Six point masses of mass m each are at the vertices of a regular hexagon of side l. Calculate the force on any of the masses.

0 2 Views | Posted 4 months ago
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  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    This is a Long Type Questions as classified in NCERT Exemplar

    Explanation- consider a diagram having vertices A,B,C,D,E and F

    AC= AG+GC=2AG

    = 2lcos300= 2l 3 2

    = 3 l = A E

    AD=AH+HJ+JD= lsin300+l+lsin300=2l

    Force on mass m at A due to mass m at B is f1= G m m l 2  along AB

    Force on mass m at A due to mass m at C is f2= G m m 3 l 2 = G m 2 3 l 2  along AC

    Force on mass m at A due to mass m at D is F3= G m m ( 2 l ) 2 = G m 2 4 l 2 along AD

    Force on mass m at A due to mass m at E is F4= G m m 3 l 2 = G m 2 3 l 2 along AE

    Force on mass m at A due to mass m at F is F5= G m 2 l 2  along AF

    Resultant force due to F1 and F5 is F1= f 1 2 + f 5 2 + 2 f 1 f 2 c o s 60

    = 3 G m 2 3 l 2 = G m 2 3 l 2 along AD

    So net force along AD = F1+F2+F3= G m 2 l 2 + G m 2 3 l 2 + G m 2 4 l 2 = G m 2 l 2 1 + 1 3 + 1 4

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  G M m ( R + x ) 2 = G M m ( R x ) R 3

->R3 = (R + x)2 (R – x)

->R3 = (R2– x2) (R + x)

->x2 + Rx – R2 = 0


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Stress = F A = T A = W A

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( v e ) A = ( v e ) B M 1 R 1 = M 2 R 2

R is not correct.

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