A uniform plank of a see-saw is 8 m long and is supported in the centre. A boy
weighing 50 kgf sits at a distance of 2.5 m from the fulcrum. Where must another
boy weighing 40 kgf sit, so as to balance the plank?
Answers
Explanation:
50×2.5=40x
125=40x
125/40=x
x=3.125
Boy weighing 40 kg must sit 3.125m from tbe fulcrum to balance the planck.
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Concept:
To balance the see saw at the fulcrum, we must ensure that the centre of mass of the see saw should be at the fulcrum.
Given:
Length of plank of see saw = 8 m
Weight of boy 1 = 50 kg
Distance of boy 1 from fulcrum = 2.5 m
Weight of boy 2 = 40 kg
Centre of mass of see saw at the fulcrum as it is uniform.
Find:
Distance of boy 2 from fulcrum
Solution:
Let's assume fulcrum be at mean position, and mass of the see saw be 'S'.
Let the distance of boy 2 from fulcrum be 'a'.
According to the formula,
Msys × Rcm = ∑mr
where,
Msys = Total mass of the system
Rcm = Distance of centre of mass from mean position
∑mr = Summation of product of all the respective masses with their distance of centre of masses from the mean position.
( 50 + 40 + S ) × 0 = ( 50 × - 2.5 ) + ( 40 × a )
0 = - ( 50 × 2.5 ) + ( 40 × a )
50 × 2.5 = 40 × a
125 / 40 = a
3.125 = a
Hence, the other boy must sit at 3.125 m away from the fulcrum in the opposite direction of the first boy, as we have taken the distance of the first boy as negative and the distance of the second boy came out to be positive.
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