Answer :
Sure, let's solve this step-by-step.
Given:
- Two forces [tex]\( A + B \)[/tex] and [tex]\( A - B \)[/tex] make an angle [tex]\( 2\delta \)[/tex] with each other.
- The resultant of these forces makes an angle [tex]\( \beta \)[/tex] with the bisector of the angle between them.
We need to show:
[tex]\[ A \tan \beta = B \tan \delta \][/tex]
Step 1: Understanding the Geometry of the Problem
Let's denote the forces as vectors:
- [tex]\(\vec{F_1} = A + B\)[/tex]
- [tex]\(\vec{F_2} = A - B\)[/tex]
These two forces form an angle [tex]\( 2\delta \)[/tex] with each other. This means that if you draw the angles around the bisector, each side of the angle [tex]\( 2\delta \)[/tex] will split into two angles of [tex]\( \delta \)[/tex].
Step 2: Resultant Force
Let [tex]\(\vec{R}\)[/tex] be the resultant of [tex]\(\vec{F_1}\)[/tex] and [tex]\(\vec{F_2}\)[/tex]. The resultant [tex]\(\vec{R}\)[/tex] makes an angle [tex]\( \beta \)[/tex] with the angle bisector.
Step 3: Components Along the Bisector
Let’s find the components of the forces along the direction of the angle bisector:
- The components of [tex]\(\vec{F_1}\)[/tex] along the bisector will be [tex]\( (A + B) \cos(\delta) \)[/tex]
- The components of [tex]\(\vec{F_2}\)[/tex] along the bisector will be [tex]\( (A - B) \cos(\delta) \)[/tex]
Step 4: Perpendicular Components
Next, consider the perpendicular components:
- The perpendicular component of [tex]\(\vec{F_1}\)[/tex] will be [tex]\( (A + B) \sin(\delta) \)[/tex]
- The perpendicular component of [tex]\(\vec{F_2}\)[/tex] will be [tex]\( (A - B) \sin(\delta) \)[/tex]
Step 5: Resultant Perpendicular Component
The net perpendicular component of the resultant [tex]\(\vec{R}\)[/tex] should add up vectorially:
[tex]\[ R_{\perp} = (A + B) \sin(\delta) - (A - B) \sin(\delta) \][/tex]
[tex]\[ R_{\perp} = (A \sin(\delta) + B \sin(\delta)) - (A \sin(\delta) - B \sin(\delta)) \][/tex]
[tex]\[ R_{\perp} = 2 B \sin(\delta) \][/tex]
Step 6: Resultant Parallel Component
For the parallel component:
[tex]\[ R_{\parallel} = (A + B) \cos(\delta) + (A - B) \cos(\delta) \][/tex]
[tex]\[ R_{\parallel} = A \cos(\delta) + B \cos(\delta) + A \cos(\delta) - B \cos(\delta) \][/tex]
[tex]\[ R_{\parallel} = 2 A \cos(\delta) \][/tex]
Step 7: Relationship Between Perpendicular and Parallel Components
The resultant of these components in terms of [tex]\(\beta\)[/tex] is:
[tex]\[ \tan(\beta) = \frac{R_{\perp}}{R_{\parallel}} \][/tex]
[tex]\[ \tan(\beta) = \frac{2 B \sin(\delta)}{2 A \cos(\delta)} \][/tex]
Step 8: Simplification
By simplifying, we get:
[tex]\[ \tan(\beta) = \frac{B \sin(\delta)}{A \cos(\delta)} \][/tex]
Multiplying both sides by [tex]\(A\)[/tex], we have:
[tex]\[ A \tan(\beta) = B \frac{\sin(\delta)}{\cos(\delta)} \][/tex]
[tex]\[ A \tan(\beta) = B \tan(\delta) \][/tex]
And this completes the proof. Hence, we have shown that:
[tex]\[ A \tan \beta = B \tan \delta \][/tex]
Given:
- Two forces [tex]\( A + B \)[/tex] and [tex]\( A - B \)[/tex] make an angle [tex]\( 2\delta \)[/tex] with each other.
- The resultant of these forces makes an angle [tex]\( \beta \)[/tex] with the bisector of the angle between them.
We need to show:
[tex]\[ A \tan \beta = B \tan \delta \][/tex]
Step 1: Understanding the Geometry of the Problem
Let's denote the forces as vectors:
- [tex]\(\vec{F_1} = A + B\)[/tex]
- [tex]\(\vec{F_2} = A - B\)[/tex]
These two forces form an angle [tex]\( 2\delta \)[/tex] with each other. This means that if you draw the angles around the bisector, each side of the angle [tex]\( 2\delta \)[/tex] will split into two angles of [tex]\( \delta \)[/tex].
Step 2: Resultant Force
Let [tex]\(\vec{R}\)[/tex] be the resultant of [tex]\(\vec{F_1}\)[/tex] and [tex]\(\vec{F_2}\)[/tex]. The resultant [tex]\(\vec{R}\)[/tex] makes an angle [tex]\( \beta \)[/tex] with the angle bisector.
Step 3: Components Along the Bisector
Let’s find the components of the forces along the direction of the angle bisector:
- The components of [tex]\(\vec{F_1}\)[/tex] along the bisector will be [tex]\( (A + B) \cos(\delta) \)[/tex]
- The components of [tex]\(\vec{F_2}\)[/tex] along the bisector will be [tex]\( (A - B) \cos(\delta) \)[/tex]
Step 4: Perpendicular Components
Next, consider the perpendicular components:
- The perpendicular component of [tex]\(\vec{F_1}\)[/tex] will be [tex]\( (A + B) \sin(\delta) \)[/tex]
- The perpendicular component of [tex]\(\vec{F_2}\)[/tex] will be [tex]\( (A - B) \sin(\delta) \)[/tex]
Step 5: Resultant Perpendicular Component
The net perpendicular component of the resultant [tex]\(\vec{R}\)[/tex] should add up vectorially:
[tex]\[ R_{\perp} = (A + B) \sin(\delta) - (A - B) \sin(\delta) \][/tex]
[tex]\[ R_{\perp} = (A \sin(\delta) + B \sin(\delta)) - (A \sin(\delta) - B \sin(\delta)) \][/tex]
[tex]\[ R_{\perp} = 2 B \sin(\delta) \][/tex]
Step 6: Resultant Parallel Component
For the parallel component:
[tex]\[ R_{\parallel} = (A + B) \cos(\delta) + (A - B) \cos(\delta) \][/tex]
[tex]\[ R_{\parallel} = A \cos(\delta) + B \cos(\delta) + A \cos(\delta) - B \cos(\delta) \][/tex]
[tex]\[ R_{\parallel} = 2 A \cos(\delta) \][/tex]
Step 7: Relationship Between Perpendicular and Parallel Components
The resultant of these components in terms of [tex]\(\beta\)[/tex] is:
[tex]\[ \tan(\beta) = \frac{R_{\perp}}{R_{\parallel}} \][/tex]
[tex]\[ \tan(\beta) = \frac{2 B \sin(\delta)}{2 A \cos(\delta)} \][/tex]
Step 8: Simplification
By simplifying, we get:
[tex]\[ \tan(\beta) = \frac{B \sin(\delta)}{A \cos(\delta)} \][/tex]
Multiplying both sides by [tex]\(A\)[/tex], we have:
[tex]\[ A \tan(\beta) = B \frac{\sin(\delta)}{\cos(\delta)} \][/tex]
[tex]\[ A \tan(\beta) = B \tan(\delta) \][/tex]
And this completes the proof. Hence, we have shown that:
[tex]\[ A \tan \beta = B \tan \delta \][/tex]
