A uniform metre rule is pivoted at its mid point .A weight of 50 gf Is suspended at one end of it. Where should a weight of 100 gf be suspended to keep the rule horizontal.
Answers
According to the law of moments, total anticlockwise moments is equal to the total clockwise moments.
Hence, if we consider the 50 gf weight to be placed on the left end, then the clockwise moment will be equal to perpendicular distance between pivot and point of application of force,i.e., 50 cm multiplied by the magnitude of force ,i.e., 50 gf which will be equal to 2500 gf.
Now the anticlockwise moment should also be equal to 2500 gf. Let the 100 gf weight be placed at a distance x metres from the central pivot. Therefore , 100 gf * x metres = 2500 gf metres. After calculation , the value of x will be 25 metres from the centre
As we have one meter rule pivoted at midpoint there will be a perpendicular distance of 50 cm from the contact of force of 50gf.
Here we have moment of force of r×F.
After substituting we get moment of force=50×50gfcm.
Here this must be balanced by 100gf's moment of force.
Let us think that there would be x perpendicular distance.
On equating we get
50×50gfcm=100×xgfcm
x=50×50/100=25cm.
Hence 100gf must be applied at a distance of 25cm.