Question

A metre rule is pivoted at 50 cm mark. A force F1=20N acts at 20 cm mark and another force F2=15N acts at 75 cm mark. Calculate the moment of two forces along with their signs.

Answer 1

Answer:Let AB be the uniform metre rule, such that C is its midpoint. The metre rule is pivoted at C, such that AC=50cm and CB=50cm. 50 gf weight is suspended near A.

Let D be the point of suspension of 100 gf weight at a distance of x from C, so that the rule is horizontal.

Left hand side moment,

LHM=50×50=2500gfcm.....(1)

Right hand side moment,

RHM=x×100=100×gfcm.....(2)

To keep the rule horizontal,

LHM=RHM

2500=100x⇒x=25cm.

∴ 100 gf is to be suspended at a distance of 25 cm from mid point (or) 75 cm from the end of the rule.

Explanation:

Answer 2

Answer:

+6 Nm , -3.75 Nm

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