below shows a uniform metre rule AB pivoted at its end A at the zero mark and supported at the other end B by a spring balance when a weight of 40kgf is suspended at its 40cm mark. This rule stays horizontal . Find the reading of the spring balance when the rule is of (i) negligible mass (ii) mass 20kg.

Explanation

When the rule is of negligible mass
IN the absence of support at the end B by the spring balance, the rule will turn clockwise about the pivot A due to weight 40 kgf at the 40cm mark. To keep the rule in equilibrium (i.e., horizontal) a force F (say) is needed upwards at the end B as shown in Fig. 1.24 which is provided by the spring balance. So the reading of the spring balance will be F.

In equilibrium as shown in fig 1.23
clockwise moment about the point A
= ANticlockwise moment about the point A
or `40 kgf times 40cm= F times 100cm`
`therefore F= (40 times 40)/100 kgf= 16 kgf`
Thus the reading of spring balance will be 16kgf.
(ii) When the rule is of mass 20kg i.e., weight 20kgf.
The weight 20 kgf of the rule will act at the 50cm mark, since the metre rule is uniform. As shown in FIg.1.25, both the weight 40kgf and the weight of rule 20kgf produce clockwise moments about the point A, so a force F is needed upwards at the end B to keep the rule horizontal.

In equilibrium as shown in Fig.1.23,
Total clockwise moment about the point A
= Anticlockwise moment about the point A
or `40kgf times 40cm +20kgf times 50cm`
`=F times 100cm`
or `F= ((40 times 40)+(20 times 50))/100 kgf=26 kgf`
Thus the reading of spring balance will be 26 kgf.