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Consider a system consisting of a uniform 3.0-m long seesaw that has a mass of 40 kg and two people of masses 60 kg and 75 kg sitting on its ends. Find the position of the center of mass of the system relative to the seesaw's midpoint.

Respuesta :

Answer:

[tex]r_{cm} = 0.13 m[/tex]

Explanation:

Let the position of center of the see saw is origin

so we will have

[tex]m_1 = 40 kg[/tex]

[tex]r_1 = 0[/tex]

[tex]m_2 = 60 kg[/tex]

[tex]r_2 = -1.5 m[/tex]

[tex]m_3 = 75 kg[/tex]

[tex]r_3 = + 1.5 m[/tex]

now we will have

[tex]r_{cm} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3}{m_1 + m_ 2 + m_3}[/tex]

[tex]r_{cm} = \frac{40(0) + 60(-1.5) + 75(1.5)}{40 + 60 + 75}[/tex]

[tex]r_{cm} = 0.13 m[/tex]

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