The sum of three forces vec F_1 = 100,N,vec F | vedclass

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(B) Given that the sum of three forces is zero,$\vec F_1 + \vec F_2 + \vec F_3 = 0$,which implies $\vec F_1 = -(\vec F_2 + \vec F_3)$.This means the t

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$143$

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(B) Given that the sum of three forces is zero,$\vec F_1 + \vec F_2 + \vec F_3 = 0$,which implies $\vec F_1 = -(\vec F_2 + \vec F_3)$.This means the three forces form a closed triangle when placed head-to-tail.The magnitudes are $F_1 = 100\,N$,$F_2 = 80\,N$,and $F_3 = 60\,N$.Since $60^2 + 80^2 = 3600 + 6400 = 10000 = 100^2$,the forces form a right-angled triangle.Let $	heta$ be the angle between $\vec F_1$ and $\vec F_2$. In the vector triangle,the angle between $\vec F_1$ and $\vec F_2$ is the exterior angle at the vertex where they meet.From the geometry of the triangle,the interior angle opposite to the side of magnitude $60\,N$ is $\alpha$,where $\sin \alpha = 60/100 = 0.6$,so $\alpha = 37^o$.The angle between the vectors $\vec F_1$ and $\vec F_2$ is $180^o - \alpha = 180^o - 37^o = 143^o$.

The sum of three forces $\vec F_1 = 100\,N$,$\vec F_2 = 80\,N$,and $\vec F_3 = 60\,N$ acting on a particle is zero. The angle between $\vec F_1$ and $\vec F_2$ is nearly .......... $^o$

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