Beam / Shelf Against a Wall: Videos & Practice Problems

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

In static equilibrium problems involving beams or shelves supported against a wall, it is essential to analyze the forces and torques acting on the system. Typically, a beam is held in place by a cable or rod, with the weight of the beam acting at its center. The tension in the cable, denoted as \( T \), can be decomposed into horizontal and vertical components: \( T_x \) and \( T_y \), respectively. The beam also experiences forces from a hinge, which applies a horizontal force \( H_x \) to counteract \( T_x \) and a vertical force \( H_y \) to assist in balancing the weight of the beam.

In equilibrium, the sum of all forces in both the x and y directions must equal zero, as well as the sum of all torques. This means that if there is a force acting to the left, there must be an equal force acting to the right, and similarly for vertical forces. The relationship between the forces can be expressed as:

\( \sum F_x = H_x - T_x = 0 \)

\( \sum F_y = H_y + T_y - mg = 0 \)

Where \( mg \) is the weight of the beam, calculated as the mass multiplied by the acceleration due to gravity. In this context, if the mass of the beam is 300 kg, then \( mg = 300 \times 10 = 3000 \, \text{N} \).

To find the tension \( T \) in the cable, it is often useful to consider the torques about a point, typically the hinge, to eliminate unknown forces from the equation. The torque due to the weight of the beam and the torque due to the tension can be expressed as:

\( \tau_{mg} = mg \cdot r_{mg} \cdot \sin(90^\circ) \)

\( \tau_{T_y} = T_y \cdot r_{T_y} \cdot \sin(90^\circ) \)

Where \( r_{mg} \) is the distance from the hinge to the center of mass of the beam, and \( r_{T_y} \) is the distance from the hinge to the point where the tension acts. For a beam of length 4 meters, the distance to the center of mass is 2 meters. Setting the torques equal gives:

\( mg \cdot 2 = T_y \cdot 4 \)

From this, we can derive that:

\( T_y = \frac{mg}{2} \

Substituting the known values, we find \( T_y = \frac{3000}{2} = 1500 \, \text{N} \).

To find the total tension \( T \), we use the relationship \( T_y = T \cdot \sin(\theta) \), where \( \theta \) is the angle of the cable with respect to the horizontal. For an angle of 37 degrees, we can rearrange this to find \( T \):

\( T = \frac{T_y}{\sin(37^\circ)} \approx \frac{1500}{0.6018} \approx 2485 \, \text{N} \).

Next, to determine the net force exerted by the hinge, we need to calculate both \( H_x \) and \( H_y \). The horizontal force \( H_x \) is equal to \( T_x \), which can be found using:

\( T_x = T \cdot \cos(\theta) \approx 2485 \cdot \cos(37^\circ) \approx 2000 \, \text{N} \).

For the vertical force \( H_y \), we can use the earlier equation:

\( H_y = mg - T_y = 3000 - 1500 = 1500 \, \text{N} \).

Finally, the net force exerted by the hinge can be calculated using the Pythagorean theorem:

\( H_{net} = \sqrt{H_x^2 + H_y^2} = \sqrt{2000^2 + 1500^2} \approx 2485 \, \text{N} \).

This analysis illustrates the balance of forces and torques in static equilibrium problems, highlighting the relationships between tension, hinge forces, and the weight of the beam. Understanding these concepts is crucial for solving similar problems in physics.

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°

Answer: (D)

Beam supporting an object

In this scenario, we analyze a horizontal beam supported by a wall and a rope, with a mass of 500 kg hanging from its end. The beam itself has a mass of 400 kg and a length of 8 meters. The mass of the beam is uniformly distributed, meaning the weight acts at its center, 4 meters from either end. The gravitational force acting on the beam (denoted as little m) is calculated as:

$mg=4009.81=3924$

For the hanging mass (denoted as big M), the gravitational force is:

$Mg=5009.81=4905$

To maintain equilibrium, the sum of forces in both the x and y directions must equal zero. The tension in the rope can be resolved into horizontal (T_x) and vertical components (T_y), with the angle of the rope being 53 degrees. The equations governing the forces are:

$H=T$_x

And for the vertical forces:

$T$_y + H _y = m g + M g

To find the unknown forces, we also need to consider the torques acting on the beam. By selecting a pivot point, we can set the sum of torques to zero. The torque due to the weight of the beam and the hanging mass can be expressed as:

$\tau =rF$

Calculating the torques about a chosen point allows us to isolate the unknowns. The resulting equations lead to the determination of T_y and subsequently H_y. If H_y is found to be negative, it indicates that the initial assumption of its direction was incorrect, suggesting it acts downward instead.

After calculating the tension in the rope and resolving the forces, we find:

$H=\left(7000\right)^2+(-333)^2$

This results in a net force of approximately 7008 N acting on the hinge, illustrating the balance of forces and torques in the system. Understanding these principles is crucial for analyzing static equilibrium in mechanical systems.

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°

Answer: (D)

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.