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ICSE Class 10 Answered

A uniform half-meter beam has weights of 20gf and 30 gf suspended fromthe two ends respectly. where should the pivot be placed so to produce  a net anticlockwise moment of 10gf-cm ?
Asked by achintya.roy | 06 May, 2019, 11:09: PM
answered-by-expert Expert Answer
Anticlockwise moment = 20×(25+x) gf-cm
Clockwise moment = 30×(25-x) gf-cm

If net anticlockwise moment 10gf-cm is required, then  20×(25+x) - 30×(25-x) = 10  .......(1)

Solving eqn.(1), we get  x = 5.2 cm

Hence pivoting is done at a distance 30.2 cm from the point of suspension of 20g force
Answered by Thiyagarajan K | 07 May, 2019, 02:40: PM

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