A 500 cm meter rule is pivoted at its middle point. If a weight of 2 N is hanged from the 20 cm point, Calculate the amount of weight required to be applied at the 80 cm mark to keep it in a balanced position.
Answers
Answer:
Here you go,
According to the principle of moments, To keep an object in rotational equilibrium, the sum of anticlockwise moments and clockwise moments acting should be equal. Therefore, the amount of weight to be hanged from the 80 cm mark must be able to generate a clockwise moment equal to the anticlockwise moment generated by the weight hanged on the left side of the meter rule.
Anticlockwise moment :
Length of lever arm = (50 – 20)
= 30 cm
= 0.30 m
Since the length of the lever arm is the distance from its mid-point, where it is balanced
Force applied = 2 N
Anticlockwise moment = lever arm x force applied
= 0.30 x 2 N
= 0.6 Nm
Clockwise moment:
length of lever arm = (80 – 50)
= 30 cm
= 0.30 m
Since the length of the lever arm is the distance from its Force applied.
Let it be ‘F’.
Thus, clockwise moment = F x 0.30
= 0.30 F Nm
Clockwise moment = Anticlockwise moment
0.30 F = 0.6
F = 2 N
A weight of 2 N needs to be hanged from 80 cm point to keep the meter rule balanced.
I hope this helps you
Follow me and mark me as brainliest
Explanation:
500cm=______m,900cm=_____m