Question

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?

Answer 1

Answer:

The distance of the father from the center is  $d_f = \frac{1}{2} \ m$

Explanation:

From the question we are told that

    The distance of the son from the center is  $d_s = 2 \ m$

 

Let the mass of the son be  $m_s$

     then the mass of the father is  $m_f = 4m_s$

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

         $\tau_s = \tau_f$

Where  $\tau_s$ is the torque of the son which is mathematically represented as

        $\tau_ s = m_s * d_s * g$

while $\tau_f$ is the torque of the father which is mathematically represented as

        $\tau_f = m_f * d_f * g$

=>    $\tau_f = 4 m_s * d_f * g$

 So

         $4 m_s * d_f * g = m_s * d_s * g$

substituting values

        $4 * d_f * = 2$

 =>    $d_f = \frac{1}{2} \ m$