assume the mass of the rule is at the 50 cm mark.
massrule*(50-20)=22.5g*20
solve for massrule
force on knife edge=20+massrule
A uniform half metre rule is balanced horizontally across a knife edge placed 20cm from A. A mass of 22.5g is hung from end A.
1. What is the mass of the rule
2. What is the force exerted on the rule by the knife edge?
4 answers
1000g_1kg
22.5g_x
x_22.5g^1kg/1000g
m_0.0225kg
22.5g_x
x_22.5g^1kg/1000g
m_0.0225kg
half of meter rules is 50 cm
from point A to pivot = 20 cm and rest part is 30 cm
by using law of lever we get
20 cm*22.5 g= 30 cm * mass of the rule
mass of the rule = 15 g
b) force = to the weight then F= 15*9.81
from point A to pivot = 20 cm and rest part is 30 cm
by using law of lever we get
20 cm*22.5 g= 30 cm * mass of the rule
mass of the rule = 15 g
b) force = to the weight then F= 15*9.81
This is not my question