Problem Statement
Two equal weights of 10 N are attached to the ends of a thin string which passes over three smooth pegs arranged in the form of an equilateral triangle with one side horizontal. Find the pressure on each peg.
Step-by-Step Solution
Given Information
- Each weight = 10 N
- Pegs arranged in equilateral triangle
- All pegs are smooth
Step 1: Tension Calculation
Tension in string T = 10 N
(Since weights are in equilibrium)
(Since weights are in equilibrium)
Step 2: Reaction at Pegs B & C
Using formula: R2 = P2 + Q2 + 2PQ cos θ
Angle between forces = 90° + 60° = 150°
RB2 = 102 + 102 + 2×10×10×cos150°
= 100 + 100 – 200×sin60°
= 200 – 200×(√3/2)
= 200 – 173.2 = 26.8
RB = √26.8 = 5.17 N
Angle between forces = 90° + 60° = 150°
RB2 = 102 + 102 + 2×10×10×cos150°
= 100 + 100 – 200×sin60°
= 200 – 200×(√3/2)
= 200 – 173.2 = 26.8
RB = √26.8 = 5.17 N
Similarly, RC = 5.17 N (due to symmetry)
Step 3: Reaction at Peg A
Angle between forces = 60°
RA2 = 102 + 102 + 2×10×10×cos60°
= 100 + 100 + 200×(1/2)
= 300
RA = √300 = 17.32 N
RA2 = 102 + 102 + 2×10×10×cos60°
= 100 + 100 + 200×(1/2)
= 300
RA = √300 = 17.32 N
Final Pressures
RA = 17.32 N
RB = RC = 5.17 N
RB = RC = 5.17 N
Detailed Explanation
Key principles applied:
- Law of cosines for resultant forces
- Symmetry considerations for pegs B and C
- Angle determination based on equilateral triangle geometry
- Trigonometric values for 60° and 150° angles
The different angles at pegs explain the variation in reaction forces:
– Peg A handles forces at 60° angle (maximum resultant)
– Pegs B & C handle forces at 150° angle (reduced resultant)
