2 (ii) When the rule is of mass 20 kg i.e., weight 20 kgf. The weight 20 kgf of the rule will act at the 50 cm mark, since the metre rule is uniform. As shown in Fig. 1.25, both the weight 40 kgf and the weight of rule 20 kgf produce clockwise moments about the point A, so a force F is needed O ju upwards at the end B to keep the rule horizontal. B AC O - 50 cm - 40 cm 1F 20 kgf OL 40 kgf A 100 cm F O mo Fig. 1.25 In equilibrium, as shown in Fig. 1.23,
Answers
Answer:
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 kg 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
= Anti-clockwise moment about the point A
or 40kgf×40cm=F×100cm
∴F=40×40/100kgf=16kgf
Thus the reading of spring balance will be 16kgf.
(ii) When the rule is of mass 20kg i.e., weight 20kgf.
The weight 20 kg of the rule will act at the 50cm mark, since the meter 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×40cm+20kgf×50cm
=F×100cm
or F=(40×40)+(20×50)/100kgf=26kgf
Thus the reading of spring balance will be 26 kgf.