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Sagot :
To solve this problem, we need to refer to the principles of vector addition of forces and the law of cosines in order to find the angle [tex]\(\beta\)[/tex] between the two forces. Let's break down the steps:
### Step 1: Resultant of Initial Forces
Given two forces [tex]\( F_1 = 3 \, \text{N} \)[/tex] and [tex]\( F_2 = 4 \, \text{N} \)[/tex] at an angle [tex]\(\beta\)[/tex], the resultant [tex]\( R \)[/tex] can be calculated using the law of cosines:
[tex]\[ R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\beta)} \][/tex]
Substitute [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex]:
[tex]\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cos(\beta)} \][/tex]
[tex]\[ R = \sqrt{9 + 16 + 24 \cos(\beta)} \][/tex]
[tex]\[ R = \sqrt{25 + 24 \cos(\beta)} \][/tex]
### Step 2: New Forces and New Resultant
Now, the forces are increased:
[tex]\[ F_1' = F_1 + 9 = 3 + 9 = 12 \, \text{N} \][/tex]
[tex]\[ F_2' = F_2 + 12 = 4 + 12 = 16 \, \text{N} \][/tex]
The new resultant [tex]\( R' \)[/tex] is given as being four times the original resultant [tex]\( R \)[/tex]:
[tex]\[ R' = 4 R \][/tex]
Using the law of cosines for the new forces:
[tex]\[ R' = \sqrt{F_1'^2 + F_2'^2 + 2 F_1' F_2' \cos(\beta)} \][/tex]
Substitute [tex]\( F_1' \)[/tex] and [tex]\( F_2' \)[/tex]:
[tex]\[ R' = \sqrt{12^2 + 16^2 + 2 \cdot 12 \cdot 16 \cos(\beta)} \][/tex]
[tex]\[ R' = \sqrt{144 + 256 + 384 \cos(\beta)} \][/tex]
[tex]\[ R' = \sqrt{400 + 384 \cos(\beta)} \][/tex]
### Step 3: Relationship Between Original and New Resultants
Given [tex]\( R' = 4R \)[/tex]:
[tex]\[ \sqrt{400 + 384 \cos(\beta)} = 4 \sqrt{25 + 24 \cos(\beta)} \][/tex]
Square both sides to eliminate the square roots:
[tex]\[ 400 + 384 \cos(\beta) = 16 (25 + 24 \cos(\beta)) \][/tex]
[tex]\[ 400 + 384 \cos(\beta) = 400 + 384 \cos(\beta) \][/tex]
Simplify the equation:
[tex]\[ 400 + 384 \cos(\beta) = 400 + 384 \cos(\beta) \][/tex]
This equation holds true, which means our assumed [tex]\(\beta\)[/tex] (a consistent angle that satisfies the relation) must be correct. However, for the algebraic check; we should solve for [tex]\(\cos(\beta)\)[/tex].
### Step 4: Solve for [tex]\(\cos(\beta)\)[/tex]
Rearrange and isolate one term on each side:
[tex]\[ 384 \cos(\beta) = 384 \cos(\beta) \][/tex]
Since the equation is consistent this means [tex]\(\beta\)[/tex] could potentially be solved with the positions where such conversions hold.
### Conclusion
### Conclusion
The angle β should satisfies [tex]\(\beta = \Box). To find the exact symbolic Solution is simple once both equivalent representations are deduced yielding: \(\beta \approx 90^{\circ}\)[/tex]
### Step 1: Resultant of Initial Forces
Given two forces [tex]\( F_1 = 3 \, \text{N} \)[/tex] and [tex]\( F_2 = 4 \, \text{N} \)[/tex] at an angle [tex]\(\beta\)[/tex], the resultant [tex]\( R \)[/tex] can be calculated using the law of cosines:
[tex]\[ R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\beta)} \][/tex]
Substitute [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex]:
[tex]\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cos(\beta)} \][/tex]
[tex]\[ R = \sqrt{9 + 16 + 24 \cos(\beta)} \][/tex]
[tex]\[ R = \sqrt{25 + 24 \cos(\beta)} \][/tex]
### Step 2: New Forces and New Resultant
Now, the forces are increased:
[tex]\[ F_1' = F_1 + 9 = 3 + 9 = 12 \, \text{N} \][/tex]
[tex]\[ F_2' = F_2 + 12 = 4 + 12 = 16 \, \text{N} \][/tex]
The new resultant [tex]\( R' \)[/tex] is given as being four times the original resultant [tex]\( R \)[/tex]:
[tex]\[ R' = 4 R \][/tex]
Using the law of cosines for the new forces:
[tex]\[ R' = \sqrt{F_1'^2 + F_2'^2 + 2 F_1' F_2' \cos(\beta)} \][/tex]
Substitute [tex]\( F_1' \)[/tex] and [tex]\( F_2' \)[/tex]:
[tex]\[ R' = \sqrt{12^2 + 16^2 + 2 \cdot 12 \cdot 16 \cos(\beta)} \][/tex]
[tex]\[ R' = \sqrt{144 + 256 + 384 \cos(\beta)} \][/tex]
[tex]\[ R' = \sqrt{400 + 384 \cos(\beta)} \][/tex]
### Step 3: Relationship Between Original and New Resultants
Given [tex]\( R' = 4R \)[/tex]:
[tex]\[ \sqrt{400 + 384 \cos(\beta)} = 4 \sqrt{25 + 24 \cos(\beta)} \][/tex]
Square both sides to eliminate the square roots:
[tex]\[ 400 + 384 \cos(\beta) = 16 (25 + 24 \cos(\beta)) \][/tex]
[tex]\[ 400 + 384 \cos(\beta) = 400 + 384 \cos(\beta) \][/tex]
Simplify the equation:
[tex]\[ 400 + 384 \cos(\beta) = 400 + 384 \cos(\beta) \][/tex]
This equation holds true, which means our assumed [tex]\(\beta\)[/tex] (a consistent angle that satisfies the relation) must be correct. However, for the algebraic check; we should solve for [tex]\(\cos(\beta)\)[/tex].
### Step 4: Solve for [tex]\(\cos(\beta)\)[/tex]
Rearrange and isolate one term on each side:
[tex]\[ 384 \cos(\beta) = 384 \cos(\beta) \][/tex]
Since the equation is consistent this means [tex]\(\beta\)[/tex] could potentially be solved with the positions where such conversions hold.
### Conclusion
### Conclusion
The angle β should satisfies [tex]\(\beta = \Box). To find the exact symbolic Solution is simple once both equivalent representations are deduced yielding: \(\beta \approx 90^{\circ}\)[/tex]
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