Concept: Beam / Shelf Against a Wall18m
Hey guys! In this video I'm going to start solving another type of static equilibrium or complete equilibrium questions in two dimensions. In this type of problems, we're going to have a beam or a shelf-like object that's going to be held against the wall usually by a cable or by a rod. Let's check it out. In some static equilibrium or complete equilibrium problems will have shelf-like or beam-like objects being held against the wall, tension against the wall like this. First force I'm going to draw here is the mass of the beam which will happen in the middle. There is a tension here, T. It's at an angle _, so we decompose this tension into Tx and Ty. In these problems, we're going to have the hinge apply a force against the beam. Notice how there's a Tx to the left. These are equilibrium questions. All forces have to cancel on the x and the y. All torques have to cancel. If there's a force to the left, there has to be a force to the right and that's where the hinge comes in with a force on the right. We're going to call this Hx because it's the hinge force on the x axis. The hinge will always apply a horizontal force against the tension. Tension is going like this which means Tx is like this, which means Hx is like this. In fact more specifically here, I can say against Tx. The hinge will almost always apply a force Hy on the beam so the hinge will also have a y force and most of the time it's going to be helping hold it. We're going to assume that Hy is up. The reason why I say assume is because sometimes it's going to be down. In fact if you get a negative for Hy, meaning you're calculating Hy and you end up with a negative number, that means that you guessed it incorrectly. You assumed the wrong direction. If you get negative for Hy, Hy was actually down. But that's okay and what I mean by that's okay is that it doesn't mean your answer is correct, it doesn't mean that you have to start the problem again. You just realized, ÒOh well it was actually downÓ and then you fixed that. But you don't have to restart the question with it pointing down. You see how this works when I start solving the question. There's an Hy here. Just before we do an example, I want to talk about why there has to be an Hy force. You don't need to really understand this to solve the question but I just want to talk about this very briefly. You have to have a Tx so that you cancel the Hx. That's easy. But why do you use a Ty if you already have a force canceling mg? Why do you need a Ty? Why do you need an Hy, rather? Let's say if you didn't have an Hy, then Ty would have to equal mg, so that the forces cancel. The problem is Ty is farther from the axis at the hinge than mg is. If they had the same forces, this torque here would be much greater. Same forces but a bigger r vector, so this toque will be greater. ItÕs farther from the axis so this thing would spin like this, so that doesn't work. In fact what we need is we need Ty to be less than mg. Ty needs to be less than mg so that they actually, because it's farther, so that they balance in torque. But now it's not enough to hold it on the y-axis so there needs to be another force in the y-axis that's why Hy exists. Hopefully that makes sense. If it doesn't, do worry about it too much. You don't really need that in solving questions. We're just going to assume that there's always an Hy up. Let's solve this problem here. We have a beam of mass = 300. I'm going to put that in the middle here. ThatÕs mg. I'm going to use gravity as 10, so 300*10 is 3000 mg. It's 4 meters in length that means that this distance here is 2 meters and this distance here is 2 meters and I have a tension it's held horizontally against a vertical wall by a hinge, got a hinge here, and a light cable as shown. ThereÕs gonna be a tension here. This is T. It's going to split into a Ty and a Tx. IÕm gonna put Tx right here. The angle here is 37 degrees and I want to know what is the (a) tension on the cable and (b) the net force that the hinge applies on the beam. Before we talk about (b), let me draw the hinge forces. Again, there's going to be always an Hx to cancel out the Tx and we're going to assume an Hy up as always. Net force on the hinge is the combination of Hy and Hx. You have Hy this way, Hx this way. They're going to combine to form a H net and that's what we want to know. H net will simply be, we're looking for H net, the net force by the hinge will simply be the Pythagorean of its components, Hx^2 + Hy^2. I like asking this question here because to find Hx, you basically have to know how to calculate everything else. It's sort of a comprehensive question that guarantees that you know almost everything about this question. I got all my forces here. Remember, once I decompose T into Tx and Ty, I don't really have a T. Same thing here. We're going to use the component forms, Tx and Ty, Hx and Hy and not the total vectors. As with any complete equilibrium questions, I'm going to start by writing sum of all forces equals zero, sum of all torques equals zero. Sum of all forces in the x axis equals zero, sum of all torques in the y-axis equals zero. There are only two forces in the x-axis. I have Hx Tx. They cancel so I can write that one equals the other. On the y-axis, I have two forces going up and that's going to cancel with the one force going down. Ty and Hy going up equals mg. First thing we're looking for is T. By the way, the way we're going to find T is by finding Tx or Ty and then we're going to find T. Remember that Tx = T cos_. I could technically replace this with Tcos_ but I don't want to do that. We're going to leave everything in component form because I think it's much easier this way. We're going to leave them as Tx and Ty and then once you find Tx, you're going to be able to get T because you know the angle or you can get Ty first and then find T. Really when we're looking for T, we're really looking for Tx or Ty whichever one we can find in the easiest way. I cannot find Tx because I don't know Hx. I cannot find Hy because even though I know mg, I don't know Hy. These two equations by themselves are insufficient, so I'm going to have to write a third equation. The third equation will be sum of all torques equals zero about a point. We want to make sure that we pick a clever points to solve this. If we're looking for T, we're first looking for either Tx or Ty. Both of which act here, both of which act at this point here. What we want to do is we wanna write a torque equation at a point away from that point where the T is so that T's will show up on my torque equation. I have a few points that I could pick here, one at the hinge, two at mg and three, at the end where the tension happens. Remember, you always want to write the sum of all torques equals zero equation at a point where there's a force that's why it's either one, two or three. You want to write it at a point where the force you're looking for is not at. I don't want to write this point. This is a bad point. I want to write it at one or two. Here, one of the two is actually better than the other. Point one is the best point because there are two forces, Hy and Hx. If I write it here, I will have fewer torques. Point one is the best point to write. Hopefully you're convinced that point one is the best one. I want to sort of go off to the side real quick and talk about a different point here which is the fact that no matter which point you pick, one two or three, there will never be a torque by Hx and there will never be a torque by Tx. That's because these guys are pointing in the direction of these points. Both of these guys will never produce a torque, at least not if the hinge is horizontal. Rally no matter what, you only had one torque to worry about here which is Hy and mg. What that means is that you could have really picked either one of them. But I had already convinced to about number one, so let's just keep going with that. It really doesn't matter at the end of the day. There are two torques about point one. Let's draw point one right here. Remember, Hy does not produce a torque because it's on the axis. Hx does not produce a torque because it's on the axis and because the r vector would make a 90-degree angle anyway. If it had some length, I have mg halfway here and then I have Ty. Tx does not produce a torque because the r vector, IÕm just going to draw this real quick here but IÕm going to delete it. The r vector between the angle between the r vector for Tx and the Tx itself makes an angle of 180. The sine of 180 is zero. Think of a door. If you push this way in the door, it doesn't spin. YouÕre really left with just these two guys which is awesome. mg is pushing down so it does this and Ty pushes up so it does this. They go in opposite directions. The torque of mg is clockwise negative, the torque of Ty is counterclockwise positive. I can write that torque mg = torque Ty. I'm going to expand both sides of this equation. This is going to be MgRsin_. This is going to be Ty r sin_. _ is not necessarily the 37. Be careful. The r vector for mg looks like this. It has a length of half of the distance of the length, which is length of the bar which is 2. Then this vector here is the entire thing, 4 meters. Both of them notice make a 90-degree angle here. The sine is actually 90, not 37 because we're not talking about T, we're talking about Ty. That's one of the advantages of doing this in component form, leaving the Tx and Ty, not using T. Both of them are 90, so this is going to be 1 and 1. The distances are Mg 2 and Ty 4. I have that Ty = 2 Mg / 4. In other words, Mg / 2. This by the way should make sense. Some of you may have thought about this when I was talking about up here that Ty is double the distance, so it can have the same magnitude as mg. In fact, it has to have half of the force of mg so that the torques balance. We just sort of proved that here. In a question like this where the tension is all the way at the end and you have a horizontal bar with no masses on it, so the standard most simplest form you're going to have the Ty is Mg / 2. I know mg so Ty is going to be 1500 N. I got see Ty. That's good news because remember, once we know Ty, weÕd be able to find T by using the equation that connects the two. Ty is defined as Tsin_, so T is Ty / sin_ so it's 1500 / sin37. If you do this, you get a T of 2500 N. We got that. T is 2500 N up here. Now we want to know H net. To know H net, I need Hx and Hy. Let's look for Hx and let's look for Hy in no particular order. Notice from this equation over here that Hx is Tx. I don't know Tx but I can get it. Now I can get it. Hx Tx which is T cos_ which is 2500, we just got that, cosine of 37. If you plug all of this in, you get that this is 2000 N. That's good. All I got to do is find Hy. If you look around, you will see that there is an Hy equation right here, trying to get an arrow there. That's going to be hard. There's an Hy equation right here. Now that I know Ty and I know mg, I'm going to be able to find Hy. No need to write a new equation. We already got that. Ty + Hy = mg. Hy is mg Ð Ty. mg is 3000, Ty is 1500, so Hy is 1500. This shouldn't be surprising that Hy is1500, same thing as Ty. Ty was holding half of the mg so Hy had to hold the other half of the mg. Now I can combine to find H. H will be the square root of Hx^2 + Hy^2. If I do this, I get Hy is 1500 and then Hx is 2000. If you combine this, you get 2500 N. This is the same as this, which is not a coincidence. It's not a coincidence. This type of problem, IÕm going to call this the base case because it's the simplest problem when you have the beam with the tension at the edge, no mass is on top. It's pretty interesting situation because this is very symmetric. There's a lot of symmetry in this problem. The symmetry is that Ty will equal Hy. Obviously Tx always equals Hx. Because the yÕs and the xÕs are both the same, IÕm going to have that T will equal h and the angle that T makes with the horizontal will be the same angle that H makes with the horizontal. You see this angle right here, that angle here is going to be 37, that angle will be _ of H. In this problem if you were to calculate it, you would get it 37. There's a ton of symmetry here. Kind of interesting to notice. That's it for this one. Hopefully it's made sense and let me know if you have any questions.
Problem: A beam 200 kg in mass and 6 m in length is held horizontally against a wall by a hinge on the wall and a light rod underneath it, as shown. The rod makes an angle of 30° with the wall and connects with the beam 1 m from its right edge. Calculate the angle that the Net Force of the hinge makes with the horizontal (use +/– for above/below +x axis).13m
Example: Beam supporting an object14m
Hey guys! Let's check out this example of a horizontal beam that's being held against the wall with the rope. It also has a 500-kg mass at its end here. The beam has mass of 400. We got the 500 there, so I'm going to call 400 little m and the 500 will be big M. The beam has a length l = 8 meters. I'm going to assume uniform mass distribution on the beam which means that the 400-kilograms acts right here. I have little mg will be 4000. I have big mg right here which is going to be 5000. This is a distance of half the length, 4 meters. This is going to be 4 meters as well. Notice how I have a tension here. IÕm going to actually delete that T there. Because of equilibrium on the 500, this tension here equals 5000 equals the mg. It's just a connecting force. You can just think that the bar is essentially being pulled right here with mg. IÕm going to do that so instead of having mg and T, I just have the same one force. The truth is if you had two mgÕs, little mg and big Mg, right here. IÕm also going to have a tension here. This tension, we're actually going to write it out. This tension gets split into Tx and Ty. This is going to be our Tx right here and this is going to be our Ty right here. The angle between the cable and the horizontal is 53, so this angle right here is of 53 degrees. It says it connects 2 meters from the right edge of the beam. This distance here is 2 meters, which means that this distance here will also be 2 meters. A 500-kilogram object hangs from the right edge of the beam. We see that. We wanna calculate the magnitude of the net force the hinge applies on the beam. What is H net? Remember, the hinge will apply an x-force and a y-force. What we're going to do is we're going to have an H force this way, Hx, to cancel out the Tx. Hy, we donÕt know if it's up or down so we're going to assume Hy is up. Let me write this down here. We're going to assume Hy to be up and let's see what we get. Once I find Hx and Hy, I'll be able to calculate H net which is this. H net will be the square root of Hx^2 + Hy^2. What we're really looking for in this problem is we're looking for Hx and Hy first so that we can find the magnitude of the net force here. We'll be able to continue that there. How are going to find Hx and Hy? Like every torque, every equilibrium problem, we're going to write that sum of all torques equals zero and the sum of all forces equals zero.
Let's start with forces because it's simpler. Sum of all forces in the x equals zero and sum of all forces in the y axis equals zero. There are only two forces in the x-axis, Hx + Tx so they must cancel each other. Hx = Tx. On the y-axis, I have two forces up and two forces down. I have Ty + Hy = little mg + big Mg. I don't know Ty or Hy but I know the 2 mgÕs. They're going to be Ty + Hy, they are 4000 and 5000 so it's 9000 total. This is the best do so far. I'm looking for Hx but that requires knowing Tx and IÕm looking for Hy but that requires knowing Ty. We're going to have to write a third equation. At this point, you can think of this question as not even you looking for Hx and Hy but you looking for Tx and Ty. We're going to write a third equation that has to be a torque equation, sum of all torques equals zero. We're going to pick the best possible points. We're looking for Tx and Ty so we want to pick a point away from here so that we can actually find Tx and Ty. We could also pick this one or this one. In here, it really doesn't matter. I'm going to do this about this point right here. Let's actually just do it over here. The sum of all torques about this point here, let's call this point one, two, three and four. These are all the possible candidates. We're going to go with this one because I want to find Ty and Tx. I want to make sure my torque equation includes Ty and Tx, so I don't want to pick point number three. That's a bad choice if I want to find Ty and Tx. I want to make sure I picked the point where I will cancel the most forces. In this case, it's Hy and Hx, though those two forces really only Hy will give me a torque. Hx doesn't give you a torque. Really, all of these points, Tx also doesn't give you a torque. Really all of these points only have each one force that would give you a torque. They're all sort of similarly just as good or just as bad hardly when we look at it. We're just going to go with torque about point one. It would have worked with other points. Don't sweat that too much. Torque about point one means I have one here and I have 4000 pulling this down. This is little mg. I have 5000 big Mg over here and then I have Ty. Ty we don't know. Both of these guys are producing a torque in this direction. If this is torque of little mg, this is torque of big Mg, both of these guys are clockwise so they're both negative and then this is going to provide a counter clockwise torque positive. I'm going to be able to write that the torque of Ty = torque of little mg + torque of big Mg. I'm going to expand this equation here. It's going to be Ty r sin_. Notice that all the angles are going to be 90. This is the r vector for the little mg. This is the r vector for big Mg and this is the r vector for Ty. In all of these, the angles are 90 degrees, so sin90. mg r sin90 + big Mg r sin 90. Sine of 90 is 1, so these guys go away. The distance from the axis we chose to Ty is this whole thing here. If you look here, I got a 4 with a 2 so that's a 6. Little mg is just a 4 and for big Mg, itÕs the entire length, it's 8. If you start plugging this in, you get 6 Ty = little mg is 4000*4, big mg is 5000*8. This is going to be 56000. I want to divide that by 6, so Ty will be 56000 / 6 which is 9333 N. That's Ty.
Once I know Ty, I can find Hy. I'm going to plug it into this equation right here. Ty + Hy = 9000. This means that Hy will be 9000 Ð Ty or 9000 Ð 9333, which means Hy is negative 333 N. I got a negative. What the heck does that mean? Remember how we assumed Hy to be up. We got a negative number. This means we assume this incorrectly. All we got to do here is we're going to say Hy is actually down but nothing changes. The magnitude is still 333. Nothing to redo or undo. We're just going to keep going. I now have Hy. I just need Hx. I now have Hy and I have Ty as well. I just need Hx. To find Hx, I'm going to need Tx. How do I get Tx? From Ty, I can get Tx. I can go from Ty to T and then from T to Tx. Once I have Tx, Tx is the same as Hx. Let's do that. Ty = T sin_. I can figure out the T. I can find out the T = Ty / sin_. In this case, the _ we're talking about is the angle that T makes with the x axis which is 53. Ty was 9333 divided by the sine of 53. The sine of 53 is 0.8. If you do this, you get 11666 N. Now we got Ty. We got T. We're going to find Tx. Tx = T cos_. I now have T so I can plug it in, 11666 cos53. If you do this, you get 7000. Tx = 7000, which means Hx = 7000 as well. Now that I have Hy and I have Hx, I can plug it into that equation up top. Here I'm going to get the square root of Hx which is 700^2 and then Hy -333. It doesn't matter that it's negative because it's going to get squared anyway. This is going to be 70008. Most of the vector came from here, which makes sense. It's a much bigger number. 70008 N is the net force for the hinge and this is the final answer. One point that I want to make here is in this problem, the reason why we had a negative Hy is because there's actually way more mass to the right of the cable. If you want to calculate this, the center of mass of the cable, the center of mass of the system that's being held which is beam plus the 500 kilograms is actually somewhere over here to the right of the tension. This is the only time where you're going to have an Hy that is going down. But it's not worth trying to figure that out, trying to guess. It's better to just always go with the Hy going up because chances are Hy will be up. I just wanted to show you an example where it doesn't go up, basically to show you that nothing weird happens. Also, we got to see how it works when we have sort of an object hanging from it, so you have two objects in total. That's it for this one. Let me know if you have any questions and let's keep going.
Problem: A beam 200 kg in mass and 4 m in length is held against a vertical wall by a hinge on the wall and a light horizontal cable, as shown. The beam makes 53° with the wall. At the end of the beam, a second light cable holds a 100 kg object. Calculate the angle that the Net Force of the hinge makes with the horizontal (use +/– for above/below +x axis).11m
A uniform metal rod, with a mass of 5.0 kg and a length of 2.0 m, is attached to a wall by a hinge at its base. A horizontal wire is bolted to the wall above the base of the rod and holds the rod at an angle of 30° above the horizontal. The wire is attached to the top of the rod.
(a) Find the tension in the wire.
(b) Find the horizontal and vertical components of the force exerted on the rod by the hinge. Use g = 10 m/s2.
A horizontal, nonuniform beam of mass M and length ℓ is hinged to a vertical wall at one side, and attached to a wire on the other end. The bar is motionless and a wire exerts a force T at an angle of φ with respect to the vertical. If the mass of the beam is 7 kg, the tension in the wire is 35 N, sin φ = 3/5 and cos φ = 4/5, how far is the center of mass of the beam from the hinge? The acceleration due to gravity is 10 m/s2.
1. 0.32 ℓ
2. 0.5 ℓ, of course
3. 0.4 ℓ
4. Not enough information is given.
5. 0.18 ℓ
6. 0.16 ℓ
7. 0.375 ℓ
8. 0.666667 ℓ
9. 0.3 ℓ
10. 0.625 ℓ
A solid bar of length L has a mass m 1. The bar is fastened by a pivot at one end to a wall which is at an angle θ with respect to the horizontal. The bar is held horizontally by a vertical cord that is fastened to the bar at a distance x from the wall. A mass m2 is suspended from the free end of the bar. Find the tension in the cord.
1. T = (m1 + 1/2 m2) (L/x) gcosθ
2. T = (m1 + m2) gsinθ
3. T = (m1 + m2) gcosθ
4. T = (m1 + m2) (L/x) (g/2)
5. T = 0
6. T = (m1 + 1/2 m2) (L/x) g
7. T = (m1 + 1/2 m2) (L/x) gsinθ
8. T = (1/2 m1 + m2) (L/x) gsinθ
9. T = (1/2 m1 + m2) (L/x) g
10. T = (1/2 m1 + m2) (L/x) gcosθ
A uniform sign is supported at P as shown in the figure. If the sign is a square 0.7 m on its mass is 7.0 kg. What is the magnitude of the horizontal force that P experiences?
A) 24 N
B) 98 N
C) 34 N
D) 0 N
A uniform bar has mass 3.00 kg and length 6.00 m. The lower end of the bar is attached to the wall by a frictionless hinge. The bar is held stationary at an angle of 53.1° above the horizontal by a horizontal cable that runs from the upper end of the bar to the wall. For an axis at the hinge, what is the magnitude of the torque produced by the weight of the bar?
A uniform beam 4.30 m long and weighing 2500 N carries a 3200 N weight 1.50 m from the far end, as shown in the figure below. It is supported horizontally by a hinge at the wall and a metal wire at the far end. What are the horizontal and vertical components of force exerted on the beam by the hinge?
A uniform beam 4.30 m long and weighing 2500 N carries a 3200 N weight 1.50 m from the far end, as shown in the figure below. It is supported horizontally by a hinge at the wall and a metal wire at the far end. How strong does the wire have to be? That is, what is the mximum tension it must be able to support withot breaking?
A uniform bar (I = 1/3 ML2 for an axis at one end) has mass M = 5.00 kg and length L = 6.00 m. The lower end of the bar is attached to the wall by a frictionless hinge. The bar is held stationary at an angle of 60° above the horizontal by a cable that runs from the upper end of the bar to the wall. The cable makes an angle of 37° with the bar. What is the tension in the cable?
A uniform bar with mass 50 kg and length 4.0 m is attached to a wall by a frictionless hinge. The bar is held in a horizontal position by a light rope that is attached at the end of the bar. The other end of the rope is attached to the wall. The rope makes an angle of 30° with the wall. What is the magnitude of the resultant force that the hinge exerts on the bar?
A uniform bar with mass 50 kg and length 4.0 m is attached to a wall by a frictionless hinge. The bar is held in a horizontal position by a light rope that is attached at the end of the bar. The other end of the rope is attached to the wall. The rope makes an angle of 30° with the wall. What is the tension in the rope?
A uniform bar has mass 30 kg and length 6.0 m. One of the bar is attached to a vertical wall by a frictionless hinge. A light horizontal cable connects the other end of the bar of the wall and holds the cable at an angle of 37° above the horizontal. If the cable is cut, what is the magnitude of the angular acceleration of the bar just after the cable is cut?
A uniform bar has mass 30 kg and length 6.0 m. One of the bar is attached to a vertical wall by a frictionless hinge. A light horizontal cable connects the other end of the bar of the wall and holds the cable at an angle of 37° above the horizontal. What is the tension in the cable?
One end of a uniform bar that is 6.0 m long is attached to a vertical wall by a frictionless hinge. The bar is held at an angle of 60° above the horizontal by a horizontal rope that is attached to the other end of the bar, as shown in the sketch. If the tension in the rope is 120 N, what is the mass m of the bar?
One end of a uniform bar that has weight w is attached to a wall by a fricitonless hinge. The bar is held in a horizontal position by a cable tha makes an angle of 37° with the bar. The tension in the cable is 80.0 N. What is the weight of the bar?