Question 164: Laws of Motion

Question

Two forces $$\overline{\mathrm{F}}_1$$ and $$\overline{\mathrm{F}}_2$$ 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 $$\cos ^{-1}\left(\frac{1}{n}\right)$$. The value of $$|n|$$ is _______.

Accepted Answer

Let's denote the magnitude of the smaller force as $$F$$, hence the magnitude of the larger force is $$3F$$. The resultant force $$\vec{R}$$ is equal in magnitude to the larger force, which means $$|\vec{R}| = 3F$$. When two forces $$\vec{F}_1$$ and $$\vec{F}_2$$ act on a body, the magnitude of their resultant $$\vec{R}$$ can be found using the law of vector addition:
$$|\vec{R}| = \sqrt{|\vec{F}_1|^2 + |\vec{F}_2|^2 + 2|\vec{F}_1||\vec{F}_2|\cos	heta}$$,
where $$	heta$$ is the angle between $$\vec{F}_1$$ and $$\vec{F}_2$$. Given that in our case $$|\vec{R}| = 3F$$, $$|\vec{F}_1| = F$$ and $$|\vec{F}_2| = 3F$$, by substituting these values into the equation, we get:
$$3F = \sqrt{F^2 + (3F)^2 + 2(F)(3F)\cos	heta}$$
$$9F^2 = F^2 + 9F^2 + 6F^2\cos	heta$$
Simplifying this equation by subtracting $$10F^2$$ from both sides gives:
$$-F^2 = 6F^2\cos	heta$$
Dividing both sides by $$-F^2$$ gives:
$$-1 = -6\cos	heta$$
Therefore, $$\cos	heta = \frac{1}{6}$$.
It is given that the angle between $$\vec{F}_1$$ and $$\vec{F}_2$$ is $$\cos^{-1}\left(\frac{1}{n}ight)$$, hence comparing this with the above result, we find that $$n = 6$$. Therefore, $$|n| = 6$$.

Laws of Motion

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