Question

Question:

A person of mass 45 kg sits at one end of a sea saw of length 4 m. Where should another person of 60 kg sit on the other side in order to balance it?
[Gravity=9.8]

◇Do it with with full explanation and in detail.
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Answer 1

Answer:

Let the equilibrium is at a distance a meter from boy 

=>it is (4 - a) meter from the 45 kg person 

So m1r1=m2r2

=>a x 15 = (4 - a) x 45 

=>15a = 180 - 45a 

=>60a = 180 

=>a = 180/60 = 3 meter

Therefore the centre of mass of body will be situated 3m away from the boy

Step-by-step explanation:

Let the equilibrium is at a distance a meter from boy 

=>it is (4 - a) meter from the 45 kg person 

So m1r1=m2r2

=>a x 15 = (4 - a) x 45 

=>15a = 180 - 45a 

=>60a = 180 

=>a = 180/60 = 3 meter

Therefore the centre of mass of body will be situated 3m away from the boy

Answer 2

Answer:

The centre of mass of body will be situated 3m away from the boy

Step-by-step explanation

it's (4 - a) meter from the 45 kg person

So m1r1=m2r2

$= a x 15 = (4 - a) x 45 \\ </p><p>=15a = 180 - 45a \\ </p><p>= 60a = 180 \\ </p><p>= a = \frac{180}{60} </p><p>= 3 meter$