To solve this problem, we'll need to use vector addition and the law of cosines for the resultant of two forces.
Let:
- [tex]\( F_1 = 3P \)[/tex]
- [tex]\( F_2 = 2P \)[/tex]
- [tex]\(\theta\)[/tex] be the angle between [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex].
- [tex]\( R \)[/tex] be the resultant of [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex].
The formula for the resultant of two forces is given by:
[tex]\[ R = \sqrt{(F_1^2 + F_2^2 + 2 \cdot F_1 \cdot F_2 \cdot \cos \theta)} \][/tex]
Given:
- [tex]\( F_1 = 3P \)[/tex]
- [tex]\( F_2 = 2P \)[/tex]
Substitute these into the formula:
[tex]\[ R = \sqrt{(3P)^2 + (2P)^2 + 2 \cdot (3P) \cdot (2P) \cdot \cos \theta} \][/tex]
[tex]\[ R = \sqrt{9P^2 + 4P^2 + 12P^2 \cdot \cos \theta} \][/tex]
[tex]\[ R = \sqrt{13P^2 + 12P^2 \cos \theta} \][/tex]
When the first force is doubled, [tex]\( 3P \)[/tex] becomes [tex]\( 6P \)[/tex] and the new resultant [tex]\( R' \)[/tex] is:
[tex]\[ R' = 2R \][/tex]
Using the formula for the resultant:
[tex]\[ R' = \sqrt{(6P)^2 + (2P)^2 + 2 \cdot (6P) \cdot (2P) \cdot \cos \theta} \][/tex]
[tex]\[ 2R = \sqrt{36P^2 + 4P^2 + 24P^2 \cdot \cos \theta} \][/tex]
[tex]\[ 2R = \sqrt{40P^2 + 24P^2 \cdot \cos \theta} \][/tex]
Since [tex]\( R' = 2R \)[/tex], we can write:
[tex]\[ 2 \sqrt{13P^2 + 12P^2 \cos \theta} = \sqrt{40P^2 + 24P^2 \cos \theta} \][/tex]
Square both sides to remove the square roots:
[tex]\[ (2R)^2 = \sqrt{(R')^2} \][/tex]
[tex]\[ 4(13P^2 + 12P^2 \cos \theta) = 40P^2 + 24P^2 \cos \theta \][/tex]
Simplify and solve for [tex]\(\cos \theta\)[/tex]:
[tex]\[ 52P^2 + 48P^2 \cos \theta = 40P^2 + 24P^2 \cos \theta \][/tex]
[tex]\[ 52P^2 - 40P^2 = 24P^2 \cos \theta - 48P^2 \cos \theta \][/tex]
[tex]\[ 12P^2 = -24P^2 \cos \theta \][/tex]
[tex]\[ \cos \theta = -\frac{12P^2}{24P^2} \][/tex]
[tex]\[ \cos \theta = -\frac{1}{2} \][/tex]
The angle whose cosine is [tex]\(-\frac{1}{2}\)[/tex] is [tex]\(\theta = 120^\circ\)[/tex].
Therefore, the angle between the two forces is:
[tex]\[ \boxed{120^\circ} \][/tex]
