More 2D Equilibrium Problems: Videos & Practice Problems

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

In this scenario, we analyze an inclined beam supported by a hinge and a force, with the beam in equilibrium. The beam has a mass of 100 kg and a length of 4 meters, positioned at an angle of 30 degrees above the horizontal. The applied force is directed at 50 degrees above the horizontal. To solve for the unknown force \( F \) and the hinge forces, we start by recognizing that the sum of all forces and the sum of all torques must equal zero.

First, we break down the forces acting on the beam. The horizontal forces include the applied force \( F_x \) and the hinge force \( H_x \), which must balance each other. In the vertical direction, the forces include the vertical component of the applied force \( F_y \), the hinge force \( H_y \), and the weight of the beam \( mg \) (where \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \)). This gives us the equations:

\( F_x + H_x = 0 \) (1)

\( F_y + H_y = mg \) (2)

Next, we need to write a torque equation about a point that does not include the unknown force \( F \). Choosing the hinge as the pivot point allows us to include the torques produced by the weight of the beam and the applied force. The torque due to the weight \( mg \) is negative (clockwise), while the torque due to the applied force \( F \) is positive (counterclockwise). The torque equations can be expressed as:

\( \tau_{mg} = mg \cdot r \cdot \sin(60^\circ) \) (since the angle between the weight and the lever arm is 60 degrees)

\( \tau_F = F \cdot r \cdot \sin(20^\circ) \) (the angle between the force and the lever arm is 20 degrees)

Setting these equal gives:

\( mg \cdot 2 = F \cdot 4 \cdot \sin(20^\circ) \)

Substituting \( mg = 100 \cdot 9.81 = 981 \, \text{N} \) into the equation, we can solve for \( F \). After calculations, we find \( F \approx 1266 \, \text{N} \).

With \( F \) known, we can find the components \( F_x \) and \( F_y \) using:

\( F_x = F \cdot \cos(50^\circ) \approx 814.84 \, \text{N} \)

\( F_y = F \cdot \sin(50^\circ) \approx 970.82 \, \text{N} \)

Since \( H_x \) must equal \( F_x \) but in the opposite direction, we have \( H_x \approx -814.84 \, \text{N} \). For the vertical forces, we can find \( H_y \) using equation (2):

\( H_y = mg - F_y = 981 - 970.82 \approx 10.18 \, \text{N} \)

To find the resultant hinge force \( H \) and its angle \( \theta_H \), we apply the Pythagorean theorem:

\( H = \sqrt{H_x^2 + H_y^2} \approx \sqrt{(-814.84)^2 + (10.18)^2} \approx 814.85 \, \text{N} \)

The angle \( \theta_H \) can be calculated using the arctangent function:

\( \theta_H = \tan^{-1}\left(\frac{H_y}{|H_x|}\right) \approx \tan^{-1}\left(\frac{10.18}{814.84}\right) \approx 0.7^\circ \)

This analysis illustrates the balance of forces and torques in a static system, emphasizing the importance of understanding vector components and equilibrium conditions in mechanics.

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°

Answer: (D)

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.