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?
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
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
This is not my question
To answer these questions, we can use the concept of the lever principle. The lever principle states that for a lever in equilibrium, the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
1. To find the mass of the rule, we need to determine the mass of the rule's other half. Since the rule is balanced horizontally, the clockwise and anticlockwise moments must be equal in magnitude.
The clockwise moment is given by the formula: Clockwise moment = Distance x Force
In this case, the 22.5g mass at end A exerts a force on the rule. The distance from the knife edge to end A is 20cm or 0.2m.
So, the clockwise moment is: Clockwise moment = 0.2m x 0.0225kg (converting grams to kilograms)
Clockwise moment = 0.0045 kgm
To balance the rule, the other half of the rule must have the same mass and exert an equal but opposite moment. Therefore, the mass of the rule is also 0.0225 kg or 22.5 grams.
2. The force exerted on the rule by the knife edge can be calculated by taking the moments about the edge of the rule. The clockwise and anticlockwise moments must again be equal in magnitude.
Since the mass of the rule is 0.0225kg (or 22.5g), the force exerted by the knife edge can be calculated using the formula: Force = Mass x Acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2.
So, the force exerted on the rule by the knife edge is: Force = 0.0225kg x 9.8 m/s^2
Force = 0.2205 Newtons (N)
Therefore, the force exerted on the rule by the knife edge is approximately 0.2205 N.
1000g_1kg
22.5g_x
x_22.5g^1kg/1000g
m_0.0225kg