Steel rod is embeded into an alluminum plate as shown. The plate is 3 cm thick, 5 cm – [Free] B53
A steel rod is embeded into an alluminum plate as shown. The plate is 3 cm thick, 5 cm wide and 8 high. The circular steel rod is 2 cm diameter and 8 cm long. The center of the rod is embeded at x=2.5cm and y=6cm relative to the origin shown. Use the density of steel is 8000Kgm3 and the density of Aluminum is 2800Kgm3 Include in show work a 2D view from +x axis and +z
![Steel rod is embeded into an alluminum plate as shown. The plate is 3 cm thick, 5 cm - [Free] B53](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20640%20620'%3E%3C/svg%3E)
Answer
Center of Gravity of a Steel Rod Embedded in an Aluminum Plate
🔹 Step 1: Given Dimensions and Densities
Aluminum Plate:
Thickness = 3 cm, Width = 5 cm, Height = 8 cm, Density = 2800 kg/m³
Steel Rod:
Diameter = 2 cm ⟹ Radius = 1 cm, Length = 8 cm, Density = 8000 kg/m³
Center of rod = (x = 2.5 cm, y = 6 cm)
Thickness = 3 cm, Width = 5 cm, Height = 8 cm, Density = 2800 kg/m³
Steel Rod:
Diameter = 2 cm ⟹ Radius = 1 cm, Length = 8 cm, Density = 8000 kg/m³
Center of rod = (x = 2.5 cm, y = 6 cm)
🔹 Step 2: Volume and Mass Calculations
Volume of Aluminum Plate:
VAl = 3 × 5 × 8 = 120 cm³ = 1.2 × 10⁻⁴ m³
Mass of Aluminum Plate:
mAl = 2800 × 1.2 × 10⁻⁴ = 0.336 kg
Volume of Steel Rod (Cylinder):
VSt = π × (1)² × 8 = 8π cm³ = 8π × 10⁻⁶ m³
Mass of Steel Rod:
mSt = 8000 × 8π × 10⁻⁶ ≈ 0.201 kg
🔹 Step 3: Centroid Coordinates
Aluminum Plate Centroid:
- xAl = 1.5 cm
- yAl = 4 cm
- zAl = 2.5 cm
Steel Rod Centroid:
- xSt = 2.5 cm
- ySt = 6 cm
- zSt = 2.5 cm
🔹 Step 4: Center of Gravity (CG) Calculation
XCG:
XCG = (0.336 × 1.5 + 0.201 × 2.5) / (0.336 + 0.201) ≈ 1.874 cm
YCG:
YCG = (0.336 × 4 + 0.201 × 6) / (0.537) ≈ 4.75 cm
ZCG:
ZCG = (0.336 × 2.5 + 0.201 × 2.5) / 0.537 = 2.5 cm
✅ Final Result
Center of Gravity:
(XCG, YCG, ZCG) = (1.87 cm, 4.75 cm, 2.5 cm)