Discussion
Consider two forces \(\vec{A}\) and \(\vec{B}\) of magnitudes A and B and their resultant force \(\vec{R}\) has magnitude R. If the B is doubled, then R is also doubled. Similarly, if B is reversed, then also R is doubled. Then the ratio A : B : R is:
Answer: B
Let \(\theta\) be the angle between the vectors and \(c = \cos\theta\). Then \(R^2 = A^2 + B^2 + 2ABc\). Doubling B gives \(4R^2 = A^2 + 4B^2 + 4ABc\), and reversing B gives \(4R^2 = A^2 + B^2 - 2ABc\). Subtracting the last two: \(3B^2 + 6ABc = 0\), so \(2ABc = -B^2\).
\[ \begin{aligned} R^2 = A^2 + B^2 - B^2 = A^2 \Rightarrow R = A \\ 4A^2 = A^2 + B^2 + B^2 \Rightarrow B^2 \\ &= \tfrac{3}{2}A^2 \end{aligned} \]So \(A : B : R = 1 : \sqrt{3/2} : 1 = \sqrt{2} : \sqrt{3} : \sqrt{2}\). The ratios 2:3:2 and 2:1:3 are not consistent with R equal to A.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development