Most professors don't really do this but in case you see it sometimes the name was given to you in terms of a percentage that's instead of saying that the angle is 15 degrees. That's they're going to tell that the angle is 20 percent. The slope is a 20 percent slope. You can't use that. You have to convert that to an angle. So theta x will be the tangent. The tangent of the sorry arctangent of whatever that percentage is. OK. So for example 50 percent is point 50. So the arc tangent of point fifty or just point five. That's also kind of rare. All right. So that's the basic idea of the inclined plane all the little things and you know let's do an example real quick and then I have a practice problem for you guys.

Example #1: Intro to Inclined Planes

**Transcript**

Alright so let's see this example here. I have a ten kilograms block on top of an inclined plane. So ten the plane has a length of five meters and it makes an angle of 37 degrees with horizontal the angle it should be awesome. I want a free body diagram so let's do that real quick. Free body diagram. You can't you're supposed to draw the object as a dot and you can draw the incline or anything like that but you should do is make, draw a little axes there for the x axis and this is going to help you get all the forces in here. I recommend to draw a sketch with the plane and then the the free body diagram next to it so you can kind of match and see that you're getting it right. So this is the positive x axis because you're going to be moving this way. You're going down here with an mg. This gets decomposed into mgx along the plane mg or down the plane and mgy into the plane. Normal is out of the plane like that. This is a complete free body diagram. I might also want to probably write the y axis here as well. The positive y axis will always match with normal. Now we want to find all the forces calculate all the forces. So I'm going to use gravity, say gravity is approximately 10. So we're going to that number instead to make life a little bit easier. Mass is 10 so 10 times is one hundred, mgx remember mgx is mg sine or cosine, sine of theta right. And mgy is obviously the cosine of theta. This is backwards from what you usually do. So this is one hundred sine of 37 and 100 cosine 37. So this gives me 60. And 80. So this is a 60 and this 80. I've got three forces already, I'm missing the last force which is normal. If you look here normal simply cancels with the y axis. So normal will be. 80 as well. OK.

There are no forces cancelling even the x axis so you're definitely going to move down this way. I can draw the acceleration vector showing that I'm gonna accelerate down that that arrow just can't touch the dot. Right. So that's it. That's part (a) I've got everything now I wanna right an expression drive an expression for the acceleration of the block. Remember when we say acceleration in the block we mean the acceleration in the x axis so a is really ax. And that's what I want because we know that ay equals zero for the inclined plane. All right, how do I find acceleration in the force problem, mg mgx normal and I want acceleration so I'm gonna do F equals ma. Sum of all forces equal ma, again here you don't necessarily have to show this but if you want to be very precise you can put a little x everywhere because that's the only thing we care about here is the x axis. It's only moving in the x axis. What are the forces on the x axis, the only force is mgx. So I have mgx positive equals max. I'm just going to write ma because you know that's the only a we have. And I would just have to solve for a. One thing that a lot of people want to do when they get here is cross these two ends. But you can't do that just yet. Because if you do that you left it gx. And what is gx. Right.

There is no such thing as gx. Unless you can you want to think of gx as g sine of theta. I guess you could but why recommend you do that to avoid confusion is don't touch mgx don't cancel anything until you've replaced them mgx for what it stands for mgx is a placeholder. It's a shortcut for mg sine of theta so we don't have to write as many things. So let's expand this. Into what it is mg sine of theta equals ma and now we can cancel and we're done. a is g sine of theta right there. And this is a little equation that's probably worth remembering for your time that we don't have an inside an object going down the inclined plane that has no friction. The acceleration will be g sine of theta. If you're going down the plane it's positive g sine of theta if you go up the plane then gravity is pulling you back down so It would be negative. OK. So let's calculate that g sine of theta is going to be six meters per second squared. Ok. Part (c) says find the speed of the above, find a block speed of the bottom. So this is the final velocity. What do I know. It's released. This is kind of implied that it's from rest. It doesn't give me an initial velocity so I'm going to go with that. And I want to know the velocity at the bottom. So this is a motion problem and I have to have my variables here. Let's see Vinitial, Vfinal, a delta X and delta-T, the initial velocity is zero the final the is what I'm looking for. The acceleration I just found it's six. It's positive six. Do I have delta x I do, delta x is length here. Right. So you're going from here. To here. This is motion in a straight line even though it's down a plane. That's my x axis it's is the same thing if I move this way. I'm just tilting everything this way. So this delta x is the same thing as in this case L which is just five t is the variable that I don't have. But it's OK that I don't have to do it because I already know three things. One two three. So I can do this using the second equation to find the final velocity. So the final velocity would be the square of two, a is six delta x is five. And the final velocity will be seven point seventy five meters per second. OK. So that's it hopeful that makes sense, you absolutely have to know how to do this. This is just the beginning of how problems in inclined planes work. I have a practice problem here that I would like you to try. And just to be clear here the idea is that the block is moving at 20 and now it's going to go up the plane. OK so let's try that. I want to know how far up the plane it goes. All right let's give that a shot.

Practice: A 10 kg block moves with 20 m/s when it reaches the bottom of a long, smooth inclined plane that makes 53o with the horizontal. How far up the plane will the block reach before switching directions?

Example #2: More Inclined Problems

**Transcript**

Hey guys so I want to show you a few more inclined plane problems. Let's jump right into it. So here I have a system that is at equilibrium which means all these forces are canceling. There's no acceleration and no velocity. The masses are one kilogram for the one and two kilograms for the two. And this angle here is 53 degrees. OK. I want to know what is the tension one and tension two and notice that the surface is frictionless, no friction. So by the way I hope you can see before we start solving this that T1, T2 has to be greater than T-1 because T-2 is holding both blocks at an angle whereas T1 is only holding one of them. And this is equilibrium so forces are going to cancel. First thing you should do is write a free body diagram. So let's draw a free body diagram for the first object here. And what are the forces, mg down. Let's call this m1g. This is normal the first Block so normal one. This here down the plane is mg or m1gx. And this here is m1gy. And then I have tension one right here. This is the first object. Let's draw the second one again. Here's the inclined plane if you want. If you draw that that helps a little bit and here's block. two. Now what are the forces on. I have m2g, m2gy, m2gx I have tension 2 pull it up and I hope you realize that there's also a tension one.

So if you look at object to its being pulled down by tension one remember the direction of tension depends on the perspective of the object. So if you are one you're being pulled up by T-1 And if you're 2 you're being pulled down the plane T1 of those forces are of the same magnitude just different directions. All right so now I want to find T1 and T2 okay T1 and T2 This is an equilibrium question so no problem. So that means that the forces will cancel and you should be able to quickly look into this and realize that N1 equals m1gy and then N2 I forgot to to write N2 over here, N2 is going to be the same as this, ok. And also on the x axis forces will cancel T1 will cancel with m1gx and T2 will cancel with T1 plus m1gx because there are two forces in the x axis. Right. So I could write as many as four equations here but it's a simple equation just saying this equals this and this equals this. Right now if you try to solve you always want to start with the simplest objects simplest object has the least amount of forces on it. And here the first one is the simplest because only has two forces here. If you start here with the second object you're going to write that T-2 right there with the T2 equals T1 plus m2gx. And the problem with this is that you don't know T-2 and you don't know T1. If you go to the simple object object number 1 so that it's kind of out of order but it's ok. T1 equals m1gx. I don't know T1.

But I can found I can find m1gx ok and m1gx is m1g x goes with sine on an inclined plane. So it's m1g sine of theta, mass is just one gravity I'm going to round it at 10 and then the sine of 53 which is just point eight. So if you do this you get mgx of eight . If I calculate mgy. Well I don't mean mgy. So mgx is just 8 so that's it, mgx is T1 equals 8. And now that I know T1 I can simply plug it back in here and I'll be able to find T2, T2 is T1 which is 8 plus m2gx, m2gx is right here, m2gx is mass 2 which is 2 gravity which is going around to 10 and then the. Sine of 53. This is just 16. So it's 8 plus 16 tension 2 is 24. So the first tension is eight, the second tension is 24. Like I said it's tension two we should have expected the tension 2 to be greater. In fact if this is one and this is two you can see how T1 is holding one T2 is sort of holding three. So it makes sense that T-2 is three times larger than T-1. OK. Now I have a practice problem here and I want you guys to give this a shot. I know it looks very scary because you have two inclined planes. The angles are different and whatnot. But remember ultimately what's happening here is forces are canceling. So you can say force in this direction equals force in that direction and you might be able to get. So give us a shot to get

Practice: If m_{1} = 10 kg, Θ_{1} = 37° and Θ_{2} = 53°, what must m_{2} be so that the system is at equilibrium?

Example #3: More Inclined Problems

**Transcript**

Right so here I have one problem in four different variations of it so we can clear out with this and see how it works so I have a two kilogram crates that is on a smooth plane the makes thirty five thirty seven with a horizontal with a force F acting on its OK So basically we're going to have four situations where in all of them there is a plane but there are four different forces. OK So in all these cases the mass is two and the angle of the plane down here is thirty seven degrees now in the first situation and I'm always gonna find theacceleration in the first situation I want there to be a force that is fifteen up the plane, up the plane means this way so this force Fa is fifteen and I want to know what is the acceleration of the crates this is a force problem the way from acceleration force problems is F equals ma, sum of all forces equals ma and we know that this is along the x axis over here always OK what are the other forces that exist Well there's an mg pulling it down and there's an mgx in this way and there's an mgy and there's a normal but here are the only forces that really matter or these two they're basically competing against each other. I can calculate mgx and mgx is mg sine of thirty seven. Mass is 2 to gravity is ten sine a thirty seven is point six So there should be twelve. Okay mgx is twelve so if you write sum of all forces equals ma I have. Fifteen this way and twelve this way so I like to calculate those forces first because I am able to look at the problem and the side or figure out which way it's going to move here this object is going to move up the plane because my force pulling up is greater than my force pulling down so I'm going to say that this is the direction of positive actually it's it's kind of dictated here the direction of positive is up the plane so this is positive Fa. Negative mgx equals ma, I already have these numbers it's fifteen minus twelve equals 2a so the acceleration is three over two or one point five meters per second squared it came out as a positive and it is in fact a positive because it is going up the plane because my forcw pulling up is greater OK let's do this other one here slightly different, it's two point fifteen but now at thirty degrees above Fa clockwise. So Fa was this way I want thirty degrees above it clockwise clockwise is this way I actually meant counterclockwise I'm so sorry. Counterclockwise I want thirty degrees this way right here and it's counterclockwise so when I wanted to look this way so this is thirty right here this is thirty seven right here and this is my Fb equals fifteen What are the other forces I have mg, mgx, mgy, normal OK So notice that Fb is at an angle, Fb is at an angle of thirty degrees with the x axis ok, remember my x axis is now slanted and I am thirty degrees from that already slanted curve so I have to decompose Fb into Fbx and Fby. OK Now when you do this x goes with cosine X. always goes with the cosine if the angle's against the x axis the. My new x axis and thirty is against my x axis, my new x axis, you only flip where x goes with sine instead for mgx and mgy OK So even though this is an inclined plane problem Fbx will still be F cosine theta, the exception where flips only applies to mgx and mgy you so Fbx is Fb cosine of thirty and Fby is Fb sine of thirty.

And if you do this you should get I have it here. This is a third comes out to be thirteen and this is seven point five ok. Now remember the forces in the y axis will always can so so I can just say that these forces here will cancel with mgy, so they don't really matter the only forces I want to look at is the forces in the x axis so the forces I'm going to look into in the x axis. Aregoing to be to get out of there. The forces I'm going to look into the x axis are going to be Fbx and mg. We're saying in all these questions that going up is the direction of positive so Fbx is positive and then mgx is negative so sum of all forces equals ma I have Fbx positive minus and mgx equals ma, Fbx I just calculated is a thirteen mgx the mass didn't change the angle didn't change, mgx is still twelve over here 2a so the acceleration is point five meters per second squared. And I get it's positive because my thirteen is obviously bigger than my twelve I'm going to do one more and then I want you guys to do the last one OK I'm now going to push with five Newtons instead of fifteen I want to push with five so it's very similar to this question here except I'm pushing with the weaker force.

So now my Fc is just five I want to remind you that the mgx this way is twelve and there are other forces but they don't do anything OK so you could draw the wholething but you don't really need to for this problem specially since we're just the two so it's going to write that the sum of all forces equals mac or ma I'm sorry and then the forces are going this way is positive the forces are Fc minus mgx equals ma or five minus twelve equals 2a so a equals negative three point five meters per second squared OK So this is a situation where you're not pushing hard enough and even though you're pushing up because you're pushing or not this thing is actually accelerating down the plane OK this negative means that it's going down the plane which is the direction of negative. Down the plane. OK so I want to do this one just pause the video and give this a shot it's very similar but a little bit different OK I'm going to jump right into it but hopefully you tried it. We're now going to push down the plane with a force of five so this way I'm going to say Fd equals five what else the only other force here in the x axis is mgx So in this case we actually have both forces going down OK so when I write sum of all forces equals ma, we are saying that all these going up is positive so both of these guys are negative, I have negative Fd plus negative mgx equals ma so Fd is negative five, mgx is twelve 2a. So I gets negative eight point five meters per second squared that's the answer to this one it's negative because it's still going down that should be pretty obvious both forces are going down so they have to be negative it's going down the plane but that's it for this one, let me know if you have any questions

A block of mass m is on a frictionless plane inclined at θ with the horizontal and is pushed by a horizontal force F at a constant velocity up the plane. The acceleration of gravity is g . What is the magnitude of the normal force N the plane exerts on the block?
1. N = mg cos θ
2. N =F/ sin θ
3. N = F
4. N = mg
5. N = F tan θ
6. N = F sin θ
7. N =mg/ sin θ
8. N = mg tan θ
9. N =F/ cos θ
10. N = mg sin θ

Two blocks in contact are pushed up an inclined plane with a force 9mg, which is parallel to the plane. The force pushes on the lower block, which has mass 5m. The upper block has mass 4m. The plane is inclined at an angle θ with respect to the vertical. What is the magnitude of the force on the lower block exerted by the upper block?
1. 9mg
2. 5mg (1 − sin θ)
3. 5mg (1 − cos θ)
4. 4mg
5. 5mg cos θ
6. 5mg
7. 4mg cos θ
8. 4mg (1 − cos θ)
9. 4mg (1 − sin θ)

A block is pushed up a frictionless incline by an applied horizontal force as shown. The acceleration of gravity is 9.8 m/s2. What is the magnitude of the resulting acceleration of the block?1. 0.4082662. 0.01485183. 0.2631864. 0.4404885. 0.3608536. 0.2444767. 8.577648. 3.201979. 0.27327410. 2.00947

Three forces are exerted on an object placed on a tilted floor, as shown in the figure. Assuming the forces have magnitude F1 = 1N, F2 = 8N, F3 = 7 N, what is the component of the net force Fnet = F1 + F2 + F3 parallel to the floor?(Take the +i axis parallel to the plane and pointing downhill.)

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