• question_answer A mass m in three-dimensional space is subjected to three forces: $\theta$ ${{F}_{2}}$ and ${{F}_{3}},$ ${{F}_{1}}$ and ${{F}_{2}}$ have the same magnitude, with ${{F}_{1}}$ in the positive-x direction, and ${{F}_{2}}$ in the positive-y direction. If the mass has an acceleration of 0, which of the following statements is false? A) The magnitude of ${{F}_{3}}$ is the same as that of ${{F}_{1}}$ B) The object is in equilibrium and could be stationary C) ${{F}_{3}}$ lies in the x-y plane D) The object is in equilibrium and could be moving

 By definition, the object is in equilibrium, either static (unmoving) or dynamic (moving with a constant velocity). If the object has acceleration a=0, the net force acting on the mass must be 0 as well: ${{F}_{net}}=ma$ ${{F}_{net}}=m(0)=0$
 With force ${{F}_{1}}$and ${{F}_{2}}$ in the x-y plane, the force that will counteract them must lie in the x-y plane as well, as shown. The magnitude of that force ${{F}_{3}}$ is equal to the vector sum of ${{F}_{1}}$ and ${{F}_{2}}$ and can be calculated as follows: $\sum{{{F}_{x}}}=0={{\vec{F}}_{2}}-{{\vec{F}}_{3-x}}$ $\sum{{{F}_{y}}}=0={{\vec{F}}_{1}}-{{\vec{F}}_{3-x}}$ $\left| \,{{F}_{3}} \right|=\sqrt{F_{3-x}^{2}+F_{3-y}^{2}}$$=\sqrt{F_{1}^{2}+F_{2}^{2}}={{F}_{1}}\sqrt{2}$ both graphically and analytically, we can see that the magnitude of ${{F}_{3}}$ is not the same as that of ${{F}_{1}}.$
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