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.

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