For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Newton's 2nd Law: Advanced Examples

- Tension in a rope pulls throughout the rope. With some clever thought, we can use this fact to our advantage.
- By setting up multiple pulleys, we can cause the same tension to be applied multiple times, effectively multiplying our input force.
- Why isn't this crazy talk? We'll see why in the lesson on
*Power*when we work on*Energy*. - Making a good free-body diagram to show where all the forces are is incredibly important! Without it, we won't be able to make sense of the problem.
- Remember, if an object is experiencing an acceleration, it
__must__have a force acting on it.

### Newton's 2nd Law: Advanced Examples

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Block and Tackle Pulley System
- Example 1: A Free-hanging, Massless String is Holding Up Three Objects of Unknown Mass
- Example 2: An Object is Acted Upon by Three Forces
- Example 3: A Chandelier is Suspended by a Cable From the Roof of an Elevator
- Example 4: A 20kg Baboon Climbs a Massless Rope That is Attached to a 22kg Crate
- Example 5: Two Blocks are Roped Together on Inclines of Different Angles

- Intro 0:00
- Block and Tackle Pulley System 0:30
- A Single Pulley Lifting System
- A Double Pulley Lifting System
- A Quadruple Pulley Lifting System
- Example 1: A Free-hanging, Massless String is Holding Up Three Objects of Unknown Mass 4:40
- Example 2: An Object is Acted Upon by Three Forces 10:23
- Example 3: A Chandelier is Suspended by a Cable From the Roof of an Elevator 17:13
- Example 4: A 20kg Baboon Climbs a Massless Rope That is Attached to a 22kg Crate 23:46
- Example 5: Two Blocks are Roped Together on Inclines of Different Angles 33:17

### High School Physics Online Course

I. Motion | ||
---|---|---|

Math Review | 16:49 | |

One Dimensional Kinematics | 26:02 | |

Multi-Dimensional Kinematics | 29:59 | |

Frames of Reference | 18:36 | |

Uniform Circular Motion | 16:34 | |

II. Force | ||

Newton's 1st Law | 12:37 | |

Newton's 2nd Law: Introduction | 27:05 | |

Newton's 2nd Law: Multiple Dimensions | 27:47 | |

Newton's 2nd Law: Advanced Examples | 42:05 | |

Newton's Third Law | 16:47 | |

Friction | 50:11 | |

Force & Uniform Circular Motion | 26:45 | |

III. Energy | ||

Work | 28:34 | |

Energy: Kinetic | 39:07 | |

Energy: Gravitational Potential | 28:10 | |

Energy: Elastic Potential | 44:16 | |

Power & Simple Machines | 28:54 | |

IV. Momentum | ||

Center of Mass | 36:55 | |

Linear Momentum | 22:50 | |

Collisions & Linear Momentum | 40:55 | |

V. Gravity | ||

Gravity & Orbits | 34:53 | |

VI. Waves | ||

Intro to Waves | 35:35 | |

Waves, Cont. | 52:57 | |

Sound | 36:24 | |

Light | 19:38 | |

VII. Thermodynamics | ||

Fluids | 42:52 | |

Intro to Temperature & Heat | 34:06 | |

Change Due to Heat | 44:03 | |

Thermodynamics | 27:30 | |

VIII. Electricity | ||

Electric Force & Charge | 41:35 | |

Electric Fields & Potential | 34:44 | |

Electric Current | 29:12 | |

Electric Circuits | 52:02 | |

IX. Magnetism | ||

Magnetism | 25:47 |

### Transcription: Newton's 2nd Law: Advanced Examples

*Hi, welcome back to educator.com, today we are going to be talking about Newton's second law in some really advanced examples.*0000

*Single dimension is just an introduction to talking multi-dimensionally, now we are going to get the chance to really structure our muscles to see how much we can do with Newton's second law.*0006

*There are going to be no new specific concepts or major ideas, but it is just going to be the chance to see really advanced complicated problems.*0016

*We will just be exploring a bunch of examples in this lesson.*0025

*But first I have got an interesting thing to talk to you about.*0029

*This is going to be the block and tackle pulley system.*0031

*You might have heard of block and tackle before.*0035

*It is used on sailing ships, sometimes on non-sailing ships like motor driven ships.*0037

*It is a way to be able to use tension to give you mechanical advantage over what you are lifting.*0042

*You can put less force into your lifting, than it is the actual weight.*0048

*We will talk about how that is possible, it is pretty interesting.*0052

*With clever construction, we actually cause the tension in our rope to give us more strength than we first expect.*0055

*First consider a really basic single pulley lifting system.*0061

*You have got some weight pulling down, some force of gravity pulling down on our object.*0065

*If we want to lift this, we going to need a tension pulling here, because this rope is just right here, so it is going to be whatever tension is in that rope.*0070

*So we are going to have to pull with a force, tension left be equivalent to the force of gravity, for us to be able to lift that block.*0078

*There is only one direct connection and it is that rope.*0086

*But, there is a clever thing that we can do.*0090

*Let us consider a double pulley lifting system like this one right here.*0092

*So, in this one, we do not have just one rope, we have two ropes.*0098

*This seems kind of weird, because there is this one connection here, but really what it is, it is the two sides of the pulley.*0103

*We have got the two different sides of the pulley.*0111

*So, whatever tension we put in here, is actually going to be translated all the way through, so we are going to have a tension here, and a tension here, so we are going to get double the lift of that tension that we put in.*0113

*We might have the force of gravity pulling down here, but we can put in a tension equal to one half the force of gravity, and be able to counter out the force of gravity.*0124

*We can put it in a static equilibrium, with only half the force of gravity.*0133

*This is really shocking, it seems sort of like magic.*0138

*The reason is why you will have to work with the fact that, if you want to lift that, you are going to have to pull double the rope as we would in the single pulley lifting system, to this rope has to go a certain amount, and this rope to go up a certain amount.*0141

*In the other system, we only had to pull up, if we want pull it up a metre, we only had to pull the rope a metre.*0157

*In this one, if we want to pull it a metre, we have to pull the rope a metre here, and a metre here, which means 2 m .*0161

*We will talk about this a little bit later, we will get the chance to revisit this really quickly when we are talking about work and energy, but you can certainly understand that there is a bit of trade off here.*0167

*Let us consider an even more complicated system.*0179

*Consider a quadruple pulley lifting system.*0182

*This time, instead of having to just double up once, we are going to have tension here, tension here, tension here, tension here.*0184

*So we are actually going to get the 'T' to show up 4 times.*0193

*For one force of gravity, whatever the force of gravity (the weight is of our object), we can actually put in, T = (1/4)F _{g}, and we will be able to offset that, put everything into static equilibrium.*0197

*So, by using these 4 pulleys, we are able to spread that tension in the rope and cause it to occur multiple times.*0211

*It seems sort of fake, -(and this pulley should actually be connected up here, otherwise it will just fall to the ground)-, and unbelievable, but it does actually work, and you will possibly get the chance to see it if you are in a Science museum, or if you are on an old sailing ship, if you get the chance to play with block and tackle, that is what this set up is called.*0221

*You can actually see mechanical advantage in motion.*0239

*Whatever the advantage you are looking at, you have to commensurate more amount of rope, but it could be potentially really useful.*0241

*You would be able to move really heavy things, which you otherwise would not be able to move.*0249

*The point of all this, is mot to give us a new formula or something new to do in our problems, it is just to make us realize how important it is to be able to carefully think through what we are doing.*0252

*If we carefully consider all the forces being involved, with a really good free body diagram, we can see really interesting things happen that we might not notice at first.*0261

*So, drawing really good diagrams is really important and it is something you got to make sure you do, because otherwise you might miss out on really key piece of information.*0271

*On to the examples.*0279

*We are going to have a bunch of examples today, and this is the first one.*0282

*Say we have got a free hanging mass-less string, and it is holding up three objects of unknown mass.*0285

*We know the tensions in each section of the string though.*0291

*First off, let us name these masses.*0294

*We will call them mass-1 (M _{1}), mass-2 (M_{2}) and mass-3 (M_{3}), and we will say that going up is the positive direction.*0296

*What are the tensions?*0307

*The tension in the top section, T _{1} = 98 N , tension in the middle section, T_{2} = 49 N, tension in the third one, T_{3} = 9.8 N .*0309

*What are the masses of the objects?*0334

*We would not be able to figure it out, but luckily we have got gravity on our side, so we can use gravity to get the information we need.*0335

*We do a free body diagram for each one of these.*0342

*This one is being pulled up with a tension of 98 N, but it is also being pulled down by T _{2} = 49 N, and also being pulled down by the force of gravity on it.*0345

*This one is being pulled up by 49 N (because that is what is above it), pulled down by 9.8 N (because that is what is below it), and also being pulled down by its own force of gravity, so F _{g2}.*0358

*Finally we have got this one, pulled up by 9.8 N, pulled down by F _{g3}, but it does not have any tension pulling it down.*0372

*Let us start figuring out what these masses are.*0386

*Everything is in static equilibrium, so we know that the acceleration across the board is going to be zero for everything.*0388

*Let us look at M _{3} first.*0394

*We know that F _{net,3}, (if it has no acceleration, we know that F_{net} is equal to zero.), so F_{net,3} = 9.8 N - F_{g3} = 0 (everything inside of it), so, F_{g3} = 9.8 N = M_{3}g , M_{3} = 1 kg .*0396

*That is basically the method we will use for all of these.*0439

*F _{net,2} = 49 N + 9.8 N - F_{g2} = 39.2 N - F_{g2} , since this is in static equilibrium, F_{net,2} = 0, F_{g2} = 39.2 N = M_{2}g , M_{2} = 39.2/9.8 = 4 kg .*0443

*For the last one, we have that F _{net,1} = 98 N - 49 N - F_{g} = 49 - F_{g,1} = 0, F_{g1} = 49 N = M_{1}g = 49 N, M_{1} = 5 kg .*0504

*That is how we do it.*0550

*Also, one thing we could have done from the very beginning is, we could have figured it out that there is 10 kg total, because if we consider it as a one big system, then we would have, the tension on the system (in static equilibrium) = 98 N - F _{g,system} = 0.*0551

*So, 98 N = F _{g,system} = M_{system}g , M_{system} = 98/g = 10 kg, because if we were to add up (5+1+4) = 10.*0580

*So there are two different ways of doing it, we could have figured out that the total is going to be 10 kg, and if we have two weights, figure out the third, they all work, they are all different ways of doing it.*0611

*Figure out which one works best for you, and also the specific kind of question you are looking at.*0618

*On to the next example.*0622

*Example 2: An object rests on a frictionless surface (we are looking top-down), and is being acted upon by 3 forces.*0625

*Since we are looking down on it, it is not going to be affected by gravity, so we do not have to worry about gravity for this one.*0634

*It is being affected by these 3 forces, but it is in static equilibrium, it is not moving.*0640

*F _{1} = 250 N (at an angle 147 degrees), F_{2} = 150 N and F_{3} is pulling directly down.*0645

*We want to find out what is F _{3}.*0655

*First, let us start working on figuring out what the components to F _{1} are.*0659

*We know that this is static equilibrium.*0663

*So, we know that F _{net} = (0,0) as a vector, both x and y components are going to be nothing, they are going to cancel out to nothing.*0667

*What does F _{1} break down into?*0679

*What is this angle right here?*0681

*The whole angle is 147 degrees, and that is a right angle, that means that this is going to be 57 degrees right here.*0683

*So, we use trigonometry, and we get that (F _{1} is over here), 250 × sin(57) , and here we have got 250 × cos(57) .*0691

*We get, F _{1,x} = 136 N , F_{1,y} = 210 N .*0711

*So we know that 136 N pulling this way, 136 N pulling up as well.*0724

*Now, what about F _{2}, we know its magnitude = 150 N, but we do not know what its angle is yet, that is an interesting thing we were not given.*0730

*That means that there is a possibility, but we do know for sure in static equilibrium, the x's have to cancel out.*0740

*F _{3} do not have any x component, it is pointing directly down.*0745

*That means that whatever F _{2} is, whatever its angle turns out to be, we know that F_{2,x} = F_{1,x} = 136 N, otherwise they will not cancel out, it will have an acceleration, it could not be in static equilibrium.*0751

*So, F _{2,x} must be 136 N, and that is an interesting thing.*0768

*But, there is two possibilities, either F _{2} could point up, or F_{2} could point down.*0773

*So if we have got two possibilities, that is going to give us two possible answers.*0779

*We do not know which one it is going to be yet, we can work on that in just a moment.*0783

*Let us finish the whole problem out.*0786

*Finishing this out, we know from what we did before, that we have got that, F _{1,x} = 136 N, and F_{1,y} = 210 N .*0792

*So, we know F _{2,x} has to be 136 N as well.*0818

*So, 136 N is what we have got, possibility 1, possibility 2.*0831

*What has to be the other side?*0836

*Now we can use the Pythagorean theorem.*0840

*We know that, 150 ^{2} = 136^{2} + F_{2,y}^{2}*0842

*We get, sqrt(4004) = F _{2,y}, we will get +/- when taking the square root if we are doing it as algebra.*0863

*Sometimes you can get rid of that, but in this case we know that both (+) and (-) are both possibilities.*0882

*So, F _{2,y} = +/- 63.3 N .*0888

*So there is two possibilities.*0896

*We know that it is either pulling up at 63.3 N and canceling out the x, or itis pulling down at 63.3 N canceling out the x.*0898

*We know that 210 N,( because we know that F _{3} is going to cancel out the rest of the y component), 210 N + F_{2,y} + F_{3} = 0 .*0907

*You add up all the forces in the y component, we get zero because acceleration of y is zero, just like the acceleration of x is zero, how we got 136 N being pulled both ways.*0924

*So, because the acceleration of the y is zero, we are able to get this formula right here.*0937

*Now we are going to split into the two different possibilities.*0945

*We have got 210 N + 63.3 N + F _{3} = 0 and 210 N - 63.3 N+ F_{3} = 0.*0948

*The possibilities of when F _{2} is pulling up, or when it is pulling down, it is going to be pulling either a little bit up or little bit down.*0963

*It is mainly pulling to the side, because of how strong it has to the side, the F _{1}'s x component, but it has to pull up or down just like we talked about up here.*0971

*210 + 63.3 + F _{3} = 0, 273.3 + F_{3} = 0, F_{3} = -273.3 N, or F_{3} = -146.7 N .*0981

*They are both answers, both are possibilities, we do not have enough information to be sure of which one of the two possibilities the answer is, but we have the answer for what the question is asking for here, and we could if we were given a little more information, if we knew that F _{2} is pointed up or pointed down, we will be able to figure out whether or not, we would able to figure out which of these answers we would go with.*1010

*But these are the two possibilities for the third force.*1031

*Third example: We have got a chandelier that is suspended by a cable from the roof of an elevator, in a fancy hotel.*1034

*There is a bottom for our elevator, right there.*1041

*Assuming that the cable is mass-less, and the chandelier weighs 10 kg, what is going to be the tension in the cable when the elevator is at rest?*1054

*Also, what happens when the elevator starts moving around?*1061

*So the elevator is moving up, do you think the tension is going to increase or decrease?*1064

*If the elevator is moving up, have you ever stood in an elevator and paid attention what the floor felt like under your feet?*1069

*As the elevator starts to move up, you feel slightly heavier.*1076

*That is because the floor of the elevator has to push against you.*1079

*The floor of the elevator actually pushes against you to be able to give you an acceleration.*1083

*It then just maintains the standard normal force that you are used to so you do not go through the floor, but to be able to accelerate you, it has to put a force on you.*1087

*So, it puts an acceleration through the floor.*1096

*That acceleration comes by putting in extra force.*1100

*What happens when the elevator wants to go down?*1103

*If you are trying to go down in an elevator, you will notice that you experience a brief moment of weightlessness, not weightlessness, but of a less weight, as you start to accelerate down towards the Earth.*1105

*That is because, normally, you got some normal force pulling you down, but as we accelerate down, we need less normal force, but now we are being allowed to move towards the centre of the Earth.*1114

*The normal force is receding, because it is allowing you to gain a velocity, it is allowing you to get some acceleration for a brief period of time.*1127

*Finally, what would the acceleration be necessary, if the elevator were to have no tension in the cable?*1135

*What is the tension when it is at rest?*1142

*When is the tension when there is an acceleration of 1 m/s/s?*1145

*What is the acceleration when it is -1 m/s/s?*1150

*We will have positive as going up.*1153

*1 m/s/s will increase the tension because we are going to have to pull that chandelier along.*1156

*When the acceleration is going down, when it is -1 m/s/s, we will need less tension, because we are going to be able to allow it to go in the direction gravity would normally be taking it, so we just need to give it less force.*1161

*Finally, what would have to be the acceleration for the elevator to have no tension in the cable whatsoever?*1176

*We are going to start off with slightly simplified diagrams, so we have got some tension pulling up (elevator is not the important thing here), the elevator is attached up here, we have got some mass of 10 kg which is being pulled down by the force of gravity.*1182

*Force of gravity = 10 × 9.8 = 98 N, so being pulled down by 98 N.*1205

*If the acceleration = 0, what is the tension going to be?*1213

*If a = 0, tension is going to have to cancel out gravity.*1218

*So, T - F _{g} = mass × acceleration = 0, because we are completely still, T = F_{g} = 98 N.*1226

*So, if the acceleration = 0, if things are static, 98 N.*1243

*What if the elevator was moving up?*1249

*That is going to require more tension, so we will expect to see the tension grow.*1253

*Let us look at this again.*1257

*If acceleration =1, then T - F _{g} = Ma, T - 98 N = 10 kg × 1 m/s/s, so, T = 98 + 10 = 108 N .*1259

*So if you want to make that chandelier accelerate at 1 m/s/s, you are going to need to give it another 10 N past what it takes to stay off gravity, because you have to keep holding off gravity with what would normally be a normal force, but in this case is tension, because it is not resting on anything, it is being pulled up by a cable.*1287

*What if we had an acceleration of -1?*1310

*Very easy, just the same thing.*1315

*We have got, T - F _{g} = Ma, T - 98 N = 10 × (-1), T = 98 - 10 = 88 N, so we need a little less tension, 10 N less tension to be able to allow it to get that bit of acceleration, because gravity will take it the rest of the way, gravity puts in 98 N, and the tension needs to only bring it 10 N of down force, so that it will be accelerated at 1 m/s/s down.*1318

*Finally, if we want no tension whatsoever, what would the acceleration have to be?*1360

*If T = 0, T - F _{g} = Ma, - F_{g} = Ma, -98 = 10 × a, a = -9.8 m/s/s, same thing as gravity.*1368

*If we want to have no tension, we need to let the elevator go into complete free-fall.*1397

*You just drop the elevator, and there is going to be no tension, because the cable is going to be accelerating with gravity, the chandelier is going to be accelerating with gravity, the elevator is going to be accelerating with gravity, gravity is going to be taking over all the work.*1401

*So, the only way to get rid of the tension completely is to let gravity do all the work, it does not have to be resisting anything, there is no longer any resistance, you are letting gravity take over.*1412

*Example 4: A 20 kg baboon climbs a mass-less rope attached to a 22 kg crate.*1428

*The 22 kg crate on the ground is over a frictionless tree limb, so the baboon climbs the rope with an acceleration of 3 m/s/s, what acceleration will the crate have?*1446

*What is the maximum acceleration that the baboon can achieve without moving the crate?*1455

*Our baboon is pulling up, he is putting in some tension, just like if you are climbing a rope to a ceiling, you would be putting in a tension into the rope as it pulls you up, as you use your arms, and the rope resists it, you would be able to yourself up the rope.*1460

*That is exactly what our baboon is doing here, he is climbing the rope just as you might have to in gym class.*1476

*He is climbing, so he is putting in a tension.*1482

*At the same time, that tension is going to pull in here, and it is going to start pulling on the crate.*1484

*So either it is going to lower its effect of weight so much that the tension is actually going to lift it, so the box will be easier for somebody else to come along and lift, or it is going to get so much tension in the rope that it is actually pulled off the ground, as the baboon is climbing.*1490

*That is our basic idea, let us start working on it.*1506

*We will just consider our tree limb to a perfect pulley, mass-less and frictionless, and our baboon is going to climb up things, so if the acceleration of the baboon is 3 m/s/s, what will the acceleration of the crate be?*1511

*He is going to climb up with some acceleration, which is going to cause some tension in the rope.*1524

*So, over here, that tension gets canceled out, pulled the other way, rope is always pulling off position, that tension is going to be the same here.*1533

*Free body diagram for the monkey, he is going to be pulled up, by the tension he is putting into the rope, and he is also being pulled down by his weight.*1540

*So his force of gravity = 20 × g =196 N.*1551

*What is the tension?*1560

*We do not know what the tension is yet, but we do know that his acceleration = 3 m/s/s.*1562

*So, we can figure out what the tension must be.*1570

*If we make up positive, and keep in mind, that is going to turn out to be that 'down' from the crate's point of view is positive, that is really strange, because we are actually wrapping it, flipping it all the way, we have to keep in mind that positive stays the same around the pulley, so positive 'up' for the baboon means negative 'up' for the crate.*1572

*Positive goes into the ground from the point of view of the crate.*1595

*T - F _{g} = M_{baboon} × a_{baboon} , we do not know what the tension is, but we know everything else.*1598

*T - 196 = 20 × 3, T = 256 N.*1609

*So, 256 N, what would be the acceleration on the crate?*1623

*The crate has got two things happening on it.*1626

*It has got T, and its own force of gravity = 22 × g = 215.6 N .*1629

*So, it experiences F _{g} - T = M_{crate} × a_{crate}, (Tension is negative from the crate's point of view, positive from the baboon's point of view.)*1642

*So, F _{g} - T = 215.6 - 256 = 22 × a_{crate} , (F_{g} and T are not equal, which means we are going to have some lift off.)*1687

*Solving, a _{crate} = -1.84 m/s/s .*1701

*So, the baboon is able to achieve an acceleration of 3 m/s/s, the crate is also going to have to achieve 1.84 m/s/s .*1717

*What about when we want to have the baboon climb, but not cause any movement?*1735

*If we want to have the baboon climb, but not cause any movement, then we are going to have to look for what is the maximum acceleration for the baboon is going to happen when we get the most tension, because tension is what enables it.*1740

*Keep in mind, unlike our block problems from before, they are not rigidly connected, the baboon is climbing hand over hand, going up the rope, he is not attached to a single fixed point, whereas the two blocks are attached to a fixed point, when one of them moves, the other one has to move, but in this case the baboon is able to climb up it.*1764

*It still puts a tension, otherwise nothing would be able to move him around, he would not have an external force to cause him to actually go up.*1787

*But he does not connect directly to the crate, so they are going to be decoupled, our accelerations are decoupled.*1794

*The maximum acceleration for the baboon will be when there is the most tension in the rope, without moving crate.*1801

*What is the maximum tension we can have without moving the crate?*1812

*We have to cancel out the baboon, otherwise he is going to start falling.*1816

*So we have got, T = 196 + T _{a}, (196 is what which cancels weight, T_{a}('a' for acceleration) is what is able to give more than that.)*1824

*Most of this tension could be, is equal to the force of gravity on the crate, because otherwise the crate is going to start to lift off the ground.*1856

*196, that which cancels out the baboons weight, and the acceleration he gets in addition to canceling out his weight, cannot exceed the weight of the crate.*1864

*What is the weight of the crate equal to?*1872

*The weight of the crate = 215.6, so we get the acceleration part of the tension, T _{a} = 215.6 - 196 = 19.6*1874

*What is the maximum acceleration our baboon can have?*1903

*The maximum acceleration the baboon can have is going to be, we have already canceled out, we have got our T, we have got our force of gravity pulling down on the baboon + Tension, we are going to split that tension into the one that cancels weight + the tension that actually does acceleration.*1906

*These two things cancel one another out, and that leads to, equal to, the mass of the baboon × acceleration.*1933

*So we have got, T _{a} = M_{baboon}a_{baboon}, 19.6 = 20 × a , a = 19/20 = 0.98 m/s/s, is the maximum acceleration, which is still pretty good, 1 m/s/s, that is a tenth of the acceleration due to gravity, which seems kind of low, but if you try to climb a rope faster than that, that would be pretty impressive.*1946

*Final example: Two block of masses, M _{1} = 5 kg, M_{2} = 10 kg, are roped together on an incline, both having different angles.*1993

*The first box is on an incline of 50 degrees, and second one is on an incline of 20 degrees.*2007

*They also each have different masses as we said before.*2015

*Assuming the pulley and rope mass-less, and the pulley is frictionless, what acceleration will they experience?, what will be the tension in the rope?*2022

*Very first thing to do, we have force of gravity pulling down on both, but force of gravity is not what really matters, what matters is how much they get canceled out, and how much is parallel.*2031

*We are going to, remember, whatever the force of gravity is perpendicular, F _{normal}, they are going to cancel each other out.*2044

*So, while they are still present, for our purposes, we are not going to pay attention to them, because they are not going to have any effect on what we are doing.*2055

*Force of gravity perpendicular over here, gets canceled out by the normal force, because the normal force is able to cancel out whatever perpendicular.*2061

*So the normal force cancel the force of gravity perpendicular to the side of the triangle, once again, we are not going to worry about these for the purpose of our problems.*2072

*The only thing that we are really concerned about, is what is the parallel component.*2082

*Up here, we have got a 20 degree angle, over here we have got a 50 degree angle.*2089

*So, sin(50) × F _{g}, and over here, sin(20) × F_{g}.*2098

*We have not figured out F _{g} yet, but that will take no time at all.*2113

*F _{g} = 5 × 9.8 = 49 N, over here, F_{g} = 10 × 9.8 = 98 N.*2116

*If you want to figure out what the parallel components are, we get, sin(50) on this one, it is going to be equal to, the force of gravity on the first block, parallel to the triangle (down the side of the triangle) is equal to 37.5 N .*2132

*Over here, we are going to get that the force of gravity on the second block, parallel, is going to be equal to 33.5 N.*2158

*So, we have got what the two parts are parallel.*2170

*Which direction can we expect to move in?, we know we are going to have to move in this direction because there is more force in the system pulling this way, than this way; this is the smaller one, so this is the side that is going to win.*2172

*Let us finish this up.*2190

*So, we know that the parallel force, which is the only force we have to worry about, because we know that gravity perpendicular and normal force cancel one another out, we know that this force of gravity, F _{g,1}(parallel) = 37.5 N, F_{g,2}(parallel) = 33.5 N.*2193

*Now we want to find out what is our acceleration, what is the tension in the rope.*2220

*So, there are going to be tensions pulling both this way.*2225

*As we have talked about before, we can actually treat this whole thing as one big system.*2227

*So, we have got one big system.*2232

*We can figure out the tensions, come up with a bunch of equations, but there is no reason to, because we know that what are the external forces acting on it, the tensions are not external, those are the things that links our system, that makes our system.*2234

*So, the tensions are not going to cancel out whether we can treat it as a system, they are what enables us to use it as a system.*2246

*So we only have to pay attention to what is working outside, if we treat it as a one big system.*2251

*What forces are acting on the system?*2255

*That is going to be, = M _{system} × a, unlike the baboon and crate problem, we got a rigid connection here, if one thing moves, the other has to move, so we know that we can use the same acceleration for the two of them.*2260

*What forces are here?, let us make to the right as positive, we got 33.5 - 37.5 = (10+5)a, solving, -4 = 5a, a = -0.27 m/s/s, which makes sense, because as we have talked about before, we are going to see an acceleration pulling things that way.*2274

*In this case, it is a very small acceleration, but it is still pulling it along.*2312

*This is what our acceleration is, it is parallel, it is not just some general acceleration, this is the magnitude, and what we know, we know if we were to consider it as a vector, it is the amount parallel to the triangle and pulling to the left.*2318

*It is going to be parallel from both of their point of views.*2332

*They are going to have very different acceleration vectors.*2334

*This one is only going to come up, at a 20 degree incline, while this one is going to go down at a 50 degree incline, so one is rising , the other is falling, but as we talked about before, the only thing effecting this is the parallel component, because remember, the normal force cancel out the perpendicular component of gravity.*2337

*The only thing we really have to worry about here is, what the total acceleration is, and it is up to us, if we had a problem where we had to consider which direction it is moving, then we would have to work on it, but in this case, it is enough to say that it is -0.27 m/s/s, or really 0.27 m/s/s, and then talk about which direction it is moving in.*2354

*Now, what if we were to solve for the tension is?*2375

*If we want to solve for what the tension is, we can look at a single block on its own.*2378

*So, let us consider looking at the smaller block.*2385

*The smaller block is being pulled down with 37.5 N, it is being pulled up with the tension (which we will figure out in a second).*2388

*Another important thing is, because the way we got this constructed, tension is parallel.*2403

*As we have talked about before, it is possible for the rope to be pulling at a different angle, and cause things to not be parallel to the surface, but in this case, we have set up our problem in such a way, or the problem is given to us in such a way that the rope is going to run parallel to the surface, so we do not have to worry about this.*2409

*So, it is just going to be these two things acting directly, without having to worry about putting in any more angles.*2423

*We already figured out that this one was the parallel component, so we are good to go from here.*2427

*This is the negative direction, so tension is pulling in the positive direction, T- 37.5 = M _{little block} × a_{little block} = 5 kg × (-0.27 m/s/s), solving, T = 36.15 N, which makes sense, because the tension in the rope is going to be have to be less than the force of gravity parallel, otherwise it is going to go in the direction of the tension.*2432

*The larger force is the one that will win out, which is going to have the effect, if they are directly opposed.*2493

*So, the tension is smaller than the force caused by gravity.*2497

*That gives us everything for this problem, we have got it solved.*2505

*I hope you learned something, there is a bunch of different possibilities in really advanced examples that you can do with forces, but just by breaking it down into good free body diagram, carefully analyzing the forces involved, putting down all the things that you know, you can work your way through it, and you can figure out what you are going to do.*2507

*I hope you enjoyed it, and we will see you at educator.com for the next lesson.*2520

2 answers

Last reply by: Peter Ke

Thu Apr 28, 2016 12:35 PM

Post by Peter Ke on February 27, 2016

I don't understand why there's no possibilities for F1 y-axis to go up and down while there is 2 possibilities for F2 y-axis to go up and down, why is that?

1 answer

Last reply by: Professor Selhorst-Jones

Thu Nov 29, 2012 11:04 PM

Post by Abdelrahman Megahed on November 29, 2012

Why can't you just say Fnetx=0; F2cos(theta)-F1cos(57)=0 and then solve for (theta) and finally reapply this in the Fnety=0 eq which would be Fnety=0=F2sin(theta)+F1sin(57)-F3=0, ??