Question

Two particles each of mass 'm' are placed at $A$ and $C$ are such $AB=BC$ . The gravitational force on the third particle placed at $D$ at a - distance $L$ on the perpendicular bisector of the line $AC$ is

• A. $\frac{G{m}^2}{\sqrt{2}{L}^2}alongBD$
• B. $\frac{G{m}^2}{\sqrt{2}{L}^2}alongDB$
• C. $\frac{G{m}^2}{\sqrt{2}{L}^2}alongAC$
• D. none of these

Answer 1

Answer: (B)

Since $AB=BC=DB=L$, we have $AD=\sqrt{2}L$
Now force between particles placed at A and D is: $F=Gmm/(\sqrt{2}L{)}^2=$ $G{m}^2/2L^2$
Similar force particle at C will exert on particle at D.
So total force on particle $D$ will be ${}_{2}F$.
Now the resultant component of these two forces along DB will be: $2Fcos45^{\circ }$
Since $AB=BC=BD=L$ and $\angle DBC=\angle DBA=90^{\circ }$, we get $\angle ADB=$ $\angle BDC=45^{\circ }$
So, the resultant force on mass $m$ at $D$ will be $\frac{G{m}^2}{\sqrt{2}{L}^2}$