Respuesta :
Answer:
The mass of the chain is 75.7 grams
Explanation:
Let the mass of chain be M
Mass of stick = m = 0.24 kg
We are given that When a small chain is suspended from one end, the balance point moves 12.0 cm toward the end with the chain
x= 12 cm = 0.12 m
Length of stick L = 1m
Balancing the moment of force about the pivot we have
[tex]mgx-M(\frac{L}{2}-x)g=0\\0.24 \times 9.8 \times 0.12=M(\frac{1}{2}-0.12)(9.8)\\\frac{0.24 \times 9.8 \times 0.12}{(\frac{1}{2}-0.12)(9.8)}=M[/tex]
0.0757=M
75.7 grams=M
Hence the mass of the chain is 75.7 grams
When balanced, the clockwise moment about the balance point equals the
anticlockwise moment.
- The mass of the chain is approximately 75.789 g
Reasons:
The mass of the meter stick, [tex]m_s[/tex] = 0.24 kg
Change in the position of the balance point = 12.0 cm towards the end with the chain
Required:
The mass of the chain
Solution:
Initial position of balance point = 50 cm from each end
New position of balance point from the end with the chain = 50 - 12 = 38
Therefore;
New position of balance point from the end with the chain, [tex]d_{chain}[/tex] = 38 cm
Position where the mass of the stick acts = The middle (50 cm from each end)
Length from the center of mass of the stick to the balance point, [tex]d_s[/tex] = 12 cm
When the meter stick is balanced, we have;
[tex]m_s[/tex] × [tex]d_s[/tex] = [tex]m_{chain}[/tex] × [tex]d_{chain}[/tex]
Where;
[tex]m_{chain}[/tex] = The mass of the chain
Which gives;
0.24 kg × 12 cm = [tex]m_{chain}[/tex] × 38 cm
[tex]m_{chain} = \dfrac{0.24 \, kg \times 12.0 \, cm}{38 \, cm} = \dfrac{36}{475} \, kg \approx 0.075789 \, kg[/tex]
- The mass of the chain, [tex]m_{chain}[/tex] ≈ 0.075789 kg = 75.789 g
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