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Two children sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the first kid has a mass of 36 kg and sits at a distance of 1.2 meters from the pivot, where should the second kid of mass 24 kg sit to balance the see-saw?

Answer :

cjmejiab

To solve the problem it is necessary to apply the concepts of Torque and equilibrium.

The Torque is defined as:

[tex]\tau= F*d[/tex]

Where

F= Force

d = Distance

In this particular case, the force is caused by the weight of both children.

In turn, when there is equilibrium, the two torques must be equal therefore

[tex]\tau_1 = \tau_2[/tex]

[tex]m_1gd_1 = m_2_gd_2[/tex]

[tex]m_1d_1 = m_2d_2[/tex]

Replacing with our values

[tex](36)(1.2)=(24)d_2[/tex]

Re-arrange to find [tex]d_2[/tex]

[tex]d_2 = \frac{(36)(1.2)}{24}[/tex]

[tex]d_2 = 1.8m[/tex]

Therefore the distance that the second kid should sit to balance the see-saw is 1.8m from the pivot.

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