More 2D Equilibrium Problems: Study with Video Lessons, Practice Problems & Examples

In analyzing an inclined beam held at equilibrium, we consider forces and torques acting on it. The beam's weight (mg) and an applied force (f) create a balance of forces in both x and y axes, leading to equations such as $f=h\_x$ and $f\_y+h\_y=\mathrm{mg}$. By applying torque equations about strategic points, we can derive the magnitude and direction of the net force, ultimately leading to a comprehensive understanding of mechanical equilibrium.

Inclined beam against the floor

Hey, guys. So in this example, we have an inclined beam that's held against the floor. So you have a force f here, and it also has the hinge here supporting it. So let's check it out. We have a 100-kilogram beam, so mass equals 100. It has a length of 4 meters and it's held at equilibrium. This means all the forces are 0, add up to 0, and all the torques add up to 0 by a hinge down here and by a force that you apply f right here. The beam is held at an angle of 30 degrees above the horizontal. So this angle here is 30. I'm going to put the 30 in here so that I can draw the mg in here and your force is directed at 50 degrees above the horizontal. So the distance from your force all the way to the horizontal right here is 50. Now if this is 30, this is 30 as well. So we're going to split up that 50 into you got a 30 here. And if the bottom is 30, the whole thing is 50, it means that the top over here is 20. So 20 plus 30 equals 50. We want to find f. We also want to find the magnitude and direction of the net force. So this f can get split into f_x and f_y. And if the f_x is this way, remember, the forces have to cancel. There's only one force, in the x-axis, which is to the right. So the hinge must pull this thing back to the left to hold it in place. So I'm going to say that the hinge has an h_x. We're going to assume that the vertical hinge force will be up. So we're going to assume that I have an h_y that is up. And if this assumption is correct, our total h_net will look like this. Okay? So I have the net h force h_net and then I have the angle which is theta_h. So we're looking for f and we're looking for h_net and theta_h. Cool? So let me just highlight the stuff we're looking for. And the way we're going to solve this is by writing that the sum of all forces equals 0 on the x-axis, the sum of all forces equals 0 on the y-axis. And if necessary, which it will be, we're going to write torque equations or at least one torque equation. Okay? One key difference in how I'm going to solve this versus some of the previous questions we solved is that instead of working with f_x and f_y in their component forms, I'm actually just going to work with f. And the reason is in this particular case, it's going to be simpler. So if you had a question, where you had a bar like this held by a rope, you had a tension that split into tension_y and tension_x. Tension_x produced no torque. Torque of tension_x equals 0. So tension_x was kind of useless. It didn't really do much. So it's easier to think of t since x was useless. It was easier to think of t as just t_y and then keep these two separate, the x and y instead of working with just the total vector t. So these questions were a little bit simpler so it was better to do that. Here, we got a bunch of angles. I got the 30, I got the 50, whatever. So it's going to be simpler to just work with the complete form of the vector, the vector form and not the components, the individual components. We're going to work with the entire vector. It doesn't matter. You could have done it the other way and it would have worked just as well. But you just have one more force because you have f_x and f_y. That's 2 forces as opposed to just having f which is 1. So I'm going to do that. It would have worked the other way but that's how we're going to roll for this one. Alright. So, the forces in the x-axis are h_x and f_x. So I can write that f_x and h_x are equal to each other because they cancel each other. The forces in the y-axis are f_y h_y, and then mg. So I can write that f_y plus h_y equals mg. Now notice that I don't know f_x. I don't know h_x. I don't know f_y. I don't know h_y. I know mg. So we don't know. There's a ton of stuff we don't know here. So this is not going to be enough. I'm going to have to write a third equation which is going to be a torque equation. Sum of all torque equals 0 about some reference axis. Now remember, the way you want to do this is the reason why we're writing this equation is because we're looking for f. So you want to write your torque equation away from this point, right? So let's say we got points 1, 2, 3. Point 3 is the worst point to write the torque equation about because if you were to do this if you were to write the sum of all torques on point 3, these guys, you're basically treating this as the axis. And these forces

A 200 kg, 10 m-long beam is held at equilibrium by a hinge on the floor and a force you apply on it via a light rope connected to its edge, as shown. The beam is held at 53° above the horizontal, and your rope makes an angle of 30° with it. Calculate the angle that the Net Force of the hinge makes with the horizontal (use +/– for above/below +x, and use g=10 m/s².)

• A. θ = −65.9°
• B. θ = −21.1°
• C. θ = +21.1°
• D. θ = +65.9°

Here’s what students ask on this topic:

How do you determine the net force acting on an inclined beam in equilibrium?

To determine the net force acting on an inclined beam in equilibrium, you need to consider both the forces and torques acting on the beam. Start by resolving all forces into their x and y components. For the beam to be in equilibrium, the sum of forces in both the x and y directions must be zero. This can be expressed as:

$\sum F_x=0$

$\sum F_y=0$

Next, apply the torque equilibrium condition about a strategic point to simplify calculations. The sum of torques about any point must also be zero:

$\sum \tau =0$

By solving these equations, you can find the magnitudes and directions of the forces acting on the beam, ensuring it remains in equilibrium.

What is the significance of choosing the right point for calculating torque in 2D equilibrium problems?

Choosing the right point for calculating torque in 2D equilibrium problems is crucial because it simplifies the equations and reduces the number of unknowns. The ideal point is one where multiple forces act, as this can eliminate several torque terms, making the calculations more manageable. For example, if you choose a point where a force acts, the torque due to that force will be zero, as the lever arm is zero. This approach helps in isolating the target variable and solving for it efficiently. The goal is to write the torque equation about a point that includes the target variable, ensuring it appears in the equation for solving.

How do you resolve forces into their components in 2D equilibrium problems?

To resolve forces into their components in 2D equilibrium problems, you need to break down each force into its x and y components using trigonometric functions. For a force F acting at an angle θ from the horizontal, the components can be found as follows:

$F_x=F\cdot \cos \left(\theta \right)$

$F_y=F\cdot \sin \left(\theta \right)$

These components can then be used in the equilibrium equations for forces in the x and y directions. This method allows you to analyze the problem in a more straightforward manner by dealing with scalar quantities rather than vector quantities.

Why is it important to consider both force and torque equilibrium in 2D problems?

Considering both force and torque equilibrium in 2D problems is essential because it ensures the object is in complete mechanical equilibrium. Force equilibrium ensures that the object does not translate, meaning it does not move linearly in any direction. This is achieved by setting the sum of forces in both the x and y directions to zero:

$\sum F_x=0$

$\sum F_y=0$

Torque equilibrium ensures that the object does not rotate, which is achieved by setting the sum of torques about any point to zero:

$\sum \tau =0$

By considering both conditions, you ensure that the object remains stationary and stable, which is the essence of mechanical equilibrium.

How do you choose the reference axis for writing torque equations in 2D equilibrium problems?

Choosing the reference axis for writing torque equations in 2D equilibrium problems involves selecting a point that simplifies the calculations. The best reference axis is typically a point where multiple forces act, as this can eliminate several torque terms. The steps are:

1. Identify the target variable you need to solve for.

2. Choose a point away from the target variable to ensure it appears in the torque equation.

3. Select a point with the most forces acting on it to cancel out as many torques as possible, reducing the number of terms in the equation.

This strategic choice helps in isolating the target variable and solving the problem more efficiently.