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A father and his young son get on a teeter-totter. The son sits 2 m fromthe center, but the father has to sit closer to balance. Where does the father have to sit to balance the teeter-totter if he weighs 4 times as much as his son?

Respuesta :

okpalawalter8

Answer:

The distance of the father from the center is  [tex]d_f = \frac{1}{2} \ m[/tex]

Explanation:

From the question we are told that

    The distance of the son from the center is  [tex]d_s = 2 \ m[/tex]

 

Let the mass of the son be  [tex]m_s[/tex]

     then the mass of the father is  [tex]m_f = 4m_s[/tex]

Now for the teeter-totter to be balanced the torque due to the weight of the father must be equal to the torque due to the weight the son, this is mathematically represented as

         [tex]\tau_s = \tau_f[/tex]

Where  [tex]\tau_s[/tex] is the torque of the son which is mathematically represented as

        [tex]\tau_ s = m_s * d_s * g[/tex]

while [tex]\tau_f[/tex] is the torque of the father which is mathematically represented as

        [tex]\tau_f = m_f * d_f * g[/tex]

=>    [tex]\tau_f = 4 m_s * d_f * g[/tex]

 So

         [tex]4 m_s * d_f * g = m_s * d_s * g[/tex]

substituting values

        [tex]4 * d_f * = 2[/tex]

 =>    [tex]d_f = \frac{1}{2} \ m[/tex]      

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