three balls ,a,b,c are kept in a straight line. the sepration between a c is 1 metre , and b is kept at the midpoint between them. the masses of a,b,c are 100 g ,200g,300g,respectively . find the net gravitation force on a,b,c .

We have three balls a,b and c placed in straight line.
Given,
ma=100 g=0.1 kgmb=200g=0.2 kgmc=300g=0.3 kgab=0.5 m=bcac=1mNow using relation,F=Gmm'r2Force on a=Force on a due to b +Force on a due to c=Gmambab2+Gmamcac2=6.67×10-11[0.1×0.20.52+0.1×0.312]=6.67×10-11[0.020.25+0.031]=7.34×10-10NSimilarly force on b,Force on b due to c -Force on b due to a(Forces are in opposite direction)=Gmbmcbc2-Gmbmaba2=6.67×10-11[0.2×0.30.52-0.2×0.10.52]=6.67×10-11[0.060.25-0.020.25]=1.06×10-11NForce on c=Force on c due to a +Force on c due to b=Gmcmaac2+Gmcmbcb2=6.67×10-11[0.3×0.112+0.3×0.20.52]=6.67×10-11[0.03+0.060.25]=1.8009×10-11N

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