Question

Three masses 2m,m & m are placed at the vertices of an equilateral triangle of side ‘a’. The force on mass 2m, due to other masses is ……………..

Answer 1

Answer:

superb Question. I think the strategy in solving this is to Involve : Gravitation forces : G m1 m2/ d square and find Resultant

f Net = 4Gmsqaure/ L square x cos 30ndegrees

Answer 2

Answer:

$\huge\bold{Given: }$

• Three masses 2m,m & m are placed at the vertices of an equilateral triangle of side ‘a’. The force on mass 2m

$\huge\bold{To \: Find: }$

• other masses is

$\huge\bold{Solution: }$

Draw a perpendicular AD to the side BC.

$\sf∴AD=AB \sin(60) = \frac{ \sqrt{3} }{2} 1$

Distance AO of the centroid O from A is

$\sf \frac{2}{3} AD \\ AO = \sf\frac{2}{3} \binom{ \sqrt{3} }{2} 1 = \frac{1}{ \sqrt{3} }$

$\sf \: By \: symmetry, AO=BO=CO = \frac{1}{ \sqrt{3} } </p><p></p><p>$

Force on mass 2m at O due to mass m at A is

$\sf \: F _{oa} = \frac{Gm(2m)}{(1/ \sqrt{3} ) {}^{2} } = \frac{6Gm {}^{2} }{ {1}^{2} } along \: OA$

Force on mass 2m at O due to mass m at C is

$\sf \: F _{oa} = \frac{Gm(2m)}{(1/ \sqrt{3} ) {}^{2} } = \frac{6Gm {}^{2} }{ {1}^{2} } along \: OC</p><p></p><p>$

$\sf \: Draw \: a \: line \: PQ \: parallel \: to \: BC \: passing \: through \: O. \\ \sf Then \:  ∠BOP=30=∠COQ</p><p>Resolving \:  FˉOB \:  and  \: FˉOC \:  into \: two \: components.</p><p></p><p>$

Components acting along OP and OQ are equal in magnitude and opposite in direction. So, they will cancel out while the components acting along OD will add up.

∴ The resultant force on the mass 2m at O is

$</p><p> \sf \: FR=F _{OA} −(F _{OB} \: sin30+F _{OC} \: sin30)</p><p></p><p>$

$\pink{ \sf \: = \frac{6Gm {}^{2} }{ {1}^{2} } - ( \frac{6Gm {}^{2}}{ {1}^{2} } \times \frac{1}{2 } + \frac{6Gm {}^{2}}{ {1}^{2} } \times \frac{1}{2} ) = 0}$