Three forces F_1=10,N,F_2=8,N,and F_3=6,N are | vedclass

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(A) Given that the particle is in equilibrium under the action of three forces $F_1$,$F_2$,and $F_3$. Since $F_2$ and $F_3$ are perpendicular,their re

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$2$

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(A) Given that the particle is in equilibrium under the action of three forces $F_1$,$F_2$,and $F_3$. Since $F_2$ and $F_3$ are perpendicular,their resultant force $F_{23}$ is given by $F_{23} = \sqrt{F_2^2 + F_3^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\,N$. For the particle to be at rest,the resultant of all forces must be zero,meaning $F_1$ must be equal and opposite to the resultant of $F_2$ and $F_3$. When $F_1$ is removed,the only forces acting on the particle are $F_2$ and $F_3$. The net force acting on the particle becomes the resultant of $F_2$ and $F_3$,which is $10\,N$. Using Newton's second law,$F_{net} = ma$,we have $10\,N = 5\,kg 	imes a$. Therefore,$a = \frac{10}{5} = 2\,m/s^2$.

Three forces $F_1=10\,N$,$F_2=8\,N$,and $F_3=6\,N$ are acting on a particle of mass $5\,kg$. The forces $F_2$ and $F_3$ are applied perpendicular to each other such that the particle remains at rest. If the force $F_1$ is removed,then the acceleration of the particle is $......\,m/s^2$.

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