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2 years ago

Three masses are attached to a uniform meter stick, as shown in Figure 12.9. The mass of the meter

Engineering

1 answer:

sp2606 [1]2 years ago

8 0

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

The mass of that balances the system is $\underline {316.\overline 6}$ 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

$m_3 = \dfrac{9,500}{30} = 316. \overline 6$

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

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

The normal reaction at the fulcrum is 5,804.25 N

Learn more about the moment of a force here:

brainly.com/question/19464450

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