Question

Two forces 3N and 2 N are at an angle θ such that the resultant is R. The first force is now increased to 6N and the resultant become 2R. The value of θ is

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Two forces 3N and 2 N are at an angle θ such that the resultant is R.

The first force is now increased to 6N and the resultant become 2R.

To Find,

Value of θ.

Here, taking

F₁ = 3 N

F₂ = 2 N

1st case,

R² = 9 + 4 + 2 × 3 × 2 Cos θ ...... (i)

2nd case,

(2R)² = 36 + 4 + 2 × 6 × 2 Cos θ .... (ii)

Dividing eq (ii) by (i), we get

⇒ 4 = 40 + 24 + Cos θ/30 + 12 Cos θ

by Cross multiplying, we get

⇒ 52 + 48 Cos θ = 40 + 24 Cos θ

⇒ 24 Cos θ = - 12

⇒ Cos θ = - 12/24

⇒ Cos θ = - 1/2

⇒ θ = 120.

Hence, the required value of θ is 120.

Answer 2

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R^2 = A^2 + B^2 + 2AB Cos theta A=3N B= 2 N

R^2 = 13 + 12 Cos theta

Case 2 4R^2 = 40 + 24 Cos theta

put the value of R^2

here we got

4 ( 13 + 12 Cos theta ) = 40 + 24 Cos theta 52 +48 Cos theta = 40 + 24 Cos theta

Now 24 Cos theta = 40-52

Cos theta = -12 / 24

Cos theta = -1/2

Cos 120 degree = -1/2

Theta = 120 degree

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