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Concept #1: Equilibrium in 2D - Ladder Problems

Transcript

Hey guys! In this video, we're going to start solving equilibrium questions that are two-dimensional and IÕm going to give an example of a classic problem in equilibrium that is the latter problem. Let's check it out. So far we've solved equilibrium problems that were essentially one-dimensional, meaning all the forces acted in the same axis, either you had all the forces in the x-axis or all the forces in the y-axis, most of them in the y-axis. Even if you had something at an angle like this, let's say you had something like this, that's still essentially one-dimensional because the angles were the same when we wrote the torque equation and they cancel. In all the problems weÕve solved so far, sin_ in the torque equation never really mattered because it either canceled or it was just the sin90 which is 1. Now we're going to finally solve some problems where that's not the case. We're going to actually have to worry about the angle. More advanced problems as it says here will include problems in two dimensions in two axes. Some of them in some of these cases, we may need to decompose the forces. Some of them will be decomposed. Remember however the torques are scalars so we will never need to decompose them. We're going to decompose forces in problems, but we don't have to decompose torques because torques are scalar. They may be positive or negative but they are scalars. They don't have a vector direction. Let's check out this problem here.

We have a ladder of mass 10 kilograms, so m = 10 and it's uniformly distributed. What that means is that the mg of the ladder will act right at the middle so I'm going to do this here and say this is 2 meters and this here is 2 meters as well. The ladder has length 4, that's why I did 2 and 2. It rests against a vertical wall while making an angle of 53 with the horizontal shown. This is 53. This by the way, because this is 90, if this is 53, this is 90 Ð 53 which is 37. Let's just put that there. When I calculate a bunch of stuff, these are all the things you might see in a classic ladder question. I want to find the normal force at the bottom of the ladder and a bunch of other stuff. But before I read the list, let's talk about what forces we have here. You have an mg that pulls you down. Obviously there's a normal here that pushes you up because the latter is resting against the vertical wall. There's also normal right here and normal is always perpendicular to the surface, so normal is going to be like this. There are two normal forces here. There's normal bottom and thereÕs normal at the top. Notice that obviously we want this ladder to be in complete equilibrium so all forces cancel and all torques cancel. But if you look at what we have right now, there's a force going to the left but there's no forces going to the right. At least I haven't drawn them yet. That means this ladder would not be an equilibrium. There has to be a force going to the right and that force will be friction over here. Because we want the ladder not to move, this is static friction. There's enough forces that everything cancel. These are all the forces you're going to have. I can write that the sum of all forces in the x-axis equals zero. What this means is that the forces in the x cancel each other, so I'm going to have normal top equals friction static. That's friction at the bottom. Sum of all forces on the y-axis equals zero, so this means that N bottom equals mg. The two cancel. N bottom = mg. I can write more equations. Now I can write torque equations. Sum of all torques at any point p equals zero. There are three points here what I might want to write this. There's the point one here at the bottom, two at the middle and three at the top. These are points where forces happen. Remember, you want to write your torque equations about the point where a force happens so you have fewer terms when you write out the torque equation. Let's see.

Here we want to find the normal force at the bottom of the ladder. This one is the easiest thing to find that's why I put it here first. Normal at the bottom is just mg. Here we have mg, so N bottom is m = 10 and n, we're going to use 10 as well for g and this is going to be 100 N. That's easy. This is just 100 N. What about the normal force at the top of the ladder? This whole thing here is done. The normal force at the top of the ladder to solve for that, I would need to know static friction. I don't have enough information just yet to find static friction, so I don't have _. I do have normal at the bottom. I don't have _ and I wouldn't be able to use that anyway. Normal at the top, to find that I'm going to have to write a torque equation because this here is not enough. We're going to write a torque equation. If you want normal at the top, you want to write a torque equation with your axis being somewhere else. The reason being you want your N top to show up on the equation. If you put the axis of rotation here, then there will be no torque due to normal at the top and it won't be part of the equation. We're going to carefully select our axis to be one of the other two points. IÕm going to pick the bottom here and IÕm going to pick that point because that point is actually the best of the two to pick. Remember, you want to write the torque equation about a point with as many forces a possible so that you have as few terms as possible. There are two forces acting at point one, only one force acting at point two. Point one is definitely the best one to write a torque equation about it. Sum of all torques at point one equals zero. There are two torques there. I'm going to draw this. Here's point one. Here's the ladder. I have mg acting here and then I have N top acting here. The r vector from here to mg looks like this; from the axis to mg looks like this, and then from the axis to N top looks like this. The bottom one has a length of 2 and the top one has a length of 4, half and the total distance. The angle that I'm supposed to use is not the 53 right here. But instead IÕm supposed to use the 37 right here. That's what we're going to use. We're going to use _ = 37. For this one here, IÕm supposed to use this angle and if this is 53, then this is 53 as well. They have different rÕs and different _Õs. You have to be very careful here. Hopefully that makes sense. Let's keep going here. I'm going to say that this torque is going this way, torque of mg, and the torque of normal is going that way and I'll show you that just a second. Imagine your ladder looks like this. mg is pushing down so it's trying to cause a rotation in this direction and this direction right here. This is clockwise so it's negative. Normal at the top is pushing this way. It's trying to cause a rotation this direction and this is positive. They're cancelling each other so I can write the torque of N top equals torque of mg. IÕm going to expand both equations. Torque N top is N top. It's r vector and sin_, mg r vector sin_. The r vector for N top is 4 meters long. For mg, itÕs 2 meters long. The sine for N top right here between N top and its r vector is 53 and this one is 37. I have mg. I can find the sines so I'll be able to find N top. N top will be m which is 10, g which is 10, times 2 sin37 is 0.6 divided by 4 divided by sin53 which is 0.8. If you multiply all of this, you get 37.5 N. As soon as I get this, I will be able to know. As soon as I know N top, I know friction static as well. Friction static is the same thing as N top so it's going to be 37.5 just the same. This was N bottom. This is N top and this is friction. If I scroll up over here, 37.5. Normal force at the top 37.5 N. Frictional force at the bottom of the ladder, 37.5 N as well. I want to know the minimum coefficient of static friction needed so I'm looking for _ static min. Just ignore the min. All you're really looking for is _ static. Plug it into here. WeÕre going to say friction static is 37.5. You know that the equation for friction static is _ static normal. This gets replaced with _ static normal = 37.5. Do I use normal bottom or normal top, what do you think? Friction is over here so you use normal bottom. _ static will be 37.5 divided by normal bottom. Normal bottom was 100, so _ is going to be 0.375. Remember, _ is supposed to be a number between 0 & 1. I got a number between 0 and 1 so I have higher confidence that this is correct. This is in fact the right answer. That is _ which by the way this is Part D. Here we found this is A and then somewhere here we found B and C together. The coefficient here is 0.375 unit less. I want to know the total contact force at the bottom of the ladder. What's the deal with that? Look at the bottom of the ladder. There are two forces, N bottom and friction static. One thing you might be asked is for the total force which is simply the vector addition between these two. It's going to look something like this. The total contact force is just a combination of the two forces at the bottom. Part E, total contact force. You're going to have N bottom which is 100. You're going to have friction static 37.5. Then the total contact force, IÕm going to call this F bottom is what we're looking for. This is just basic vector addition. We're going to use the Pythagorean theorem to combine these two. F bottom is just the square root of 100^2 because of this number plus this squared, 37.5^2. If you do this, you get 107 N which is the magnitude of this force. IÕm going to just solve for one more thing. It wasn't asked here but I could have asked also what is the direction of the bottom force of the total force at the bottom, so _ bottom. Remember, this is just vector stuff, is the arctangent of the y-force divided by the x-force. The y-force is 100. The x-force is 37.5. If you do the arctangent of this, you get 69 degrees, which means that 69 degrees is right here. That's it for this one. These are all the classic things you would be asked on a basic ladder question. Hopefully this makes sense and let me know if you have any questions. Let's keep going.

Practice: A ladder of mass 20 kg (uniformly distributed) and length 6 m rests against a vertical wall while making an angle of Θ = 60° with the horizontal, as shown. A 50 kg girl climbs 2 m up the ladder. Calculate the magnitude of the total contact force at the bottom of the ladder (Remember: You will need first calculate the magnitude of N,BOT and f,S).

Example #1: Minimum angle and friction for ladder

Transcript

Hey guys! In this example, IÕm going to solve a literal ladder question, meaning a ladder question with no numbers and only variables. This is just going to get a little hairy so unless you know that your professor likes this kind of stuff or may like this kind of stuff, you might want to skip this one and save some brainpower. If you do need it though, let's check it out. We have a ladder of mass M that is uniformly distributed, meaning the mg acts in the middle of the mass of the object and it has length L. The whole thing is L, mg acts in the middle over here. IÕm gonna draw it. That means that this piece here is L/2 and this over here is L/2 as well. It says it rests against a vertical wall while making an angle. I don't have the angle here. This is _ with the horizontal as shown. This is _ right here. We want to derive an expression for a bunch of stuff. Before I do that, let's draw some of the other forces. This is resting on the floor so it's got a normal at the bottom. ItÕs resting against the wall so it has a normal this way. Normal at the top always perpendicular, normal force. There also has to be a friction at the bottom here to cancel out normal at the top. First question, we want to write an expression for the minimum coefficient of friction of static friction needed for the ladder to stay balanced at an angle of _. The idea is that given _, given an angle _, what is the minimum coefficient of friction that you need for that _ to work? You can imagine that if you have a ladder that is very steep, let's say this angle here is 80 degrees, it doesn't take a lot of friction to hold this ladder as opposed to if you have a ladder like this, it's trying really hard not to fall like this. It's trying not to slide this way here which would cause it to fall. As you change the angle, you can see how the angle and the _ are related. In fact the smaller my angle, the more coefficient of friction I need to have. Let's find an expression for _ static. We're going to start this problems like we start all static equilibrium problems which is by writing sum of all forces equals = 0 and sum of all torques = 0. Sum of all forces in the X axis equals 0. This means that the forces in the X cancel the forces in the X here are N top and friction static. Sum of all forces in the Y axis equals 0. It means that the force in the y axis will cancel as well. This is N bottom = Mg. Before I write torque equations, let's see maybe I don't even need it. I'm looking for _ static. I hope you see that _ static I will find it in here somewhere. I'm going to start at this equation by expanding friction into _ and normal. Friction static will be _ static normal, friction at the bottom so this is gonna be normal at the bottom and equals N top. We're looking for _ static. These literal questions, these questions have to derive an expression means that the answer has to be in terms of the given variables. I don't know what M is in terms of numbers if it's a 3 or a 50, but I'm given M since it says mass M, that means it's a given which means I can use that. It's allowed to be in the final answer in the expression. M, L and _ and then stuff like g and constants and other universal constants could show up in the answer. N top and N bottom are not given so they cannot show up in the answer. I'm going to have to replace them with other stuff. Notice here that N bottom is equal to Mg. M and g are both are both variables that could be in the final answer. Right away you want to replace that but you don't have N top. You're going to have to find an expression for N top. IÕm gonna write here the _ static = N top / N bottom, N bottom is Mg, so I have to get this and I'll be able to continue.

Let's go find an expression for N top. To do this, I'm not going to be able to use these two equations because I would use them I'm stuck I need more. I'm going to have to write a torque equation. Sum of all torques at some point equal zero. I'm going to have to pick a good point to write this torque equation. There are three points here that I could use: one, two or three. I want to make sure I don't use point three over here as the axis of rotation that's because when you write your torque equation, the point that you pick to be your axis will have no torques on it because a force acting on the axis of rotation doesn't produce a torque. If you were to pick point three, we're not going to do that. Normal top would have no torque therefore N top is not going to show up in the equation so you can't solve for it. You want N top to show up in the equation, so you want to pick either points one or two. Out of these two choices, one is a better point to pick because there's two forces acting on it. Remember, you always want to pick the point with the most forces so that you cancel the most things and you have the fewest number of terms on your equation. We're going to pick one. Then if you pick one, let's draw one over here as the axis of rotation and then here's our ladder. I have mg in the middle and then I have N top here. These are the only two forces that will produce a torque. This is my r vector here. This is my r vector for Mg and it's going to be r = L/2, itÕs the halfway point. Then this is going to be r equals the entire length. This angle here is the given angle _. This angle here is the other angle, itÕs the complementary angle. I'm going to call this _Õ and I'm going to say that _ + _Õ = 90 so you can think of _Õ as 90 degrees minus _. This angle over here is _. This angle here is the same angle here. When I write the torque for mg, I'm going to use this angle but when I write the torque for N top, IÕm gonna use this angle. The torque due to mg causes tries to cause a rotation this way. A torque due to normal tries to cause a rotation this way. Here's your ladder, mg is pushing this way, N top is pushing that way. TheyÕre opposite to each other. This one is negative, this one is positive so I can write that torque N top = torque Mg. I can expand these equations. Torque N top is going to be N top r sin_ = Mg r sin_. They each have their rÕs and their _Õs. The r for N top is all the way at the end, L. For Mg itÕs L/2. The angle for N top is _, that's the angle that I want. It's between the axis of rotation of the force, so that's _. For Mg, it's this angle right here which is _Õ, it's the other guy. _ and _Õ. Notice that the LÕs cancel and I'm looking for N top so I'm going to move everything the other way. I have N top equals Mg/2 sin_Õ/sin_. We're almost done. I can use M and g as part of my final answer. I can use _ as part of my final answer, but I can't really use sin_Õ because that's not one of the given variables so I have to replace it. There's two ways I can do this. I can rewrite sin_Õ as sin90 Ð _ or I can use a trig identity to simplify this. You should know that sin90 - _ is simply the cos _. One way that you can remember this very simply is that sin30 = cos 60, and sin60 = cos30. Complementary angles swap sines and cosines. If I have sin90 Ð_, itÕs cos_. If I had cos90 Ð _, this would have been equal to sin_. This is actually pretty good news because it means that this is going to look like this. Instead of sin_, I don't want to use this version. I actually don't even want to use this version. I want to use this version right here because it's the simplest and something interesting happens here. this is going to be cos_/sin_. You might be thinking wait isn't that tangent? Actually it's not, it's the inverse tangent though so that's good enough. sin_/cos_ obviously is tangents. It feels like a trig review, but cos_/sin_ is just 1/tan_ or your cotangent. But we're just going to write this as tangent at the bottom. This means youÕre going to have mg / 2 tan_. Now we're done. Annoying but we're done, mg and _. Everything is as it's supposed to be. I can plug this mess in here and then we can continue. _ static will be normal top which is this whole thing, mg / 2tan_ and then this whole thing is divided by mg. Look what happens. The mgÕs cancel and you're left with _ static being 1/2tan_ at the bottom. That's _ static. That's part A. All of this is part A. To find _ static, that's the relationship for _ static. Notice that as you change _, _ static will change as well.

Parts B and C will follow from here. Let me make a little bit of room here. We're going to be a little tight in space so let's write small. B, we're looking for it says find the minimum angle at which the ladder can stay balanced for a coefficient of static friction _ static. For a given _ static, what is the minimum _ that you can have? Again remember, the two of them are related so how much of an angle can _ static handle? Basically what we're going to do here, this is going to be pretty straightforward. All we're going to do is use this equation right here and solve for _. If we didn't have this equation already, we would go through a very similar process to what we did in the first part and derive that equation like torque equations and everything to get to a point where we can get an expression for _. Once we have this, all we have to do solve for _ within that equation. _ static = 1/2tan_ and I want to solve for _. IÕm gonna move tan_ up and move _ static down and it's going to look like this, tan_ = 1/2 _ static and we're looking for _ so it's the arctangent of 1/2_ static. That's the final answer for Part B. Once you have one, you can get the other. Again if you were asked to find _ minimum first, you would have gone through a similar process as what we did here. You would find this and then you would flip it around to find this. Let's do Part C. Part C says find the minimum angle at which the ladder can stay balanced for any coefficient of friction. I'm going to draw some stuff here just to illustrate a point but I'm going to delete it so you donÕt have to draw this. Imagine that at some point, the smaller the angle becomes, this is pretty easy to hold at 80 degrees. Let's say at 50 it might be a little harder. At some point it really doesn't matter. Imagine you try to hold a ladder at 10 degrees, it probably doesn't hold. It probably just falls no matter how much friction you have. There's a minimum angle of how low you can go with _ until friction just can't handle it anymore. That's what we wanna find. Again as you reduce your _, you need more and more friction. To find the minimum _, you need to figure out what angle you get for the most possible friction. The most friction can handle is when you have a coefficient of friction that is maximum. You want your coefficient of friction to be the maximum amount. The maximum amount that a coefficient of friction can have is 1. That's what we're going to do. We're going to set this to be 1 and we're going to find if _ is 1, which is the max possible, then what is _? To do this, we can simply plug it into this equation right here. You can see how B and C follow from A. _ is the arctangent of 1 / 2*1. This is the arctangent of _. If you plug this into the calculator, this is 26.6 degrees. You can go try this at home if you want. You cannot have a ladder hold against the wall or any object for that matter holding against the wall like this if your _ is less than 26.6 degrees because it will fall. Depending on the coefficient of friction between the object and the floor, this angle is going to have to be even greater than 26.6. That's it for this one. Hopefully it made sense. Let me know if you guys have any questions and let's keep going.

A uniform bar of mass M and length l is propped against a very slick vertical wall as shown. The angle between the wall and the upper end of the bar is θ. The force of static friction between the upper end of the bar and the wall is negligible, but the bar remains at rest (in equilibrium). If we take the pivot at the point where the bar touches the floor, which expression below is (Στ), where x is along the floor and y is along the wall? 1. ℓ (Fwsinθ + Mg/2 cosθ) = 0 2. ℓ (Mg/2 cosθ − Fwsinθ) = 0 3. ℓ (Mg/2 − Fw) = 0 4. ℓ (−fssinθ − ncosθ) = 0 5. ℓ (Mg/2 + Fw) = 0 6. ℓ (Fwsinθ − Mg/2 cosθ) = 0 7. ℓ (−Mg/2 cosθ − Fwsinθ) = 0 8. ℓ (fssinθ − ncosθ) = 0 9. ℓ (Fwcosθ + Mg/2 sinθ) = 0 10. ℓ (Fwcosθ − Mg/2 sinθ) = 0 
A ladder is leaning against a smooth wall. The friction between the ladder and the floor holds the ladder in place. Determine torques about the pivot point P. Let τCCW and τCW represent counterclockwise and clockwise torques about point P (the base of the ladder), respectively.
In the figure, a ladder of weight 200 N and length 10 meters leans against a smooth wall (no friction on wall). A firefighter of weight 600 N climbs a distance x up the ladder. The coefficient of friction between ladder and the floor is 0.5. What is the maximum value of x if the ladder is not to slip?A) 6.28 mB) 6.04 mC) 8.44 mD) 3.93 mE) 5.00 m
An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth, frictionless wall. If the ladder has a mass of 20 kg and it makes an angle of 60° with the ground, find the force of friction of the ground on the foot of the ladder.a. 7.8 x 102 Nb. 2.0 x 102 Nc. 50 Nd. 1.7 x 102 N