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a meter stick is hung from a string tied at the 25cm mark. A 0.50-kg object is hung From the zero end of the meter stick, and the meterstick is balanced horizontally. What is the mass of the meterstick? (a) 0.25kg, (b) 0.5kg (c) 0.75kg (d) 1.0 kg (e) 2.0kg (f) impossible to determine

Respuesta :

let m₁ = 0.25m and m₂ = 0.75m, where m is the mass of the meterstick:

The sum of all torques must be zero:
τ = 12.5 * m₁ + 25 * 0.5 - 37.5 * m₂ = 0

Replace m₁ and m₂:
12.5 * 0.25m + 25 * 0.5 = 37.5 * 0,75m

Solve for m:
3.125m + 12.5 = 28.125
25m = 12.5
m = 0.5
onyebuchinnaji

The mass of the meter stick that will balance the given mass is 0.5 kg.

The given parameters;

  • mass at the zero mark, m = 0.5 kg

The mass of meter stick concentrate at its center of gravity (50 cm mark)

The sketch of the meter stick is shown below;

      0-----------------------------------25cm------------------------------50 cm

      ↓                                                                                    ↓

      0.5kg                                                                             W

Take moment about the pivot (25 cm mark);

0.5(25) = (50 -25)W

0.5(25) = W(25)

W = 0.5 kg

The mass of the meter stick is 0.5 kg.

Learn more here:https://brainly.com/question/14439374

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