Ring on a two-segment string — a 90 cm string is tied between points A and B at the same level, 60 cm apart. A 120 g ring slides on the string. A horizontal force P is applied so that the ring is vertically below B in equilibrium. Find P (in gram-force).

Introduction / Context:This statics problem couples geometry with equilibrium of a particle acted on by string tensions, weight, and an external horizontal force. Such configurations model pulley/string support systems and help build skill in resolving concurrent forces using geometry-derived direction cosines.

Given Data / Assumptions:

• String length L = 90 cm connects points A and B at the same level, AB = 60 cm.
• A ring of weight W = 120 g-force slides frictionlessly on the string.
• A horizontal force P acts on the ring so that its final position is vertically below B.
• String is light and inextensible; tensions are equal in both segments (T).

Concept / Approach:

Let the ring be vertically below B at distance y. One segment (B–ring) is vertical; the other (A–ring) is slanted. Use the length constraint to find y and the slanted segment length. Then write horizontal and vertical equilibrium of the ring with two equal tensions T, the horizontal force P, and weight W.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Use geometry first to get y = L − AB′ where AB′ is the projected length; the 5–12–13 direction ratios confirm exact arithmetic, giving P = 80 g without rounding errors.

Why Other Options Are Wrong:

40 g and 60 g underestimate the necessary balance of the horizontal component. 100 g overestimates; 120 g equals weight, not the required horizontal balance.

Common Pitfalls (misconceptions, mistakes):

Assuming unequal tensions in a single frictionless string; forgetting that the B–ring segment is vertical and contributes no horizontal component; mixing centimetres and metres inconsistently.

Final Answer:

80 g