Question

Two children are balanced on a seesaw that has a mass of 15.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m from the pivot. The center of mass of the seesaw is 0.213 m from the pivot (on the side of the first child).

(a) If the second child has a mass of 30.4 kg, how far (in m) is she from the pivot?(b) What is the supporting force (in N) exerted by the pivot?

Answer 1

The distance of the second child from the pivot is 1.47m and the support at the pivot is 699.72N. The calculation steps can be found in the attachment below. Thank you for reading.

Step-by-step explanation:

Answer 2

The three external forces acting on the system are the weights of the two children and the supporting force of the pivot.

Let us examine the torque produced by each. Torque is defined as

`\tau = r Fsin\theta`

Here
`\theta = 90\°`, so
`sin\theta = 1` for all three forces.

The torque exerted by three forces are

`\tau_1 = r_1w_1`

`\tau_2 = r_2 w_2`

`\tau_s = r_p w_s`

Now, the condition for equilibrium is that the sum of the torques on both children is zero, therefore

`\tau_s \tau_1 = \tau_2`

`r_sw_s + r_1 w_1 = r_2 w_2`

Weight (w) is mass times the acceleration due to gravity. Then

`r_sm_s g + r_1m_1 g = r_2m_2 g`

`r_2 = ((r_s+m_s)+(r_1m_1))/(m_2)`

Replacing we have,

`r_2 = ((0.213*15)+((1.6)(26)))/(30.4)`

`r_2 = 1.47m`

As expected, the heavier child must sit closer to the pivot to balance the seesaw.

PART B) We start considering the equilibrium in the system, then

`\sum F = 0`

`F_p w_1+w_2+w_3`

`F_p = m_1g+m_2g+m_sg`

`F_p = (26)(9.8)+(30.4)(9.8)+15*9.8`

`F_p = 699,72N`

The supporting force exerted by the pivot is almost 700N