Sections | |||
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Intro to Forces + Newton's Laws | 16 mins | 0 completed | Learn |

Force Problems with Motion | 28 mins | 0 completed | Learn |

Forces with Multiple Objects | 26 mins | 0 completed | Learn |

Vertical Forces & Equilibrium | 34 mins | 0 completed | Learn |

More 1D Equilibrium | 16 mins | 0 completed | Learn |

Vertical Forces & Acceleration | 26 mins | 0 completed | Learn |

Landing & Jumping Problems | 19 mins | 0 completed | Learn |

Forces in 2D | 35 mins | 0 completed | Learn |

2D Equilibrium | 25 mins | 0 completed | Learn |

Concept #1: 2D Equilibrium Problems

**Transcript**

Hey guys so in this video I want to talk about two dimensional work equilibrium problems which are problems where forces are gonna cancel but at least one force is going to be an angle not saying flatten on the x and the y but at an angle so when we cancel all the forces we have to do this both in the x and the y axis. Let me show you. So the first thing we're going to have to do is decompose all the forces because like I said at least one of the forces will be at an angle and then because we're in equilibrium in both dimensions I can say that the sum of all forces in the x equals zero and the sum of all forces in the y equals zero. Ok So, here I have let me call this T1 and I can call this T2. The other forces acting here are there's actually one more actually mg and just real quick I'm gonna use gravity as as 10 instead of 9.8 just around to 50. And this T2 here has to be decomposed into T2(x) and T2(y). When I do this I want the angle not to be up here but I want the angle to be down here. However it turns out that this is exactly the same angle because this reflects over here so 37 is the angle that I would use. So I set up the sum of all forces equals zero in the x axis and the y axis. But the way that I like to think of these problems that I think makes more sense it's more visual it's to think that's in these statements are correct that all the forces to the left equal all the forces to the right because all the forces pulling this way equal all the forces pulling this way. And we've done this before. Same thing here all the forces up equals all the forces now. And also I have absolute value signs here. And the reason for that is that when I write this way I don't have to worry about signs. And I'll show why really quick. So I'm going to start with sum of our forces in the x axis equal zero. And there are only two forces in the x axis T1 and T2(x) so you could do it this way you could say T1 is to the left so I'm gonna make it negative T1 plus T2(x) to the right so I'm gonna make it positive T2(x) equals zero. And then I'm going to move T1 to the other side of the equation so that it becomes positive and then I get T2(x) equals T1, T2(x) is a force on the right and T1 is the force on the left. So left equals right and they're both positive. But instead of going through all of this I could have just looked at the drawing here and said Hey these two have to be the same. Because this is at equilibrium and I would have arrived at this much faster. But notice that they're both positive because magnitudes are the same. So I can do this in the y axis as well. And then this would mean that this equals this. Sum of all forces in y equals zero. But I'm going to skip some steps here and just say top equals bottom. So you want to write this you want to say left cancels with right, up cancels with down and then we start trying to solve this. So if you go here I don't have T1 and I don't know T2(x) either but I can expand T2(x) into T2 cosine of theta now you always want to do this you only want to expand. On a component of a vector if you know at least one of these two variables. Right now I know zero out of two variables. And if I expand I know one out of three variables and that's better. Even though it miight not seem better because I'm still missing two things. I'm in fact a little closer so. What you don't want to do is expand this. If you don't know any of these three then you will know zero out of three which is certainly worse than zero out of two things. OK. So if you know one of the two you want to expand and I know this. So I can write T1 equals T2 cosine of 37, cosine of 37 is point eight so T1 equals Point eight T2, I'm stuck here I can go anywhere else. So I'm gonna have to continue over here and hopefully the y axis workout. And I do know mg I know mg is 50. And again because I know. One of the two. Variables that T2(y) can be changed into T2 sine of theta I know theta then I'm going to replace I'm gonna expand T2(y) so T2 Sine of 37 equals 50 if I move things around T2 I find to be eighty three point three. That's the first answer. And obviously to get the second answer you just have to plug this back here. So T1 is point eight times eighty three point three which is sixty six point seven Newtons and that is my second answer. Alright so you gonna say left equals right top equals down trying to solve for what you're looking. Let's check out the next one. So here I have a 1 metre long string connected to the center of a two kilogram ball. So the length of the string is this. The mass of the ball is two kilograms put over here. And it is 30 centimeters in diameter so diameter equals point three meters, remember in physics we're only going to use radius so let me convert that real quick. Which rests against a smooth. This means no friction vertical wall and shown. So I want to know tension in the string the force of the wall on the ball. So let me draw a free body diagram here real quick, let me draw these forces here real quick first. So I got a T here. Tx here. And we got a Ty here. There's also an mg here. OK.

So I drew a T it's an angle. Ideally I want this angle to be down here and I can split it up into Tx and Ty those are sort of the obvious forces and at this point you might wonder is there anything else. Well this ball is being pulled to the right by the tension because of the way that's it is hanging T gets decomposed into Tx and Ty so we got a Tx in the right which means the ball is pushing against the wall therefore the wall has to push back against the wall. Right.

Another way of thinking about this or visualize is that there has to be a force by the way this force is called normal is the fact that this ball is in equilibrium it sits there. So there has to be a force cancelling the force to the right. Otherwise it will be moving to the right. Right. F equals ma give us the idea that if there is only one force to the right than it has to accelerate to the right. If it's not accelerating because there's either no forces or there are enough forces so that they cancel. So these two have to be the same. Once I draw this I can draw if you want to draw free body diagram and it would look like this. Same things a little simpler. And I can say up equals down, left equals right. So Tx equals N and Ty equals mg now let's start over here. Remember I only want to expand Tx into Tcosine of theta if I know theta and I don't know they. So that's kind of a bad idea if I do this. Look I am missing three variables. Right. And then here Tsine of theta. I do know mg. The mass is 2 so I'm gonna use gravity to be 10 so I can round this nicely to 20. So Tsine of theta equals 20. So this equation I have a little bit more information but I'm still missing two things. OK. I don't know Tx and I don't know Ty either. So we're really stuck here. And the only way to get out of here is going to be to find this theta here. And to find theta I'm going to use the fact that I know this length and I actually know this length over here as well. If you use these numbers that were given to you 30 centimeters or more precisely actually the radius and you use the 1 meter length of the rope. You'll notice that I have a little. Little triangle that gets formed right here. Right and if I draw a triangle. This is one meter. This is the radius right there. This is half of the diameter right there. And I can then find this angle which is what we want. OK. I can do this using SOHCAHTOA. So. Notice that I have this is over this over here is the opposite side I'm sorry the adjacent angle this is the hypotenuse. So I have a hypotenuse I should be using the cosine. So I can say cosine of theta, cosine of theta equals adjacent over hypotenuse, adjacent is r, hypogenous is l but we really want is theta, So theta is the arc cosine of r which is point fifteen over l which is one. So this is just the r cosine of point 15. And the r cosine of point 15 is eighty one point four. Which makes sense because this height's much the hypotenuse is much bigger than r. So this is a very steep angle like this. All right so now that I know that I can just plug it in here. And that's going to sort of unlock things. So T cosine of eighty one point for equals N I'm still missing 2 here so I'm kind of stuck but notice that here T sine of 81 point four equals 20. You only have one or none.

So I'm going to be able to find T here and plug it over here. So T equals 20 divided by the sine of eighty one point four. And if I do this very carefully. I get that this is twenty point two Newtons. Once I know T, I can go back in here to find N, N is T cosine of eighty one point four. And if you do this you plug in these numbers and the calculator, you get that N is three point zero two. And by the way this is answer of the part A which was asking for the tension in the string and part B asks for the force of the wall on the ball. This is whoops. This is normal. Right. So I got both answers by doing that. Not how these questions are going to work, up equals down do you go down, left equals right, no matter how many forces you have that's what happens if you have to measure equilibrium. And you just have to write the equivalences, the equations and then figure out a way to get the answers by playing with the two equations. That's it.

Practice: For the system below, what is the tension on the longer rope? (Hint: Forces cancel at the middle (red) point).

Practice: Two light ropes of same length are connected to the center of two identical 3 kg balls, holding them in the air against each other. If the force between the two balls is 10 N, calculate the tension on the ropes (it is the same for both). Calculate the angle between the two ropes (shown).

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Concept #1: 2D Equilibrium Problems

Practice #1: 2D Equilibrium

Practice #2: 2D Equilibrium

In the figure below the left-hand cable has a tension T1 and makes an angle of 42° with the horizontal. The right-hand cable has a tension T3 and makes an angle of 41° with the horizontal. A W1 weight is on the left and a W2 weight is on the right. The cable connecting the two weights has a tension 30 N and is horizontal. The acceleration of gravity is 9.8 m/s2. Determine the mass M2.
1. 6.6
2. 5.32
3. 7.58
4. 3.77
5. 2.66
6. 4.7
7. 9.73
8. 4.96
9. 4.53
10. 8.27

In the figure below, a block of mass M hangs in equilibrium. The rope which is fastened to the wall is horizontal and has a tension of 38 N. The rope which is fastened to the ceiling has a tension of 59 N, and makes an angle θ with the ceiling. The angle θ isA) 33°B) 40°C) 45°D) 50°E) 65°

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