A uniform half meter rule AB is balanced horizontally on a knife edge placed 5cm from B with a mass of 80g at B .Find the mass of the rule
Answers
Mass of the ruler = 100g = 0.1 kg
Length of the ruler l = 1m
Moment of the unknown mass of the fulcrum is equal to mg(0.2)
Moment of the mass of the ruler of the fulcrum is equal to (0.1)g(0.1)
Because the system is in equilibrium so mg(0.2) = (0.1)g(0.1)
=> m = 0.05kg = 50g
The 50g mass is moved to mark 10cm
The moment of the fulcrum because of the above stated mass is (0.05)(g)(0.3)
= 0.147 Nm
The movement of the ruler about the fulcrum is (0.1)g(0.1) = 0.098 Nm
The ruler will tilt towards the mass placed at 10cm mark
To make the system equal let another mass 50g or 0.05kg be placed
at xm from the fulcrum
Moment of the 50g mass at 10 cm mark is equal to moment of the ruler about the fulcrum + moment of the new 50g mass about the fulcrum
=>0.147 = 0.098 + (0.05)(g)(x)
=> 0.049 = 0.49x
=> x = 0.1m = 10cm from the fulcrum
Thus, the new 50g is placed at 50cm mark of the ruler
PLS MARK BRAINLIEST