A uniform meter rule weight 150gf and 250gf hang from the points A&B of the meter rule such that OA is 40 cm and OB is 20 cm. A distance from O where a 100gf weight should be placed to balance the meter rule.
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pivoted point O = 50cm
weight at A = 150gf
weight at B = 250 gf
OA = 40 cm
OB = 20 cm
According to principle of movements anticlockwise movements = clockwise movements
A × OA = B × BA
150 × 40 = 250 × 20
6000 != 5000
AS anti clockwise movements is greater than clockwise moments so weight of 100 gf should be changed towards right side of pivoted point
let distance of weight of 100gf from pivoted point be x
6000 = 5000 + 100x
100x = 1000
x = 10
So weight of 100gf should be hanged at 60 cm to bring the rule in equilibrium....
PLEASE MARK AS BRAINLIEST
weight at A = 150gf
weight at B = 250 gf
OA = 40 cm
OB = 20 cm
According to principle of movements anticlockwise movements = clockwise movements
A × OA = B × BA
150 × 40 = 250 × 20
6000 != 5000
AS anti clockwise movements is greater than clockwise moments so weight of 100 gf should be changed towards right side of pivoted point
let distance of weight of 100gf from pivoted point be x
6000 = 5000 + 100x
100x = 1000
x = 10
So weight of 100gf should be hanged at 60 cm to bring the rule in equilibrium....
PLEASE MARK AS BRAINLIEST
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