The wooden beam shown in the figure is 200mmx200mm and 4m long. It is hinged at A and remains in equilibrium at θ with the horizontal. Find the inclination θ. Sp. gr. of wood = 0.6.

Problem Statement

The wooden beam shown in the figure has a cross-sectional dimension of 200 mm x 200 mm and a length of 4 m. It is hinged at point A and remains in equilibrium at an angle θ with the horizontal. Given that the specific gravity of wood is 0.6, determine the inclination θ.

Solution

1. Let the length of the beam immersed under water be y.

2. Calculate the weight of the beam (W):

   W = γ_beam × V_beam = 0.6 × 9810 × (0.2 × 0.2 × 4) = 941.76 N

   W acts at the midpoint of the beam, i.e. 2 m from the hinge A.

3. Determine the buoyant force (F_B) acting on the beam:

   F_B = γ_water × V_displaced = 9810 × (0.2 × 0.2 × y) = 392.4 y N

   This force acts at the centroid of the submerged part, located at AC = 4 − y/2 from A.

4. Taking moments about the hinge A (using the horizontal component, which introduces a factor of cosθ), we have:

   941.76 × 2 cosθ = 392.4 y (4 − y/2) cosθ

   Cancelling cosθ from both sides gives:

   941.76 × 2 = 392.4 y (4 − y/2)

5. Simplify the equation:

   392.4 y (4 − y/2) = 1883.52
   ⇒ y (4 − y/2) = 4.8
   ⇒ 4y − (y²)/2 = 4.8

   Multiplying through by 2:

   y² − 8y + 9.6 = 0

6. Solve the quadratic equation:

   y = 6.5 m or y = 1.47 m

   Since the beam is only 4 m long, y = 6.5 m is not physically possible. Therefore, y = 1.47 m.

7. Using the equilibrium condition (derived from the geometry), we have:

   sinθ = 1 / (4 − y)

   Substitute y = 1.47 m:

   sinθ = 1 / (4 − 1.47) = 1 / 2.53 ≈ 0.395

8. Finally, determine the angle θ:

   θ ≈ 23.30°

Explanation

In this problem, the beam’s weight and the buoyant force due to the displaced water create opposing moments about the hinge A. The weight (941.76 N) acts through the beam’s midpoint (2 m from A), while the buoyant force (392.4 y N) acts through the centroid of the submerged portion, located at 4 − y/2 from A.

Setting up the moment equilibrium (and canceling the common cosθ factor) leads to a quadratic equation in y. The physically acceptable solution, y = 1.47 m, is then used to relate the geometry of the beam to its equilibrium position via the relation sinθ = 1/(4 − y), which yields an inclination of approximately 23.30°.

Physical Meaning

The equilibrium of the beam is determined by the balance between its weight and the buoyant force acting on the submerged portion. Physically, this means:

• The weight of the beam creates a moment about the hinge, tending to rotate the beam downward.
• The buoyant force (which depends on the volume of water displaced) creates an opposing moment.
• The position of the submerged centroid (4 − y/2 from the hinge) is critical, as it determines the lever arm for the buoyant force.

The equilibrium condition ensures that the restoring moments balance the overturning moment due to the beam’s weight. If the buoyant force were too small (or if the submerged depth were insufficient), the beam would not achieve a stable equilibrium and could tip over. Here, the derived angle θ ≈ 23.30° represents the precise balance point where these forces counteract each other.