Question

When two forces of magnitude P and Q are perpendicular to each other, their resultant is of magnitude R. When they are at an angle of 180° to each other their resultant is of magnitude R/√2. Find the ratio of P and Q.

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Answer 1

Answer ⇒ P/Q = 2 + √3

Explanation ⇒ Given,

P² + Q² + 2PQCos90 = R²

∴ P² + Q² + 0 = R²

∴ R² = P² + Q²

R² = (P + Q)² - 2PQ

Now, When θ = 180,

P² + Q² + 2PQCos180 = R²/2

∴ P² + Q² - 2PQ = R²/2

∴ R² - 2PQ = R²/2

∴ 2PQ = R²/2

∴ PQ = R²/4

Now,

R² = (P + Q)² - 2(R²)/4

∴ R² = (P + Q)² - R²/2

∴ (P + Q)² = 3R²/2

∴ P + Q = √3/√2 × R

Also, P - Q = R/√2

Solving it, we will get,

2P = R(√3 + 1)/√2

∴ P = R(√3 + 1)/2√2

Now, 2Q = (√3 - 1)/√2

∴ Q = (√3 - 1)/2√2

∴ P/Q = (√3 + 1)/(√3 - 1)

∴ P/Q = (√3 + 1)²/2

∴ P/Q = (4 + 2√3)/2

∴ P/Q = 2 + √3  

Hope it helps.

Answer 2

Answer:

P² + Q² + 2PQCos90 = R²

∴ P² + Q² + 0 = R²

∴ R² = P² + Q²

R² = (P + Q)² - 2PQ

Now, When θ = 180,

P² + Q² + 2PQCos180 = R²/2

∴ P² + Q² - 2PQ = R²/2

∴ R² - 2PQ = R²/2

∴ 2PQ = R²/2

∴ PQ = R²/4

Now,

R² = (P + Q)² - 2(R²)/4

∴ R² = (P + Q)² - R²/2

∴ (P + Q)² = 3R²/2

∴ P + Q = √3/√2 × R

Also, P - Q = R/√2

Solving it, we will get,

2P = R(√3 + 1)/√2

∴ P = R(√3 + 1)/2√2

Now, 2Q = (√3 - 1)/√2

∴ Q = (√3 - 1)/2√2

∴ P/Q = (√3 + 1)/(√3 - 1)

∴ P/Q = (√3 + 1)²/2

∴ P/Q = (4 + 2√3)/2

∴ P/Q = 2 + √3