Two forces and are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between and is The value of is

Question

Two forces ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ is $\cos^{- 1} (\frac{1}{n}) .$ The value of $| n |$ is ___________ . [2024]

Smart Achievers · Accepted Answer

(6) $let F_{1} = F_{0} then, F_{2} = 3 F_{0} also, F_{R} = 3 F_{0}$

$F_{R}^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$

$∴ 3 F_{0} = \sqrt{F_{0}^{2} + 9 F_{0}^{2} + 6 F_{0}^{2} \cos θ}$

$3 F_{0} = F_{0} \sqrt{10 + 6 \cos θ}$

$9 = 10 + 6 \cos θ$

$\cos θ = - \frac{1}{6} \Rightarrow θ = \cos^{- 1} (\frac{1}{- 6})$

$n = - 6 \Rightarrow | n | = 6$

Solution

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Solution

Q. Two forces F→1 and F→2 are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between F→1 and F→2 is cos-1(1n). The value of |n| is ___________ . [2024]

Ans. (6) let F1=F0 then, F2=3F0 also, FR=3F0 FR2=F12+F22+2F1F2cosθ ∴3F0=F02+9F02+6F02cosθ 3F0=F010+6cosθ 9=10+6cosθ cosθ=-16⇒θ=cos-1(1-6) n=-6⇒|n|=6

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