1. **Crane Stability**: The crane’s stability depends on the position of the combined centre of gravity of the crane and load relative to the crane’s base.

2. **Overturning Moment**: The load creates a clockwise moment about the crane’s base = Weight of load × horizontal distance from base. Moving the load closer reduces this distance, reducing the overturning moment.

3. **Restoring Moment**: The crane’s own weight creates an anticlockwise restoring moment about the same point. For stability: Restoring moment ≥ Overturning moment.

4. **Critical Analysis**: Moving the load closer shifts the combined centre of gravity toward the base, increasing stability margin. However, this must be balanced against practical lifting requirements and the crane’s working radius limitations.

5. **Real-world Factors**: Ground conditions, wind effects, and dynamic forces during lifting also affect stability, making position control crucial for safe operation.

Step 1: Identify distances from pivot
— Distance of beam’s center of gravity from pivot = 2.0m — 1.5m = 0.5m (to the right)
— Distance of unknown weight from pivot = 1.5m (to the left)

Step 2: Apply principle of moments
Taking clockwise moments as positive:
Clockwise moment = Anticlockwise moment
W × 1.5 = 60 × 0.5
1.5W = 30
W = 20N

Answer: A weight of 20N must be placed at the left end.

Step 1: Identify the perpendicular distance
Since the force is applied perpendicular to the wrench, the perpendicular distance equals the wrench length = 0.25m

Step 2: Calculate the moment
Moment = Force × perpendicular distance
Moment = 80N × 0.25m = 20 Nm

Answer: The moment about the bolt is 20 Nm.

Note: This turning effect helps the mechanic overcome the resistance of the bolt threads.

Step 1: Calculate weights
— Weight of plank = 8 × 9.8 = 78.4N (acts at center = 1.5m from left end)
— Weight of person = 70 × 9.8 = 686N (acts at 1.0m from left end)

Step 2: Apply equilibrium conditions
Let F₁ = force from left prop, F₂ = force from right prop

Vertical equilibrium: F₁ + F₂ = 78.4 + 686 = 764.4N … (1)

Step 3: Take moments about left prop (at 0.5m)
Distances from left prop:
— Person: 1.0 — 0.5 = 0.5m
— Plank center: 1.5 — 0.5 = 1.0m
— Right prop: 2.5 — 0.5 = 2.0m

Moment equilibrium:
Clockwise moments = Anticlockwise moments
686 × 0.5 + 78.4 × 1.0 = F₂ × 2.0
343 + 78.4 = 2F₂
F₂ = 421.4/2 = 210.7N

Step 4: Find F₁ using equation (1)
F₁ = 764.4 — 210.7 = 553.7N

Answer:
— Left prop force = 553.7N
— Right prop force = 210.7N

Verification: Check moments about right prop to confirm answer.

Part 2 — Analysis with moved pivot:
If pivot moves 0.2m toward 25N weight:
— New distance of 25N weight from pivot = 0.8 — 0.2 = 0.6m
— New distance of rod’s center from pivot = 0.2m (opposite side)
— Rod creates moment = 18N × 0.2m = 3.6 Nm

New equilibrium equation:
(25 × 0.6) + (18 × 0.2) = 30 × d’
15 + 3.6 = 30d’
d’ = 18.6/30 = 0.62m

The 30N weight must be moved closer to the new pivot position to maintain equilibrium.

Step 2: Calculate required restoring moment
Total restoring moment needed = 75,000 Nm
Crane’s contribution = 45,000 Nm
Counterweight contribution needed = 75,000 — 45,000 = 30,000 Nm

Step 3: Calculate minimum counterweight
Counterweight × 8m = 30,000 Nm
Counterweight = 30,000/8 = 3,750N

Step 4: Safety analysis with 20% reduction
Reduced counterweight = 3,750 × 0.8 = 3,000N
New restoring moment = (3,000 × 8) + 45,000 = 69,000 Nm
Safety margin = 75,000 — 69,000 = 6,000 Nm deficit

Critical Evaluation:
Reducing the counterweight by 20% creates a dangerous situation where the overturning moment exceeds the restoring moment by 6,000 Nm. This would cause the crane to tip over. Safety regulations typically require a significant safety factor (often 25-50% above minimum) to account for:
— Dynamic loading effects
— Wind forces
— Ground conditions
— Operational variations
Therefore, the proposed reduction is extremely unsafe.

Part (a): Moment due to bridge weight
Weight = 8000 × 9.8 = 78,400N (acts at center of mass = 12m from pivot)
Horizontal distance of center of mass from pivot = 12 × cos(30°) = 12 × 0.866 = 10.39m
Moment = 78,400N × 10.39m = 814,416 Nm (clockwise)

Part (b): Required cable tension
For equilibrium: Cable moment = Bridge moment
T × 20 = 814,416
T = 40,721N

Analysis of tension vs. angle:
As bridge angle θ increases:
— Horizontal distance of center of mass = 12 × cos(θ)
— Moment arm decreases with cos(θ)
— Required tension = (78,400 × 12 × cos(θ))/20 = 47,040 × cos(θ)

At θ = 0° (horizontal): T = 47,040N
At θ = 30°: T = 40,721N
At θ = 60°: T = 23,520N
At θ = 90° (vertical): T = 0N

Conclusion: Tension decreases as the bridge rises, following a cosine relationship. This is why drawbridges are easier to lift as they approach vertical position.

Step 1: Force advantage needed
Mechanical advantage = Load/Effort = 15,000N/400N = 37.5

Step 2: Distance ratio
Distance ratio = Handle movement/Load movement = 2.4m/0.3m = 8

Step 3: Lever arm ratio calculation
For equilibrium: Load × load arm = Effort × effort arm
Effort arm/Load arm = Load/Effort = 37.5
Therefore: Effort arm = 37.5 × load arm

Step 4: Practical design considerations

If load arm = 0.04m, then effort arm = 1.5m
This gives a reasonable handle length while achieving required mechanical advantage.

Trade-off Analysis:

Advantages:
— High mechanical advantage reduces required human force
— Manageable handle length (1.5m)
— Achieves required lifting distance

Disadvantages:
— Large handle movement (2.4m) for small lift (0.3m)
— Slower operation due to distance ratio
— Requires more space for operation

Safety Considerations:
— Include 50% safety factor: design for 600N maximum force
— Add locking mechanism at different heights
— Ensure stable base to prevent tipping
— Use hardened steel for wear resistance
— Include overload protection

Final Design Recommendation:
Use compound lever system or gear reduction to improve distance ratio while maintaining force advantage. This would reduce handle movement while keeping force requirements reasonable.

Counterexamples:

2D Examples:
1. **Boomerang:** L-shaped or V-shaped boomerangs have their center of gravity in the empty space between the arms
2. **Horseshoe:** The center of gravity lies in the gap between the prongs
3. **Picture frame:** Rectangular frame’s center of gravity is in the empty center space

3D Examples:
1. **Hollow sphere:** Center of gravity at geometric center, but interior is empty
2. **Ring or torus:** Center of gravity at the center of the hole
3. **U-shaped channel:** Center of gravity in the open space

Physics Principles:

Mathematical Basis:
Center of gravity position: r̄ = (Σmᵢrᵢ)/Σmᵢ

For objects with complex geometry, this calculation can place the center of gravity outside the material boundaries.

Physical Interpretation:
— Center of gravity is a mathematical construct representing the average position of weight distribution
— It doesn’t require physical material to exist at that point
— It’s the point where the entire weight appears to act for rotational calculations

Engineering Implications:

Stability Analysis:
— Objects with external centers of gravity have unique stability characteristics
— May be inherently unstable in certain orientations
— Require careful consideration in design for safety

Practical Applications:**
1. **Aircraft design:** Some aircraft configurations have centers of gravity outside the fuselage
2. **Crane design:** Jib cranes often have overall center of gravity outside the mast structure
3. **Automotive:** Some race cars designed with center of gravity optimized for performance

Design Considerations:**
— Must ensure adequate support structures
— May require active stability control systems
— Need to consider dynamic loading conditions
— Important for transportation and installation procedures

Conclusion:**
The student’s statement reflects a common misconception. Understanding that center of gravity can exist outside physical boundaries is crucial for advanced engineering design and helps explain the behavior of many everyday and engineered objects.