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Question: There are two fields of forces which are represented by \[\vec F = ay\hat i\] and \(\vec F = ax\hat ...

There are two fields of forces which are represented by F=ayi^\vec F = ay\hat i and F=axi^+byj^\vec F = ax\hat i + by\hat j where aa and bb are constants. Find out whether these forces are conservative or not.

Explanation

Solution

In vector algebra, when a force is represented in the form of F=Pi^+Qj^\vec F = P\hat i + Q\hat j it means the components of force vector in X and Y directions are PP and QQ respectively. A conservative force is one whose work done on a body is independent of the path taken.

Complete step by step answer:
We know that, if a force is conservative then its work done must be independent of path and hence, if a force field F\vec F is conservative its curl is always zero which means ×F=0\vec \nabla \times \vec F = 0.Let us check the first given force field F=ayi^\vec F = ay\hat i as its YY and ZZ components are zero.

Checking the ×F=0\vec \nabla \times \vec F = 0 for force F=ayi^\vec F = ay\hat i we get,
\vec \nabla \times \vec F = \left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ {\dfrac{\partial }{{\partial x}}}&{\dfrac{\partial }{{\partial y}}}&{\dfrac{\partial }{{\partial z}}} \\\ {ay}&0&0 \end{array}} \right)
Finding determinant of above matrix we get,
×F=0+(ak^)\vec \nabla \times \vec F = 0 + ( - a\hat k)
×F=ak^\therefore \vec \nabla \times \vec F = - a\hat k
Hence, the value of ×F0\vec \nabla \times \vec F \ne 0 for the given force F=ayi^\vec F = ay\hat i.Hence, force F=ayi^\vec F = ay\hat i is not a conservative force.

Similarly, Let us check for the force F=axi^+byj^\vec F = ax\hat i + by\hat j
We have,
\vec \nabla \times \vec F = \left( {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ {\dfrac{\partial }{{\partial x}}}&{\dfrac{\partial }{{\partial y}}}&{\dfrac{\partial }{{\partial z}}} \\\ {ax}&{by}&0 \end{array}} \right)
Again, finding determinant of above matrix, we have
×F=0+0+0\vec \nabla \times \vec F = 0 + 0 + 0
×F=0\therefore \vec \nabla \times \vec F = 0
Hence, the value of ×F=0\vec \nabla \times \vec F = 0 for the given force F=axi^+byj^\vec F = ax\hat i + by\hat j. Hence, the force F=axi^+byj^\vec F = ax\hat i + by\hat j is a conservative force.

So, force F=axi^+byj^\vec F = ax\hat i + by\hat j is a conservative force while force F=ayi^\vec F = ay\hat i is not a conservative force.

Note: Remember the operator \vec \nabla is known as del operator and its defined as xi^+yj^+zk^\dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k and this del operator is widely used in differential calculus in order to find the divergence and curl of a given field the term ×F\vec \nabla \times \vec F for any vector field F\vec F is known as curls of the field and if this value is zero it’s said to be vector field is irrotational.