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

Correct Answer: A

Solution :

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 \[{{F}_{1}}.\]