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A 0.106 kg meter stick is supported at its 40 cm mark by a string attached to the ceiling. A 0.627 kg object hangs vertically from the 6.05 cm mark. A second mass is attached at another mark to keep it horizontal and in rotational and translational equilibrium. If the tension in the string attached to the ceiling is 22.3 N, find the value of the second mass. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of kg

Respuesta :

Explanation:

The given data is as follows.

          Mass of the stick, (M) = 0.106 kg at 0.4 m

               [tex]m_{1}[/tex] = 0.627 kg (at 0.605 m)

          Tension, (T) = 22.3 N

              [tex]m_{2}[/tex] = ?

Therefore, for equilibrium,

         T = [tex]m_{1}g + Mg + m_{2}g[/tex]

        22.3 N = [tex](0.627 + 0.106 + m_{2}) \times 9.8[/tex]

            22.3 N = 7.183 + [tex]9.8m_{2}[/tex]

               [tex]m_{2}[/tex] = 1.54 kg

Thus, we can conclude that value of the second mass is 1.54 kg.