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Difficulty: MediumLoad Distribution and Balancing

A uniform 1010-foot wooden beam weighing 5050 pounds is placed on a flat surface. In four different scenarios, exactly 150150 pounds of additional weight is placed on the beam in various configurations. Arrange the scenarios in order of the beam's overall balance point (center of gravity), measured from the LEFT end of the beam, from the GREATEST distance to the LEAST distance.

  1. 1Two 7575-pound weights: one placed 44 feet from the left end, and one at the extreme right end (1010 feet)
  2. 2A 5050-pound weight placed 22 feet from the left end and a 100100-pound weight placed 88 feet from the left end
  3. 3A 100100-pound weight placed 22 feet from the left end and a 5050-pound weight placed 88 feet from the left end
  4. 4A single 150150-pound weight placed 33 feet from the left end

Answer

The correct order from greatest to least distance from the left end is: the two 7575-pound weights (furthest right), the 5050/100100-pound weights, the 100100/5050-pound weights, and finally the single 150150-pound weight (closest to the left).
Because the total system weight is identical (200200 lbs) in every scenario, the relative position of the center of gravity is determined purely by the sum of the moments (weight ×\times distance) of the added weights from the left end. Calculating these moments yields 10501050, 900900, 600600, and 450450 ft-lbs respectively. A larger moment means the center of gravity is pulled further to the right, which translates to a greater overall distance from the left end.

Step-by-Step Solution

1
Identify the components that determine the overall center of gravity (balance point).
The center of gravity's distance from an origin (the left end) is found by dividing the total turning moment (torque) by the total weight of the system.
This is the fundamental principle of moments for finding the balance point of a loaded beam.
2
Observe the constants across all four scenarios.
In every scenario, the total system weight is identical at 200200 lbs (5050 lb beam + 150150 lbs of added weight), and the beam's own moment from the left end remains constant.
Because the total weight denominator is identical across all choices, the relative position of the balance point is determined solely by the sum of the moments of the added weights.
3
Calculate the turning moment (Weight ×\times Distance) of the added weights relative to the left end for each scenario.
Two 7575-lb weights: (75×4)+(75×10)=1050(75 \times 4) + (75 \times 10) = 1050 ft-lbs.
5050-lb and 100100-lb weights: (50×2)+(100×8)=900(50 \times 2) + (100 \times 8) = 900 ft-lbs.
100100-lb and 5050-lb weights: (100×2)+(50×8)=600(100 \times 2) + (50 \times 8) = 600 ft-lbs.
Single 150150-lb weight: 150×3=450150 \times 3 = 450 ft-lbs.
This calculation provides a numerical value directly proportional to how far right the weight is biased.
4
Order the scenarios by their added moments from greatest to least.
The order is 1050>900>600>4501050 > 900 > 600 > 450. Therefore, the balance point distances follow the exact same order.
A greater turning moment from the left end means the center of gravity is located further to the right, which translates to a greater distance from the left end.

Key Concept

The center of gravity of a system shifts proportionally toward the location with the greatest turning moment (Force ×\times Distance). When total mass is constant, ranking moments ranks the balance points.
Estimated Time:1m 30s
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Topics

Load Distribution and Balancing

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