Respuesta :
At the equilibrium or balance point the meter stick stationary and the sum of the moments and forces acting on it is zero
- The mass of the dough is .[tex]\underline {51.\overline{51}}[/tex] grams
Reason:
Known;
Length of the stick = 1 meter
Mass of the stick, m = 0.1 kg
Point the stick will balance when the dough is at one end = 0.33 m
Required:
How much mass does the dough have
Solution:
The meter stick is balanced when the clockwise moment and the anticlockwise moment about the balance point are equal
Let M, represent the mass of the dough, we have;
The center of mass of the meter stick = The 0.5 m point
Taking moment, we have;
- [tex]\mathbf{\sum M_{clockwise} = \sum M_{anticlockwise}}[/tex]
Distance from the center of mas of the ruler to the fulcrum, d₁ = (0.5 - 0.33) m
Distance from the dough to the fulcrum, d₂ = 0.33 m
Clockwise moment, [tex]\mathbf{\sum M_{clockwise} }[/tex] = m × d₁
Anticlockwise moment, [tex]\mathbf{\sum M_{anticlockwise} }[/tex] = M × d₂
Therefore;
(0.5 - 0.33) × 0.1 = M × 0.33
Therefore, we get;
- [tex]M = \dfrac{(0.5 - 0.33) \times 0.1}{0.33} =\dfrac{17}{330} =0. 0\overline{51}[/tex]
The mass the dough has, M = 0.0[tex]\overline{51}[/tex] kg = 51.[tex]\underline{\overline{51}}[/tex] grams
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