Answer:
The mass is [tex]M_n = 0.047kg[/tex]
Explanation:
From the question we are told that
The mass of the meter stick is [tex]M_c = 0.21kg[/tex]
The distance by which the balance point moves [tex]d = 14.0 \ cm[/tex]
Before the movement of the balance point was at the middle of the meter stick i.e at the 50 cm mark
Mathematically the weight of the meter stick would be
[tex]W = M_c g[/tex]
where g is acceleration due to gravity with a value of [tex]9.8m/s^2[/tex]
[tex]W = 0.12 * 9.8[/tex]
[tex]=1.176N[/tex]
Now when the small chain is suspended from one end and the system reach equilibrium the net torque is zero and this implies that
The moment of force of the meter stick = moment of force of the necklace
[tex]W * 14 = W_n * (50 - 14)[/tex]
Where [tex]W_n[/tex] is the weight of the necklace
Substituting values and making [tex]W_n[/tex] the subject we have
[tex]W_n = \frac{1.176*14}{36}[/tex]
[tex]=0.457N[/tex]
Mathematically the weight of the chain is
[tex]W_n = M_n * g[/tex]
Substituting values and making [tex]M_n[/tex] the subject
[tex]M_n = \frac{W_n}{g}[/tex]
[tex]=\frac{0.457}{9.8}[/tex]
[tex]M_n = 0.047kg[/tex]
Answer:
0.082 kg
Explanation:
The center of the meter stick = 50 cm.
When a small chain is place at one end, the balance point is moved 14 cm towards the end with the chain.
The new balance point = 50-14 = 36 cm
Using the principle of moment,
Sum of clockwise moment = sum of anti clockwise moment.
W(36-0) = W'(50-36).................. Equation 1
Where W = weight of the chain, W' = weight of the meter stick
36W = 14W'
make W the subject of the equation
W = 14W'/36.......................... Equation 2
Given: W' = m'g, where m' = mass of the meter stick, m' = 0.21 kg
W' = 0.21(9.8) = 2.058 N
Substitute into equation 2
W = 14(2.058)/36
W = 0.8 N
But
W = mg,
Where m = mass of the chain.
m = W/g
m = 0.8/9.8
m = 0.082 kg.
Hence the mass of the chain = 0.082 kg
