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).

Difficulty: Medium

Correct Answer: 80 g

Explanation:


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:

Place A at (0, 0) m, B at (0.60, 0) m, ring at (0.60, −y) m.String length: y + sqrt(0.60^2 + y^2) = 0.90 m ⇒ y = 0.25 m (25 cm).Length of slanted segment A–ring = sqrt(0.36 + 0.25^2) = 0.65 m (65 cm).Direction cosines for A–ring: horizontal component ratio = 0.60/0.65 = 12/13; vertical = 0.25/0.65 = 5/13.Let common tension be T. Vertical equilibrium: T (up from B–ring) + T*(5/13) (up from A–ring) = W ⇒ (18/13) T = 120 ⇒ T = 120 * 13 / 18 = 86.667 g.Horizontal equilibrium: P (to the right) balances leftward component of A–ring tension ⇒ P = T*(12/13) = 86.667 * 12/13 = 80 g.


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

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