if a weight of 20 m is hung with meter rod at 20cm.Find weight of meter rod which is balanced by wedge at 40cm?
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Answer:Given:- w1=20kgf. l1=20. w2=? *l2=(50cm-40cm)=10cm
*w1×l1=w2×l2
=20×20=w2×10
=w2=20×20/10
=w2=40kgf
Thus, the weight of the Rod is 40kgf.
Explanation:
w1=weight on 1st side.
w2=weight on 2nd side.
l1=distance between pivot and point application of load on 1st side.
l2=distance between pivot and point application of load on 2nd side.
*l2 in the given question is the distance between the pivot (at 40cm mark) and point of application of load i.e. weight of the Rod which acts at its centre i.e at 50cm mark.
FORMULA for EQUILIBRIUM of Moment of Forces:- *w1×l1=w2×l2
*Weight of any UNIFORM Metre Rule acts at its centre i.e. at 50 cm mark.*
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