Question

Q-1-A: Find the magnitude of two forces such that if they act at right angles, the resultant is √10 (root ten) N. But if they act at 60 degrees their resultant is √13(root thirteen) N.



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Answer 1

Given

• If they act in right angle the resultant is √10
• If they act in 60° the resultant is √13

To Find

• Magnitude of the forces

Solution

☯ R = √A²+B²+2ABcosθ

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✭ According to the Question :

In Case 1

→ R = √A²+B²+2ABcosθ

→ √10 = √A²+B²+2AB cos 90°

→ 10 = A²+B²+2AB × 0

→ 10 = A²+B²+0

→ 10 = A²+B² -eq(1)

In Case 2

→ R = √A²+B²+2ABcosθ

→ √13 = √A²+B²+2AB × cos 60°

→ 13 = A²+B²+2AB × 0.5

→ 13 = 10 + AB

→ 13-10 = AB

→ AB = 3 -eq(2)

━━━━━━━━━━━━━━━━━━━━━━━━━

Now,

→ (A+B)² = A²+B²+2AB

→ (A+B)² = 10 + 2(3)

→ (A+B)² = 10 + 6

→ A+B + √16

→ A+B = 4 -eq(3)

Similarly,

→ (A-B)² = A²+B²-2AB

→ (A-B)² = 10 - 2(3)

→ (A-B)² = 10 - 6

→ A-B = √4

→ A-B = 2 -eq(4)

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Subtracting eq(4) from eq(3)

→ 2B = 2

→ B = 1 N

From eq(4)

→ A-1 = 2

→ A = 2+1

→ A = 3 N

Answer 2

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Let the two forces be 'A' and 'B'.

➪ Resultant ( R ) = √A² + B² + 2ABcos∅

➪ √10 = √A² + B² + 2ABcos90

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ[ cos90 = 0 ]

➪ 10 = A² + B² ㅤㅤㅤㅤ(1)

Now , when the angle between these two forces is 60° , resultant is √13.

➪ √13 = √A² + B² + 2AB cos60

➪ 13 = A² + B² + 2AB(1/2)

➪ 13 = 10 + AB

➪ AB = 3 ㅤㅤㅤㅤㅤㅤㅤ(2)

▬▬▬▬▬▬▬▬▬▬▬▬

Now ,

➣ ( A - B )² = A² + B² -2AB

➣ ( A - B )² = 10 - 2(3)

➣ ( A - B )² = 10 - 6

➣ ( A - B )² = 4

➣ A - B = 2 ㅤ ㅤ ㅤ ㅤ ㅤㅤ(3)

Also,

➣ ( A + B )² = A² + B² + 2AB

➣ ( A + B )² = 10 + 2(3)

➣ ( A + B )² = 16

➣ A + B = 4 ㅤㅤㅤㅤㅤㅤㅤ(4)

▬▬▬▬▬▬▬▬▬▬▬▬

Adding equation (3) and (4) ,

⇢ A - B + A + B = 2 + 4

⇢ 2A = 6

⇢ A = 3N

▬▬▬▬▬▬▬▬▬▬▬▬

Putting A = 3 in eqⁿ (3),

⇢ A - B = 2

⇢ 3 - B = 2

⇢ B = 3-2

⇢ B = 1N

Therefore , the two forces are 3N and 1N.