Balancing two like parallel forces by adjusting one magnitude: For two parallel forces of 20 kg and 15 kg separated by a fixed distance, by how much should the 20 kg force be reduced so that the distance of the new resultant from the 20 kg force equals the previous distance of the resultant from the 15 kg force?

Introduction / Context:
Locating the line of action of a resultant of two like parallel forces is routine in statics. For forces F₁ and F₂ separated by distance L, the distance of the resultant from the F₁ line is x = F₂L/(F₁ + F₂). This problem asks for a change in F₁ such that a symmetry of distances before and after the change is achieved.

Given Data / Assumptions:

• Initial forces: F₁ = 20 kg, F₂ = 15 kg (like, parallel).
• Separation = L (constant but unknown; it will cancel).
• After adjustment: F₁′ < 20 kg.
• Condition: distance from the 20 kg line after change equals the original distance from the 15 kg line.

Concept / Approach:
Use the centroid-like formula for the resultant position for two forces. Set the “new-from-20 kg” distance equal to the “old-from-15 kg” distance and solve for F₁′. Notice that L cancels, so the numerical answer is independent of spacing.

Step-by-Step Solution:

Verification / Alternative check:
With F₁′ = 11.25 kg, new resultant location from 20 kg line is 15L/26.25 = (4/7)L, exactly the previous distance from the 15 kg line.

Why Other Options Are Wrong:

• 5.5, 6.25, 10.5 kg: do not satisfy the balance equation; they yield incorrect positions for the resultant.

Common Pitfalls:
Reversing the “from which force” reference or using the wrong formula; forgetting that L cancels out and trying to assign a value to L.

Final Answer:
8.75 kg