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You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If 62-kg Dan sits on the left end of the board and 50-kg Tahreen on the right end of the board, where should 54-kg Komila sit to keep the board stable? Ignore the mass of the board and treat each student as point-like objects. The positive x-direction is to the right, and the origin is at the center of the chair.

Respuesta :

Answer:

Komila should sit 0.33m from the middle of the board towards tahreen.

Explanation:

We are told to treat each student as point-like objects. So i have attached a rigid body diagram to depict this.

From the diagram,

F_d is force exerted by dan

F_t is force exerted by tahreen

F_k is force exerted by komila

F_b is force of board at the mid point.

x1 is distance of dan from the centre of the chair

x2 is distance of komila from the centre of the chair

x3 is distance of tahreen from komila

We are given;

Mass of Dan;m_d = 62 kg

Mass of tahreen;m_t = 50 kg

Mass of komila;m_k = 54 kg

Now, taking moments about the centre of the chair, we have;

(F_d*x1) - (F_k*x2) - (F_t(x2 + x3)) = 0

Now,F_d = m_d*g ; F_t = m_t*g ; F_k = m_k*g

We are told that the board is 3m long. So, if we assume that the fulcrum position of the chair coincides with the midpoint of boards length, we'll have;

x1 = (x2 + x3) = 1.5

Thus, we now have;

(F_d*1.5) - (F_k*x2) - (F_t*1.5) = 0

F_d = m_d*g = 62 * 9.8 = 607.6 N

F_t = m_t * g = 50 x 9.8 = 490 N

F_k = m_k * g = 54 x 9.8 = 529.2 N

So plugging in these values, we have;

(607.6 * 1.5) - (529.2 * x2) - (490 * 1.5) = 0

911.4 - 735 = 529.2 x2

529.2 x2 = 176.4

x2 = 176.4/529.2

x2 = 0.33m

Komila should sit 0.24m from the middle of the board towards tahreen

Ver imagen AFOKE88

Komila should sit 0.33m on the left to the middle of the board to keep the board stable.

Given information:

Length of the board, l = 3 m

Mass of Dan [tex]m_d[/tex] = 62 kg

Mass of Tahreen [tex]m_t[/tex]= 50 kg

Mass of Komila,[tex]m_k[/tex] = 54 kg

Balancing the torque:

The torque acting on each person must be balanced so:

[tex](F_dx_1) - (F_kx_2) - (F_t(x_2 + x_3)) = 0[/tex]

[tex]F_d[/tex] is the force exerted by Dan

[tex]F_t[/tex] is the force exerted by Tahreen

[tex]F_k[/tex] is the force exerted by Komila

[tex]x_1[/tex] is the distance of Dan from the center of the chair

[tex]x_2[/tex] is the distance between Komila and the center of the chair

[tex]x_3[/tex] is the distance between Tahreen and Komila

Now

[tex]F_d = m_dg \\\\ F_t = m_tg \\\\F_k = m_kg[/tex]

The board is 3m long. So, if we assume that the position of the chair coincides with the midpoint of the board's length, then:

[tex]x_1 = (x_2 + x_3) = 1.5[/tex]

Thus,

[tex](F_d\times1.5) - (F_kx_2) - (F_t\times1.5) = 0\\\\[/tex]

[tex]( 62 \times 9.8\times1.5) - (50\times9.8\times x_2) - (54\times9.8\times1.5) = 0\\\\(607.6 \times1.5) - (529.2 \timesx2) - (490 \times 1.5) = 0\\\\911.4 - 735 = 529.2 x_2\\\\529.2 x_2 = 176.4\\\\x_2 = 176.4/529.2\\\\x_2 = 0.33m[/tex]

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