A father and son improvise a seesaw out of a plank and a sawhorse. The plank is uniform, has a mass one-fourth that of the father, and is 6 m long, with the sawhorse placed at the center of mass of the plank. If the son sits at the end of the plank, 3 m from the sawhorse, the father finds that to achieve balance, he has to sit 2 m from his end of the plank. What is their mass ratio Mf Ms

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oeerivona oeerivona · Answer 1 · 2020-04-10T04:15:07+02:00

Answer:

Their mass ratio, [tex]\frac{M_f}{M_s} \hspace{0.09cm}is \hspace{0.09cm}3[/tex]

Explanation:

Here we have the principle of conservation of moment

Let the mass of the father = Mf and

The mass of the son = Ms and

The mass of the plank = 1/4 Mf

With the mass of the plank being uniform, the center of mass of the plank is at the center of the plank

Taking moment about the center of the plnk we have

Sum of moment about the y axis = 0

3 m × Ms + (-1 m) × Mf = 0

∴ 3 m × Ms = (1 m) × Mf

Therefore;

[tex]\frac{M_f}{M_s} = \frac{3\hspace{0.09cm} m}{1\hspace{0.09cm} m} =3[/tex]

AFOKE88 AFOKE88 · Answer 2 · 2021-11-24T13:08:25+01:00

The mass ratio between the father and son is;

Mf/Ms = 3

Let us denote as follows;

The mass of the father; Mf

The mass of the son; Ms

Mass of plank = ¼Mf

Also;

Length of plank; L = 6 m

The son sits at 3m from the Sawhorse. Thus;

x_s = 3 m

Since the father has to sit 2 m from the end of the plank, if we take moments about the center, we have;

Mf(1) = Ms(3)

Thus;

Mf/Ms = 3

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