A 50mm * 120mm * 6n lang wooden plank is required to water with top 150 mm above the water line. This is attained by attaching 10 mm thick steel plates to each wider side of the plank at the submerged bottom. Compute the exact length of steel plates needed Use density of wood equal to 502kg / (m ^ 3) and density of steel equal to 7870kg / (m ^ 3)

To solve this problem, we need to ensure that the wooden plank floats with 150 mm of its height above the waterline after attaching steel plates to its sides. To do this, we'll follow these steps:

1. **Calculate the buoyant force required**: The weight of the water displaced must equal the total weight of the plank and the steel plates.
2. **Determine the volume of the submerged portion of the plank**: Since 150 mm is above water, the remaining \(120 \, \text{mm} - 150 \, \text{mm}\) (which gives 30 mm) is the submerged portion.
3. **Calculate the weight of the wooden plank**: Use the density of the wood and the volume of the entire plank.
4. **Determine the required weight of steel plates**: The steel plates are used to adjust the submerged depth.
5. **Compute the length of steel plates**: Using the density of steel and the weight required, compute the dimensions.

### Step-by-Step Solution:

#### 1. Buoyant force required:
The plank is floating, so the buoyant force (weight of displaced water) is equal to the combined weight of the wood and the steel plates. The submerged volume of the plank is given by:
\[
V_{\text{sub}} = 50 \, \text{mm} \times L \times 30 \, \text{mm} = 50 \times L \times 0.03 \, \text{m} = 0.0015 L \, \text{m}^3
\]
(assuming \(L\) is the length of the plank)

#### 2. Calculate the weight of the wooden plank:
The volume of the entire plank is:
\[
V_{\text{wood}} = 50 \, \text{mm} \times 120 \, \text{mm} \times L = 50 \times 120 \times L = 0.006 L \, \text{m}^3
\]
The weight of the wooden plank is:
\[
W_{\text{wood}} = \rho_{\text{wood}} \times V_{\text{wood}} = 502 \, \text{kg/m}^3 \times 0.006 L = 3.012 L \, \text{kg}
\]

#### 3. Calculate the weight of the displaced water:
The weight of the displaced water is equal to the buoyant force. The buoyant force is:
\[
F_{\text{buoyant}} = \rho_{\text{water}} \times V_{\text{sub}} = 1000 \, \text{kg/m}^3 \times 0.0015 L = 1.5 L \, \text{kg}
\]

#### 4. Determine the weight required from steel plates:
Since the weight of the wooden plank is 3.012 L kg, but the submerged portion only displaces 1.5 L kg of water, we need to add weight from steel plates to balance the plank. The required additional weight is:
\[
W_{\text{steel needed}} = W_{\text{wood}} - F_{\text{buoyant}} = 3.012 L - 1.5 L = 1.512 L \, \text{kg}
\]

#### 5. Compute the length of steel plates:
The steel plates are 10 mm thick, and they are attached to both sides of the plank. The volume of steel per unit length of the plank (for both plates) is:
\[
V_{\text{steel}} = 2 \times (0.01 \, \text{m} \times 0.03 \, \text{m} \times L) = 0.0006 L \, \text{m}^3
\]
The weight of the steel plates per unit length is:
\[
W_{\text{steel}} = \rho_{\text{steel}} \times V_{\text{steel}} = 7870 \, \text{kg/m}^3 \times 0.0006 L = 4.722 L \, \text{kg}
\]
The length of steel required to balance the weight can now be found by solving:
\[
4.722 L = 1.512 L
\]
Solving for \(L\), we get:
\[
L = \frac{1.512}{4.722} = 0.32 \, \text{m}
\]

### Final Answer:
The exact length of the steel plates needed is **0.32 meters** (or 32 cm).

Would you like any clarifications or additional details?

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Here are 5 related questions:
1. How is Archimedes' principle applied in floating body problems?
2. What role does the density of materials play in floating or sinking?
3. How would the length change if the plank was wider?
4. What would happen if we attached thinner steel plates?
5. How can we account for the added weight of screws or bolts in this setup?

**Tip**: Always check units carefully in buoyancy problems to avoid conversion errors.

A 50mm * 120mm * 6n lang wooden plank is required to water with top 150 mm above the water line. This is attained by attaching 10 mm thick steel plates to each wider side of the plank at the submerged bottom. Compute the exact length of steel plates needed. Use density of wood equal to 502kg / (m ^ 3) and density of steel equal to 7870kg / (m ^ 3).

Solve a buoyancy problem involving a wooden plank partially submerged in water and steel plates added to adjust buoyancy. This detailed step-by-step guide involves calculations with density, volume, and Archimedes' Principle. Suitable for high school students in Grades 11-12.

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