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Physics

Mechanics

Easy

Question

Three identical point masses, each of mass 1 kg lie in the x-y plane at points (0,0), (0,0.2 m) and (0.2 m, 0). The net gravitational force on the mass at the origin is

$1.67 \times 10^{- 9} (\hat{i} + \hat{j}) N$
$3.34 \times 10^{- 10} (\hat{i} + \hat{j}) N$
$1.67 \times 10^{- 9} (\hat{i} - \hat{j}) N$
$3.34 \times 10^{- 10} (\hat{i} + \hat{j}) N$

The correct answer is: $1.67 cross times 10 to the power of negative 9 end exponent left parenthesis i with hat on top plus j with hat on top right parenthesis N$

Let particle A lies at origin, particle B and C on y and x -axis respectively

$table attributes columnalign left end attributes row cell F with rightwards arrow on top subscript A C end subscript equals fraction numerator G m subscript A m subscript B over denominator r subscript A B end subscript superscript 2 end fraction i with hat on top end cell row cell equals fraction numerator 6.67 cross times 10 to the power of negative 11 end exponent cross times 1 cross times 1 over denominator left parenthesis 0.2 right parenthesis squared end fraction i with hat on top equals 1.67 cross times 10 to the power of negative 9 end exponent i with hat on top N end cell end table$
Similarly $F with rightwards arrow on top subscript A B end subscript equals 1.67 cross times 10 to the power of negative 9 end exponent j with hat on top N$
$therefore$ Net force on particle A
$F with rightwards arrow on top equals F with rightwards arrow on top subscript A C end subscript plus F with rightwards arrow on top subscript A B end subscript equals 1.67 cross times 10 to the power of negative 9 end exponent left parenthesis i with hat on top plus j with hat on top right parenthesis N$

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