A simple weighing machine is made up of a uniform bar 120cm long and mass 10kg is pivoted 30cm from one end . Find the mass required to balance 320kg mass suspended at the end of the short is the required mass is suspended at the end of the long arm
Let's assume the mass required to balance the 320 kg mass is "x" kg.
When a mass is suspended at the end of the short arm (30 cm from the pivot), we can use the principle of moments to find the mass required to balance it.
The moment of the 10 kg bar about the pivot point (30 cm from one end) is equal to the moment of the 320 kg mass about the same pivot point.
Moment of the bar = Moment of the 320 kg mass
(10 kg * g * 90 cm) = (320 kg * g * 120 cm)
900 g = 38400 g
g = 42.67 N/kg (acceleration due to gravity)
Therefore, the mass required to balance the 320 kg mass is:
x * g * 30 cm = 320 kg * g * 120 cm
x * 42.67 N/kg * 30 cm = 320 kg * 42.67 N/kg * 120 cm
x = (320 * 120) / 30
x = 1280 kg
So, the mass required to balance the 320 kg mass when suspended at the end of the short arm is 1280 kg.