A bob weighing ` 0.06 kg` hangs vertcally at the end of a sting ` 0.30 m` long. Find the force which when applien horizontally, will pull the bob ` 0. 10 m` aside from its initial psition.

Explanation

In the Fig. 2 (c ) . 8 1 , F= ? CA=0. 10 m, ltbRgt SO = SA = 0. 30 m`.
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:. ` SC= (SA^2 - CA^2 )^(1//2) = [ ( 0.3)^2 - ( 0.1)^2 ] ^(1//2)`
` = 2sqrt 2 // 10 m` ltbr As, ` ( mg)/ SC) = F/ (VA) = T/(AS)`
F= mg (CA)/(AC) = 0.06 xx 9.8 xx (0.1)/ ( 2 sqrt 2) xx 10 N`
` =0.06 xx 1//2 sqrt 2 kgf = 0/021 kg f`.