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Second Law of Motion

Two individua...

Question

Two individual forces of magnitude $F_{1}$ and $F_{2}$ act on a body of mass 1 kg as shown in the figures (i) and (i). If product of net magnitudes of forces produced in fig (i) and (ii) is 27 N then find the magnitude of forces $F_{1}$ and $F_{2}$ . (assume $F_{1}$ > $F_{2}$ )

A

3 N, 6 N

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B

6 N, 6 N

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C

6 N, 3 N

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D

3 N, 3 N

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Solution

The correct option is D 6 N, 3 N
From figure 1, net force F' $= F_{1} - F_{2}$
From figure 2 net force F" $= F_{1} + F_{2}$
Given $(F_{1} - F_{2}) (F_{1} + F_{2}) = 27 N$
$F_{1}^{2} - F_{2}^{2} = 27$
Factors of F $=$ 27 are 1, 2, 7, 3, 9
$\Rightarrow$ If $F_{1} + F_{2} = 9 N$
$F_{1} - F_{2} = 3 N$
$∴ F_{1} = 6 N$ and $F_{2} = 3 N$

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