Equilibrium in 2D: Videos & Practice Problems

To solve two-dimensional equilibrium problems, identify all forces acting on an object, ensuring they cancel out to zero in both the x and y axes. For example, a 5 kg box suspended by two cables experiences a weight force of w = -mg = -49 N. Decompose tension forces into their x and y components, apply the equilibrium rule, and solve for unknowns using equations like ΣF = 0. This method ensures a comprehensive understanding of forces in equilibrium.

2D Equilibrium Problems

In solving two-dimensional equilibrium problems, it is essential to understand that equilibrium occurs when all forces acting on an object cancel each other out. This means that the net force in both the x and y directions is zero, which can be expressed mathematically as:

\[ \sum F_x = 0 \quad \text{and} \quad \sum F_y = 0 \]

To illustrate this, consider a scenario involving a 5-kilogram box suspended by two cables. The first step is to draw a free body diagram, which visually represents all forces acting on the box. The weight force, calculated using the formula:

\[ W = -mg \]

where \( m \) is the mass (5 kg) and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), results in a weight force of:

\[ W = -5 \times 9.8 = -49 \, \text{N} \]

Next, identify the tension forces in the cables. Let’s denote the tension in the first cable as \( T_1 \) and the tension in the second cable as \( T_2 \). The tension \( T_2 \) acts at an angle, which requires decomposition into its x and y components. If the angle with respect to the horizontal is 37 degrees, the components can be expressed as:

\[ T_{2x} = T_2 \cos(37^\circ) \quad \text{and} \quad T_{2y} = T_2 \sin(37^\circ) \]

With the forces identified, the next step is to apply the equilibrium conditions. For the x-direction, the equation becomes:

\[ T_{2x} - T_1 = 0 \quad \Rightarrow \quad T_{2x} = T_1 \]

Substituting the expression for \( T_{2x} \), we have:

\[ T_2 \cos(37^\circ) = T_1 \]

For the y-direction, the equation is:

\[ T_{2y} + W = 0 \quad \Rightarrow \quad T_{2y} = -W \quad \Rightarrow \quad T_2 \sin(37^\circ) = 49 \, \text{N} \]

From this, we can solve for \( T_2 \):

\[ T_2 = \frac{49}{\sin(37^\circ)} \approx 81.4 \, \text{N} \]

Now that we have \( T_2 \), we can substitute this value back into the equation for \( T_1 \):

\[ T_1 = T_2 \cos(37^\circ) \approx 81.4 \cos(37^\circ) \approx 65 \, \text{N} \]

Thus, the magnitudes of the tension forces in the cables are approximately \( T_2 = 81.4 \, \text{N} \) and \( T_1 = 65 \, \text{N} \). Understanding these principles allows for effective problem-solving in two-dimensional equilibrium scenarios.

A chandelier is supported from the ceiling by 2 chains. Both chains make a 30° angle with the vertical. The tension in each chain is the same because of symmetry:T₁=T₂=50 N. What is the mass of the chandelier?

• A. 2.6 kg
• B. 8.8 kg
• C. 5.1 kg
• D. 4.4 kg

Answer: (B)

Traffic Signal

In this problem, we analyze a traffic signal suspended by three cables, labeled as cable 1, cable 2, and cable 3, to determine the tension in cable 1. The traffic signal has a mass of 8 kg, which exerts a gravitational force downward, calculated using the equation \( F = mg \), where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Thus, the weight of the signal is \( 8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 78.4 \, \text{N} \).

To solve for the tension in cable 1, we start by drawing a free body diagram. The system is in equilibrium, meaning the sum of forces in both the x and y directions equals zero. We denote the tensions in the cables as \( T_1 \), \( T_2 \), and \( T_3 \). Since the traffic signal is stationary, the tension in cable 3 must equal the weight of the signal, so \( T_3 = mg = 78.4 \, \text{N} \).

Next, we decompose the tensions \( T_1 \) and \( T_2 \) into their x and y components. Given the angles of the cables, we can express these components as follows:

• For cable 1 (angle = 22°):
  • \( T_{1x} = T_1 \cos(22°) \)
  • \( T_{1y} = T_1 \sin(22°) \)
• For cable 2 (angle = 60°):
  • \( T_{2x} = T_2 \cos(60°) \)
  • \( T_{2y} = T_2 \sin(60°) \)

In the x-direction, the equilibrium condition gives us:

\( T_{2x} - T_{1x} = 0 \) or \( T_2 \cos(60°) - T_1 \cos(22°) = 0 \).

In the y-direction, we have:

\( T_{1y} + T_{2y} - T_3 = 0 \) or \( T_1 \sin(22°) + T_2 \sin(60°) = 78.4 \, \text{N} \).

Now we have two equations with two unknowns, \( T_1 \) and \( T_2 \). From the first equation, we can express \( T_2 \) in terms of \( T_1 \):

\( T_2 = \frac{T_1 \cos(22°)}{\cos(60°)} \).

Substituting this expression for \( T_2 \) into the second equation allows us to solve for \( T_1 \):

\( T_1 \sin(22°) + \left(\frac{T_1 \cos(22°)}{\cos(60°)}\right) \sin(60°) = 78.4 \, \text{N} \).

After simplifying, we find that the equation can be expressed as:

\( 1.6 T_1 + 0.37 T_1 = 78.4 \, \text{N} \),

which leads to:

\( 1.97 T_1 = 78.4 \, \text{N} \).

Finally, solving for \( T_1 \) gives:

\( T_1 = \frac{78.4 \, \text{N}}{1.97} \approx 39.8 \, \text{N} \).

Thus, the tension in cable 1 is approximately 39.8 Newtons.

A sphere hangs suspended by a light string, resting against a vertical wall. The sphere has a mass of 2 kg and the string makes an 80° angle with the horizontal. What is the force from the wall against the ball?

• A. 2.2 N
• B. 176 N
• C. 3.5 N
• D. 111 N

Answer: (C)

Here’s what students ask on this topic:

What is the equilibrium condition in 2D forces?

The equilibrium condition in 2D forces states that the sum of all forces acting on an object must be zero in both the x and y directions. Mathematically, this is expressed as:

$\sum F_x=0$ and $\sum F_y=0$

This means that all horizontal forces must cancel each other out, and all vertical forces must do the same. This condition ensures that the object remains at rest or moves with a constant velocity, indicating it is in equilibrium.

How do you solve for tension in cables in a 2D equilibrium problem?

To solve for tension in cables in a 2D equilibrium problem, follow these steps:

1. Draw a free-body diagram of the object, showing all forces acting on it, including weight and tension forces.
2. Decompose any angled tension forces into their x and y components using trigonometric functions.
3. Write the equilibrium equations for both the x and y directions:

$\sum F_x=0$ and $\sum F_y=0$

4. Solve these equations simultaneously to find the magnitudes of the tension forces.

This method ensures that you account for all forces and correctly determine the tensions in the cables.

What is a free-body diagram and how is it used in solving equilibrium problems?

A free-body diagram (FBD) is a graphical representation of an object and all the forces acting on it. It is used to simplify and visualize the problem. To create an FBD:

1. Draw the object as a point or a simple shape.
2. Represent all forces acting on the object with arrows pointing in the direction of the force. Label each force (e.g., weight, tension, normal force).
3. Decompose any angled forces into their x and y components.

Using the FBD, you can write the equilibrium equations for the x and y directions and solve for unknown forces, ensuring the object is in equilibrium.

How do you decompose forces into x and y components in 2D equilibrium problems?

To decompose a force into its x and y components in 2D equilibrium problems, follow these steps:

1. Identify the angle θ the force makes with the horizontal axis.
2. Use trigonometric functions to find the components:

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

3. Replace the original force with its components in the free-body diagram.

This allows you to apply the equilibrium conditions separately in the x and y directions, simplifying the problem-solving process.

What are the common mistakes to avoid when solving 2D equilibrium problems?

Common mistakes to avoid when solving 2D equilibrium problems include:

1. Not drawing a complete and accurate free-body diagram.
2. Forgetting to decompose angled forces into their x and y components.
3. Incorrectly applying trigonometric functions (e.g., using sine instead of cosine).
4. Ignoring the equilibrium condition that the sum of forces in both x and y directions must be zero.
5. Not checking units and ensuring consistency throughout the calculations.

Avoiding these mistakes will help ensure accurate and reliable solutions to equilibrium problems.