Question

The resultant of two forces one of which is 3 times the other is 300n when the direction of the forces are reversed the resultant is 200n determine two forces and angle between them

Answer 1

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Answer 2

Answer:

The angle between the two forces = $50.133^\circ.$

Explanation:

Let the two forces be $\vec F_1$ and $\vec F_2$, such that $F_1$ and $F_2$ are their magnitudes respectively.

When,

$F_2 = 3F_1,\\|\vec F_1+\vec F_2| = 300\ N.$

We know, if $\theta$ is the angle between the two vectors, then,$|\vec F_1+\vec F_2|=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\theta}\\300=\sqrt{F_1^2+(3F_1^2)+2F_1(3F_1)\cos\theta}\\300=\sqrt{F_1^2+9F_1^2+6F_1^2\cos\theta}\\300=\sqrt{10F_1^2+6F_1^2\cos\theta}\\300=F_1\sqrt{10+6\cos\theta}\\\Rightarrow F_1 = \dfrac{300}{\sqrt{10+6\cos\theta}}.\ \ .....\ (1)$

When the direction of the forces are reversed, the angle between them will be $180^\circ -\theta$, and the resultant between the two is $(-\vec F_1) + (-\vec F_2).$

Given, $|(-\vec F_1)+(-\vec F_2)|=200\ N.$

We know,

$|(-\vec F_1) +(-\vec F_2) |=\sqrt{(-F_1)^2+(-F_2)^2+2F_1F_2\cos(180^\circ-\theta)}\\200=\sqrt{F_1^2+(3F_1)^2-2F_1(3F_1)\cos\theta}\\200=\sqrt{F_1^2+9F_1^2-6F_1^2\cos\theta}\\200=F_1\sqrt{1+9-6\cos\theta}\\200=F_1\sqrt{10-6\cos\theta}.$

Using (1),

$200= \dfrac{300}{\sqrt{10+6\cos\theta}}\times \sqrt{10-6\cos\theta}\\200\times\sqrt{10+6\cos\theta}=300\sqrt{10-6\cos\theta}\\\\\text{On squaring both the sides,}\\\\200^2\times (10+6\cos\theta) = 300^2\times (10-6\cos\theta)\\40000(10+6\cos\theta)=90000(10-6\cos\theta)\\400000+240000\cos\theta= 900000-540000\cos\theta\\(240000+540000)\cos\theta = 900000-400000\\780000\cos\theta = 500000\\\cos\theta = \dfrac{500000}{780000}=0.641\\\theta = \cos^{-1}(0.641)=50.133^\circ.$