Question

Two forces are acting at an angle of 120. The bigger force is 40n and the resultant is perpendicular to the smaller one. The smaller force is

Answer 1

Let the two Forces be P and Q and there Resultant be R. Let Q be bigger one and let R be smaller one.

Let θ ( = 120°) be the angle between the two vectors P and Q.

|R|² = |P|² +  |Q|² + 2|P||Q|Cosθ

∴ |R|² = P² + Q² + 2PQ × -1/2

∴ R² = P² + Q² - PQ

Also, Let Resultant makes an angle α ( = 90) with smaller one.

Then, $tan\alpha = \frac{PSin\theta}{Q + PCos\theta}\\tan90 = \frac{PSin\theta}{Q + PCos\theta}\\{Q + PCos\theta} = 0$

Q = -PCosθ

∴ Q = -PCos120

∴ Q = -P × - 1/2

∴ Q = P/2

∴ Q = 40/2    [Since, P = 40].

∴ Q = 20 N.

Hence, the smaller force is 20 N.

You can also calculate the Resultant force.

R² = 40² + 20² - 40 × 20

∴ R² = 1600 + 400 - 800

∴ R² = 2000 - 800

∴ R² = 1200

∴ R = 10√12

∴ R = 20√3 N.

Hence, the smaller force is 20 N and the Resultant force is 20√3 N.

Hope it helps.

Answer 2

Answer:

Let θ ( = 120°) be the angle between the two vectors P and Q.

|R|² = |P|² +  |Q|² + 2|P||Q|Cosθ

∴ |R|² = P² + Q² + 2PQ × -1/2

∴ R² = P² + Q² - PQ

Also, Let Resultant makes an angle α ( = 90) with smaller one.

Then,

Q = -PCosθ

∴ Q = -PCos120

∴ Q = -P × - 1/2

∴ Q = P/2

∴ Q = 40/2    [Since, P = 40].

∴ Q = 20 N.