From the equilibrium condition of torques:

$\overline{){{\mathbf{m}}}_{{\mathbf{1}}}{\mathbf{g}}{\mathbf{(}}\frac{\mathbf{L}}{\mathbf{2}}{\mathbf{-}}{{\mathbf{x}}}_{\mathbf{o}\mathbf{f}\mathbf{f}\mathbf{s}\mathbf{e}\mathbf{t}}{\mathbf{+}}{\mathbf{d}}{\mathbf{)}}{\mathbf{+}}{\mathbf{M}}{\mathbf{g}}{\mathbf{\left(}}{\mathbf{d}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbf{m}}}_{{\mathbf{2}}}{\mathbf{g}}{\mathbf{(}}\frac{\mathbf{L}}{\mathbf{2}}{\mathbf{-}}{{\mathbf{x}}}_{\mathbf{o}\mathbf{f}\mathbf{f}\mathbf{s}\mathbf{e}\mathbf{t}}{\mathbf{-}}{\mathbf{d}}{\mathbf{)}}}$

The magnitude of the force exerted at the pivot point is expressed as:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{(}}{{\mathbf{m}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{m}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{M}}{\mathbf{)}}{\mathbf{g}}}$

You have been hired to design a family-friendly see-saw. Your design will feature a uniform board (mass M= 11 kg: length L = 3.2 m) that can be moved so that the pivot is a distance d from the center of the board. This will allow riders to achieve static equilibrium even if they are of different mass, as most people are. You have decided that each rider will be positioned so that his/her center of mass will be a distance x_{offset} = 19 cm from the end of the board when seated as shown. You have selected a child of mass m_{1} = 23 kg (shown on the right), and an adult of mass m_{2} = 87 kg (shown on the left) to test out your prototype. Determine the distance d in meters. Determine the magnitude of the force exerted on the pivot point by the see-saw while in use in newtons.

Part (a) Determine the distance d in meters

Part (b) Determine the magnitude of the force exerted on the pivot point by the see-saw while in use in newtons

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