To solve this problem, we need to analyze the forces acting on the wooden plank when it is partially submerged in water. The plank is hinged at one end, and we want to find the angle it makes with the vertical when it reaches equilibrium. Given that the specific gravity of the plank is 0.5, we can derive the necessary relationships to find the angle.
Understanding the Forces at Play
First, let's clarify some key concepts:
- Specific Gravity: This is the ratio of the density of a substance to the density of water. A specific gravity of 0.5 means the plank is less dense than water, which will affect how it floats.
- Buoyant Force: This is the upward force exerted by the fluid on the submerged part of the plank. It can be calculated using Archimedes' principle.
- Weight of the Plank: The weight acts downwards through the center of mass of the plank.
Calculating the Weight and Buoyant Force
Let's denote the following:
- Length of the plank, L = 1 m
- Height of water, h = 0.5 m
- Specific gravity of the plank, SG = 0.5
- Density of water, ρ = 1000 kg/m³
The volume of the plank can be expressed as:
Volume of plank, V = A * L, where A is the cross-sectional area.
The weight of the plank (W) can be calculated as:
W = SG * ρ * V * g = 0.5 * 1000 * A * 1 * g = 500A * g
Now, the buoyant force (B) acting on the submerged part of the plank can be calculated as:
B = ρ * V_submerged * g = 1000 * A * h * g = 1000 * A * 0.5 * g = 500A * g
Setting Up the Equilibrium Condition
In equilibrium, the upward buoyant force must balance the downward weight of the plank. Therefore, we can set up the equation:
B = W
Substituting the expressions we derived:
500A * g = 500A * g
This shows that the forces are balanced when the plank is vertical. However, we need to find the angle θ when the plank is not vertical.
Finding the Angle θ
When the plank is at an angle θ with the vertical, the center of mass of the plank will shift, and we need to consider the geometry of the situation. The submerged length of the plank will be:
L_submerged = h / cos(θ)
The buoyant force will still be given by the volume of the submerged part:
B = ρ * (A * L_submerged) * g = 1000 * A * (h / cos(θ)) * g = 1000 * A * (0.5 / cos(θ)) * g
Setting the buoyant force equal to the weight of the plank gives us:
1000 * A * (0.5 / cos(θ)) * g = 500A * g
Dividing both sides by g and A (assuming A is not zero), we get:
1000 * 0.5 / cos(θ) = 500
Solving for cos(θ):
1000 * 0.5 = 500 * cos(θ)
1 = cos(θ)
Thus, θ = cos⁻¹(1) = 0 degrees, which we are excluding. To find the angle when the plank is at equilibrium, we can analyze the moments about the hinge.
Moment Analysis
The moment due to the weight of the plank about the hinge is:
M_weight = W * (L/2) * sin(θ)
The moment due to the buoyant force is:
M_buoyancy = B * (L_submerged/2) * sin(θ)
Setting these moments equal gives us:
500A * g * (1/2) * sin(θ) = 1000 * A * (0.5 / cos(θ)) * g * (h/2) * sin(θ)
After simplifying, we can find the angle θ that satisfies this equilibrium condition. The final result will depend on the specific geometry of the setup, but through this analysis, we can derive the necessary relationships to find θ.
In summary, the angle θ that the plank makes with the vertical in equilibrium can be determined through careful consideration of the forces and moments acting on the plank. The specific calculations will depend on the exact dimensions and properties of the plank and the water level.
