Solve all the questions in good English should the step and answer it should be very explicit including the experiment . Answer All the questions and give examples and explanation where possible

A uniform metre rule is 100 cm long. When pivoted at its centre, the pivot is at the 50 cm mark.

• Mass 1 ($m_1 = 50 \, \text{g}$) is at the 20 cm mark. Its distance from the pivot is $d_1 = 50 \, \text{cm} - 20 \, \text{cm} = 30 \, \text{cm}$. This mass will create an anticlockwise moment.
• Mass 2 ($m_2 = 100 \, \text{g}$) is at the 80 cm mark. Its distance from the pivot is $d_2 = 80 \, \text{cm} - 50 \, \text{cm} = 30 \, \text{cm}$. This mass will create a clockwise moment.

We will use the acceleration due to gravity $g = 9.8 \, \text{m/s}^2$.

• $m_1 = 50 \, \text{g} = 0.050 \, \text{kg}$

• $d_1 = 30 \, \text{cm} = 0.30 \, \text{m}$

• Force 1 ($F_1$) = $m_1 g = 0.050 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 0.49 \, \text{N}$

• $m_2 = 100 \, \text{g} = 0.100 \, \text{kg}$

• $d_2 = 30 \, \text{cm} = 0.30 \, \text{m}$

• Force 2 ($F_2$) = $m_2 g = 0.100 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 0.98 \, \text{N}$

• Anticlockwise moment ($M_{ACW}$) due to $F_1$: $$M_{ACW} = F_1 \times d_1 = 0.49 \, \text{N} \times 0.30 \, \text{m} = 0.147 \, \text{N} \cdot \text{m}$$
• Clockwise moment ($M_{CW}$) due to $F_2$: $$M_{CW} = F_2 \times d_2 = 0.98 \, \text{N} \times 0.30 \, \text{m} = 0.294 \, \text{N} \cdot \text{m}$$

The resultant moment is the difference between the clockwise and anticlockwise moments.

Since the clockwise moment is greater, the resultant moment is in the clockwise direction.

The resultant moment about the pivot is $\boxed{\text{0.147 N} \cdot \text{m clockwise}}$.

3 de faites, encore 2 aujourd'hui. Tu progresses.