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Two people of different masses sit on a seesaw. M1, the mass of person 1, is 91 kg, M2 is 45 kg, d1 = 0.8 m, and d2 = 1.1 m. The mass of the board is negligible.Calculate the magnitude of the torque about location A due to the gravitational force on person 2.What is the direction of the rotation due to this torque?Since at this instant the linear momentum of the system may be changing, we don't know the magnitude of the "normal" force exerted by the pivot. Nonetheless, it is possible to calculate the torque due to this force. What is the magnitude of the torque about location A due to the force exerted by the pivot on the board?What is the direction of this torque?Person 2 moves to a new position, in which the magnitude of the net torque about location A is now 0, and the seesaw is balanced. What is the new value of d2 in this situation?

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MechEngineer

Answer:

1/ 485.6 Nm

2/ Clock-wise

3/ 0 and no direction

4/ 1.62 m

Explanation:

1/ Torque is gravity force times the moment arm. Let's g = 9.81m/s2

[tex]T_2 = m_2*g*d_2 = 45*9.81*1.1 = 485.6 Nm[/tex]

2/ If person 2 is sitting on the right, the direction of this torque is clock-wise, since the gravity is acting downward.

3/ Assuming that location A is right at the pivot point, then the torque generated by this torque is 0, since the moment arm is 0. This has no direction.

4/ The seasaw is balanced, this means torques generated by 2 people are equal and in opposite direction

[tex]T_1 = T_2[/tex]

[tex]m_1gd_1 = m_2g_d_2[/tex]

[tex]m_1d_1 = m_2d_2[/tex]

[tex]d_2 = \frac{m_1d_1}{m_2} = \frac{91*0.8}{45} = 1.62m[/tex]

So the new location for d2 is 1.62m

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