Respuesta :
Answer:
[tex]d_{CM} = 0.78 m[/tex]
Explanation:
given,
mass of seesaw,m= 50 kg
length of seesaw, d = 3 m
mass of person on one end, m₁ = 60 Kg
mass of person on another end, m₂ = 86 Kg
position of center of mass = ?
System is in equilibrium
Clockwise moment is equal to anticlockwise moment.
[tex]m_1 g\dfrac{d}{2} + m g d_{CM} = m_2 g \dfrac{d}{2}[/tex]
[tex]m_1 d + 2 m d_{CM} = m_2 d[/tex]
[tex]d_{CM} = \dfrac{m_2 d-m_1 d}{2 m}[/tex]
[tex]d_{CM} = \dfrac{86\times 3 - 60\times 3}{2\times 50}[/tex]
[tex]d_{CM} = 0.78 m[/tex]
hence, the position of center of mass from the fulcrum is 0.78 m on 60 Kg mass side.
The position of the center of mass from the fulcrum is 0.78 m on 60 Kg mass side.
Calculation of the position of the center of mass:
Since
mass of seesaw,m= 50 kg
length of seesaw, d = 3 m
mass of person on one end, m₁ = 60 Kg
mass of person on another end, m₂ = 86 Kg
Also, The system is in equilibrium
Moreover, a Clockwise moment should be equivalent to anticlockwise moment.
So, here the position should be
= 86* 3 -60*3 / 2*50
= 0.78 m
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