Question

Three particle each of mass m are placed at the three corners of an equilateral tangle the center of the triangle is at a distance x from either corner. If a mass M be placed at the center. What will be the net gravitational force on it:

• A. Zero
• B. $3GM/x^2$
• C. $\frac{1}{36}$
• D. $GMn/x^2$

Answer 1

The correct option is A Zero

The gravitational force on mass $M$at $G$due to mass at $A$ is

$F_1=G\frac{m\times M}{1^2}=\frac{GmM}{x^2}$ Along $GA$

Gravitational force on mass $M$ at $G$ due to mass at $B$ is

$F_2=G\frac{m\times M}{1^2}=\frac{GmM}{x^2}$ Along $BG$

Gravitational force on mass $M$ at $G$ due to mass at $C$ is

$F_3=G\frac{m\times M}{1^2}=\frac{GmM}{x^2}$ Along $GC$

Draw $DE$ parallel to $BC$ passing through point $G.$ Then

$\angle EGC={30}^{\circ }=\angle DGB.$

Resolving ${\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_3$ into two rectangular components,

We have

$F_2\cos {30}^{\circ }$ Along $GD$ and $F_2\sin {30}^{\circ }$ along $GH$

$F_3\cos {30}^{\circ }$ Along $GE$ and $F_3\sin {30}^{\circ }$ along $GH$

Here, $F\cos {30}^{\circ }$ and $F_3\sin {30}^{\circ }$ are equal in magnitude and

Acting in opposite directions, and cancel out each other. The

Resultant force on mass $M$ at $G$ is

$F_1-(F_2\sin {30}^{{}^{\circ }}+F_3\sin {30}^{{}^{\circ }})$

$=\frac{GmM}{x^2}-(\frac{GmM}{x^2}\times \frac{1}{2}+\frac{GmM}{x^2}\times \frac{1}{2})$

$=0$

Hence, Net force at center mass is zero.