Three masses are attached to a uniform meter stick, as shown in Figure 12.9. The mass of the meter
stick is 150.0 g and the masses to the left of the fulcrum are m2=50.0g and m2=75.0g. Find the
mass më that balances the system when it is attached at the right end of the stick, and the normal
reaction force at the fulcrum when the system is balanced.
30 cm
+
-
40 cm
30 cm
S
mi
M₂
тз

Respuesta :

At equilibrium, the sum of clockwise and anticlockwise moments about a point is zero

The mass of that balances the system is [tex]\underline {316.\overline 6}[/tex] kg

The normal reaction force at the fulcrum is 5,804.25 N

Rason:

The mass of the stick = 150.0 g

Mass m₁ on the left = 50.0 g, location = 30 cm to the left of m₂

Mass m₂ on the left = 75.0 g, location = 40 cm to the left of the fulcrum

Mass m₃ on the right of the fulcrum. location = 30 cm to the right of the fulcrum

Required:

To find the mass of m₃

Solution:

Taking moment about the fulcrum, we have;

50 × (30 + 40) + 75 × (40) + 150 × 20 = m₃ × 30

9,500 = m₃×30

[tex]m_3 = \dfrac{9,500}{30} = 316. \overline 6[/tex]

The mass of that balances the system when it is attached at the right end of the stick, m₃ = [tex]316.\overline 6[/tex] kg

Normal reaction at the fulcrum = (50 + 75 + 150 +  [tex]316.\overline 6[/tex]) × 9.81 = 5804.25

The normal reaction at the fulcrum is 5,804.25 N

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