Question

two forces of equal magnitude are acting at a point with an angle of 60° between them.If the reasultant force is equal to 40√3 Newton find magnitude of each force​

Answer 1

Answer:

Magnitude of each force = 40 Newton

Explanation:

▸Let , magnitude of equal vector forces be A

▸Angle between the Forces , θ = 60°

▸Magnitude of Resultant of Forces , R = 40 √3 Newton

Using formula to Calculate the Magnitude of Resultant of two vectors

▶ R² = P² + Q² + 2 P Q cos θ  

( where R is the magnitude of resultant , P and Q are magnitudes of two vectors and θ is the angle b/w two vectors )

→ R² = A² + A² + 2 A . A cos θ

→ (40√3)² = 2 A² + 2 A² cos 60°

→ 4800 = 2 A² + 2 A² (1/2)

→ 4800 = 2 A² + A²

→ 4800 = 3 A²

→ A² = 4800 / 3

→ A² = 1600

→ A = √(1600)

→ A = 40 Newton

Therefore,

Magnitude of each force is 40 Newton.

Answer 2

Given ,

• Angle b/w two equal forces = 60°
• Magnitude of resultant force = 40√3 N

Let ,

The magnitude of two equal forces be " F "

We know that , the magnitude of resultant of two vectors A and B is given by

$\large \boxed{ \sf{R = \sqrt{ {(A)}^{2} + {(B)}^{2} + 2AB \cos( \theta) } }}$

Thus ,

$\mapsto \sf 40 \sqrt{3} = \sqrt{2 {(F)}^{2} + 2 {(F)}^{2} \times \frac{1}{2} } \\ \\ \mapsto \sf 40 \sqrt{3} = \sqrt{3 {(F)}^{2} } \\ \\ \mapsto \sf F = \frac{40 \sqrt{3} }{ \sqrt{3} } \\ \\ \mapsto \sf F = 40 \: newton$

$\sf \therefore{ \underline{ The \: magnitude \: of \: two \: equal \: forces \: is \: 40 \: N}}$