3P এবং 2P বলদ্বয়ের R । প্রথম বল দ্বিগুণ করলে লব্ধির পরিমাণও দ্বিগুণ হয় । বলদ্বয়ের অর্ন্তগত কোন কত ? 

Explanation:

Another Explanation (5): Let the two forces be \(\vec{P_1}\) and \(\vec{P_2}\) with magnitudes 3P and 2P respectively. Let the angle between them be \(\theta\). The magnitude of the resultant force \(\vec{R}\) is given by: \(R^2 = (3P)^2 + (2P)^2 + 2(3P)(2P) \cos{\theta}\) \(R^2 = 9P^2 + 4P^2 + 12P^2 \cos{\theta}\) \(R^2 = 13P^2 + 12P^2 \cos{\theta}\) .....(1) Now, the first force is doubled, so its magnitude becomes 6P. The new resultant force is \(\vec{R'}\) and its magnitude is 2R. Thus, \((2R)^2 = (6P)^2 + (2P)^2 + 2(6P)(2P) \cos{\theta}\) \(4R^2 = 36P^2 + 4P^2 + 24P^2 \cos{\theta}\) \(4R^2 = 40P^2 + 24P^2 \cos{\theta}\) .....(2) Substituting (1) into (2): \(4(13P^2 + 12P^2 \cos{\theta}) = 40P^2 + 24P^2 \cos{\theta}\) \(52P^2 + 48P^2 \cos{\theta} = 40P^2 + 24P^2 \cos{\theta}\) \(12P^2 = -24P^2 \cos{\theta}\) \(\cos{\theta} = -\frac{12P^2}{24P^2}\) \(\cos{\theta} = -\frac{1}{2}\) \(\theta = \cos^{-1}(-\frac{1}{2})\) \(\theta = 120^\circ\) 🎉 Therefore, the angle between the two forces is \(120^\circ\).