Question
A uniform metre rule of Mass 120g is pivoted at 60cm Mark. At what point on metre rule should a mass of 50g be suspended for it to balance horizontally
Answers
R_scott
the 120 g mass acts at the center of the metre rule (50 cm mark)
... 10 cm from the pivot
the 50 g mass needs to be on the other side of the pivot (> 60 cm mark)
120 g * 10 cm = 50 g * ? cm
... 10 cm from the pivot
the 50 g mass needs to be on the other side of the pivot (> 60 cm mark)
120 g * 10 cm = 50 g * ? cm
Jos kie kiez
To be answered
Wendy
Explain more please
Davy
Thumbs up
Vickie
Anyone are any of you real people
Vickie
Physics
A metre rule is balanced by masses of 24g and 16g suspended from its ends .find the position of the pivot??
A metre rule is balanced by masses of 24g and 16g suspended from its ends .find the position of the pivot??
Samson Opiyo
Yes
Samson Opiyo
help me solving it
Ibrahim
A uniform m metre rule of mass 110g is balance at xcm marks when a mass of 60g is lang at 10cm marks calculate. 1 the balance point x 2 the reaction at turning point