Beam / Shelf Against a Wall - Online Tutor, Practice Problems & Exam Prep

In static equilibrium problems involving beams, forces must balance in both the x and y directions. The tension in the cable can be decomposed into horizontal (t_x) and vertical (t_y) components, while hinge forces (h_x and h_y) counteract these tensions. The torque produced by these forces must also equal zero. Key equations include $\Sigma F=0$ for forces and $\Sigma \tau =0$ for torques, ensuring stability and balance in the system.

Beam / Shelf Against a Wall

Hey guys. So in this video, I'm going to start solving another type of static equilibrium or complete equilibrium questions in 2 dimensions. Now, in these 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, we'll have shelf-like or beam-like objects being held against the wall, tensioned against the wall like this. The first force I'm going to draw here is the mass of the beam, which will happen in the middle. There is tension here, \( t \), at an angle \( \theta \). So we decompose this tension into \( t_x \) and \( t_y \). In these problems, we're going to have the hinge apply a force against the beam. Notice how there's a \( t_x \) 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. So we're going to call this \( h_x \) because it's the hinge force on the x axis. So the hinge will always apply a horizontal force against the tension. The hinge will almost always apply a force \( h_y \) on the beam. The hinge will also have a y force. We're going to assume that \( h_y \) 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 \( h_y \), meaning you're calculating, and you end up with a negative number, that means that you guessed it incorrectly. You assumed the wrong direction. If you get a negative for \( h_y \), \( h_y \) was actually down, but that's okay.

Before we do an example, I want to talk about why there has to be an \( h_y \) 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 \( t_x \) so that you cancel that \( h_x \). That's easy. But why do you need a \( t_y \) if you already have a force canceling \( mg \)? Let's say if you didn't have an \( h_y \), then \( t_y \) would have to equal \( mg \) so that the forces cancel. The problem is, \( t_y \) is farther from the axis at the hinge than \( mg \) is. If they had the same forces, this torque here would be much greater. It's farther from the axis. So this thing would spin like this. So in fact, what we need is we need \( t_y \) to be less than \( mg \). 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 \( h_y \) exists. Hopefully, that makes sense.

Now let's solve this problem here. We have a beam of mass 300, \( \text{mass} = 300 \). I'm going to put that in the middle here. Right there. So that's \( mg \). I'm going to use gravity as a 10. So \( 300 \times 10 = 3,000 \). \( MG \) is 4 meters in length. That means that this distance here is 2 meters. And then this distance here is 2 meters. And I have a tension. It's held horizontally against the vertical wall by a hinge. Got a hinge here and a light cable as shown. So it's going to be tension here. This is \( t \). It's going to split into a \( t_y \) and a \( t_x \). We'll put \( t_x \) 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. </p>

Before we talk about b, let me draw the hinge forces. Again, there's always going to be an \( h_x \) to cancel out the \( t_x \) and we're going to assume an \( h_y \) up as always. Now the net force on the hinge is a combination of \( h_y \) and \( h_x \). You have \( h_y \) this way, \( h_x \) this way. They're going to combine to form an \( h_{\text{net}} \) and that's what we want to know. \( h_{\text{net}} \) will simply be the Pythagorean of its components. So \( h_x^2 + h_y^2 \). I like asking this question here because to find \( h_x \), you basically have to know how to calculate everything else. So it's sort of a comprehensive question that guarantees that you know almost everything about this question. Okay.

So I got all my forces here. Remember, once I decompose, once I decompose \( t \) into \( t_x \) and \( t_y \), I don't really have a \( t \). Same thing here. So we're going to use the component forms, \( t_x \), \( t_y \), \( h_x \) and \( h_y \), and not the total vectors. Okay? So as with any complete equilibrium questions, I'm going to start by writing the sum of all forces equals 0, the sum of all torques equals 0. So the sum of all forces in the x axis equals 0. So the sum of all forces in the y axis equals 0. And there are only 2 forces in the x axis. I have \( h_x \), \( t_x \). They cancel so I can write that one equals the other. And on the y axis, I have 2 forces going up and that's going to cancel with the one force going down. So \( t_y \) and \( h_y \) going up equals \( mg \). First thing we're looking for is \( t \), in this equation here. And by the way, the way we're going to find \( t \) is by finding \( t_x \) or \( t_y \) and then we're going to find \( t \). Okay? Remember that \( t_x \) equals \( t \cos(\theta) \). So I could technically replace this with \( t \cos(\theta) \), 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. So we're going to leave them as \( t_x \) and \( t_y \). And then once you find \( t_x \), you're going to be able to get \( t \) because you know the angle or you can get \( t_y \) first and then find \( t \). Okay? So really when we're looking for \( t \), we're really looking for \( t_x \) or \( t_y \) whichever one we can find in the easiest way. I cannot find \( t_x \) because I don't know \( h_x \). I cannot find \( h_y \) because even though I know \( mg \), I don't know \( h_y \). So these two equations by themselves are insufficient. So I'm going to have to write a third equation. The third equation will be the sum of all torques equals 0 about a point. And we want to make sure that we pick a clever point to solve this. If we're looking for \( t \), we're first looking for either \( t_x \) or \( t_y \), both of which act here. Both of which act at this point here. So what we want to do is we want to 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. Okay. So I have a few points that I could pick here. 1 at the hinge, 2 at \( mg \) and 3 at the end where the tension happens. Remember, you always want to pick torque. You always want to write the sum of all torque equals 0 equation, at a point where there's a for

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, and use g=10 m/s².)

• A. θ = −49°
• B. θ = −41°
• C. θ = +41°
• D. θ = +49°

Beam supporting an object

Hey, guys. Let's check out this example of a horizontal beam that's been held against the wall with a rope. It also has a 500-kilogram weight or mass at its end here. So the beam has a mass of 400 kg. We got the 500 there. So I'm going to call it 400 little m and the 500 will be big M. The beam has a length l equal to 8 meters. I'm going to assume uniform mass distribution on the beam, which means that the 400 kg acts right here. So I have little mg will be 4,000. I have big Mg right here, which is going to be 5,000. 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. Right? But because I'm going to actually delegate that to you there. Because of equilibrium on the 500, this tension here equals 5,000 equals the Mg. It's just a connecting force. So you can just think that the bar is essentially being pulled right here with Mg. Okay? I'm going to do that so instead of having Mg and t, I just have the same one force. Okay? So imagine that, treat this as if you had 2 mgs, little mg and big mg right here. I'm also going to have a tension. I'm also going to have a tension here. So this tension, we're actually going to write it out and this tension gets split into tx and ty. So this is going to be our tx right here and this is going to be our ty right here. Okay. The angle between the cable and the horizontal is 53 degrees. So this angle right here is 53 degrees. And it says connect 2 meters from the right edge of the beam. So this distance here is 2 meters which means that this distance here will also be 2 meters. Okay? So 4, 2 and 2. Alright? A 500-kilogram object hangs from the right edge of the beam. We see that. And we're going to calculate the magnitude of the net force, the hinge applies on the beam. So what is hnet? Remember, the hinge will apply an x force and a y force. And so 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 don't 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 hnet, which is this. Okay? So hnet will be the square root of hx+hy squared. So 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. Okay. So we'll be able to continue that there. So how are we going to find hx and hy? Well, like every torque, every equilibrium problem, we're going to write the sum of all torques equals 0 and the sum of all forces equals 0. Let's start with forces because it's simpler. Sum of all forces in the x equals 0 and sum of all forces in the y axis equals 0. There are only 2 forces in the x-axis, hx and tx, so they must cancel each other. hx equals tx. And on the y-axis, I have 2 forces up and 2 forces down. I have ty plus hy equals littlemg plus bigmg. Now I don't know ty or hy, but I know the 2 mgs. They're going to be, ty plus hy. They are 4,000 and 5,000. So it's 9,000 total. And this is the best I can do so far. I'm looking for hx, but that requires knowing tx. And I'm looking for hy but that requires knowing ty. Okay. So we're going to have to write a third equation. Alright. 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 0. 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. Okay. Now we could also pick, so we could pick this one or this one. Alright. And in here, it doesn't really matter. Let's see. I'm going to do this about I'm going to do this about this point right here. Okay? No. Let's actually just do it. Let's do it over here. So the sum of all torques about this point here. So let's call this point 1, 2, 3, and 4. These are all the possible candidates. We're going to go with this one because I want to find ty and tx. So I want to make sure my torque equation includes ty and tx. So I don't want to pick point number 3. That's a bad choice if I wanna find ty and tx. And I want to make sure, I pick the point with the most where I will cancel the most forces. In this case, it's hone and hx, hy and hx. Though those two forces really only hx would be, would give me a torque. Only hy would give me a torque. hx doesn't give you a torque. Okay? So really, all of these points, tx also doesn't give you a torque. So really, all of these points only have each one force that would give you a torque. So, you know, they're all sort of similarly, just as good or just as bad however you want to look at it. So we're just going to go with the torque of about point 1. It would have worked with other points. Don't sweat that too much. Okay? So torque about point 1 means I have 1 here and I have 4,000 pulling this down. This is little mg. I have 5,000 big Mg over here, and then I have ty. ty, we don't know. Both of these guys are providing, are producing a torque in this direction. So this is the torque of little mg. This is the torque of big Mg. Both of these guys are clockwise so they're both negative. And then this is going to provide a counterclockwise torque positive. So I'm going to be able to write that the torque of ty equals the torque of little mg plus the torque of big Mg. And I'm going to expand this equation here. It's going to be tyrsin(θ). Notice that all the angles are going to be 90 degrees. So 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.

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, and use g=10 m/s².)

• A. θ = −48°
• B. θ = −37°
• C. θ = +37°
• D. θ = +48°

Here’s what students ask on this topic:

What is static equilibrium in the context of a beam or shelf against a wall?

Static equilibrium refers to a state where a beam or shelf against a wall remains at rest, with all forces and torques acting on it balanced. This means the sum of all forces in both the x and y directions must be zero (ΣF_x = 0 and ΣF_y = 0), and the sum of all torques must also be zero (Στ = 0). In such problems, forces include the weight of the beam, tension in any supporting cables, and reaction forces at the hinge or wall. Ensuring these conditions are met guarantees the system's stability and prevents any rotational or translational motion.

How do you decompose the tension in a cable into its components in static equilibrium problems?

In static equilibrium problems, the tension in a cable can be decomposed into horizontal (T_x) and vertical (T_y) components. This is done using trigonometric functions based on the angle θ the cable makes with the horizontal. The horizontal component is given by T_x = T cos(θ), and the vertical component is given by T_y = T sin(θ). These components are then used to balance the forces and torques in the system, ensuring that ΣF_x = 0 and ΣF_y = 0.

Why is it important to consider torque in static equilibrium problems involving beams?

Considering torque in static equilibrium problems involving beams is crucial because it ensures rotational stability. Torque (τ) is the rotational equivalent of force and is calculated as τ = rF sin(θ), where r is the distance from the pivot point, F is the force, and θ is the angle between the force and the lever arm. For a beam to be in static equilibrium, the sum of all torques around any pivot point must be zero (Στ = 0). This prevents the beam from rotating and ensures it remains in a stable, balanced position.

How do hinge forces contribute to the equilibrium of a beam against a wall?

Hinge forces contribute to the equilibrium of a beam against a wall by providing the necessary counteracting forces to balance the system. The hinge exerts both horizontal (H_x) and vertical (H_y) forces. H_x counteracts the horizontal component of the tension (T_x), while H_y helps balance the vertical forces, including the weight of the beam (mg) and the vertical component of the tension (T_y). These hinge forces ensure that the sum of all forces in both the x and y directions is zero (ΣF_x = 0 and ΣF_y = 0), maintaining the beam's static equilibrium.

What are the key equations used in solving static equilibrium problems involving beams?

The key equations used in solving static equilibrium problems involving beams are:

1. Sum of all forces in the x direction: ΣF_x = 0

2. Sum of all forces in the y direction: ΣF_y = 0

3. Sum of all torques about any pivot point: Στ = 0

These equations ensure that the beam remains in a stable, balanced position without any translational or rotational motion. Forces include the weight of the beam (mg), tension in supporting cables (T), and reaction forces at the hinge (H_x and H_y), while torques are calculated based on the distances from the pivot point and the angles at which forces are applied.