Question

A seesaw has an irregularly distributed mass of 50 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 86-kg person sits on the other end.
Locate the center of mass of the seesaw (a distance from the 50-kg person).
Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

$d_{CM} = 0.78 m$

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.

$m_1 g\dfrac{d}{2} + m g d_{CM} = m_2 g \dfrac{d}{2}$

$m_1 d + 2 m d_{CM} = m_2 d$

$d_{CM} = \dfrac{m_2 d-m_1 d}{2 m}$

$d_{CM} = \dfrac{86\times 3 - 60\times 3}{2\times 50}$

$d_{CM} = 0.78 m$

hence, the position of center of mass from the fulcrum is 0.78 m on 60 Kg mass side.