The mass of at the end of the rod is $m_{1} = 0.3 mathrm{~kg}$. The length of the rod is $r_{r} = 0.8 mathrm{~m}$. The mass of at the other end of the rod is $m_{2} = 1.5 mathrm{~kg}$.

$textbf{Required: }$

Finding the distance from the mass $m_{1}$ to the balance point.

As the textbook mentions, the balance point corresponds to zero net torque assigns that

$$
begin{align*}
tau &= m_{1} ~ g ~ r_{1} – m_{2} ~ g ~ r_{2} \
&= 0 \
m_{1} ~ r_{1} &= m_{2} ~ r_{2} \
&= m_{2} ~ left( r_{r} – r_{1} right) \
&= m_{2} ~ r_{r} – m_{2} ~ r_{1}
end{align*}
$$

Rearrange and solve for the value of the mass $m_{1}$:

$$
begin{align*}
r_{1} &= dfrac{ m_{2} ~ r_{r} }{ m_{1} + m_{2} } \
&= dfrac{ 1.5 mathrm{~kg} times 0.8 mathrm{~m} }{ 1 mathrm{~kg} + 1.5 mathrm{~kg} } \
&= 0.48 mathrm{~m}
end{align*}
$$

So, the distance from the mass $m_{1}$ to the balance point is $0.48 mathrm{~m}$.

The mass of the meterstick is $m_{1} = 0.34 mathrm{~kg}$ which is balances at the center. The distance between the balance point and the end of the stick is $r_{1} = 9.5 mathrm{~cm}$.

$textbf{Required: }$

Finding the mass of the necklace.

As the textbook mentions, the balance point corresponds to zero net torque assigns that

$$
begin{align*}
tau &= m_{1} ~ g ~ r_{1} – m_{2} ~ g ~ r_{2} \
&= 0 \
m_{1} ~ r_{1} &= m_{2} ~ r_{2} \
end{align*}
$$

Rearrange and solve for the value of the mass $m_{1}$:

$$
begin{align*}
m_{2} &= dfrac{ m_{1} ~ r_{1} }{ r_{2} } \
&= dfrac{ m_{1} ~ r_{1} }{ left( dfrac{ r_{s}}{2} – r_{1} right) } \
&= dfrac{ 0.34 mathrm{~kg} times 9.5 mathrm{~cm} }{ left( dfrac{100 mathrm{~cm} }{2} – 9.5 mathrm{~cm} right) } \
&= 0.0798 mathrm{~kg}
end{align*}
$$

So, the mass of the necklace is $0.0798 mathrm{~kg}$.

The mass of the weight that attached to the left end is $m_{1} = 0.75 mathrm{~kg}$. The mass of the weight that attached to the right end is $m_{2} =m$. The length of the stick is $r_{s} = 0.9 mathrm{~m}$.

$textbf{Required: }$

(a) Finding if the mass $m$ greater than, less than, or equal to $m_{1}$.

(b) Finding the mass $m$.

Solve the proplem theoretically:

As the axis of rotation of the stick is closer to the mass $m_{1}$ that at the left end as compared to the mass $m$. So, the mass $m$ is less than $m_{1}$.

Solve the problem by the calculations:

As the textbook mentions, the balance point corresponds to zero net torque assigns that

$$
begin{align*}
tau &= m_{1} ~ g ~ r_{1} – m_{2} ~ g ~ r_{2} \
&= 0 \
m_{1} ~ r_{1} &= m_{2} ~ r_{2} \
&= m ~ left( r_{s} – r_{1} right) \
end{align*}
$$

Rearrange and solve for the value of the mass $m_{1}$:

$$
begin{align*}
m &= dfrac{ m_{1} ~ r_{1} }{ r_{2} } \
&= dfrac{ m_{1} ~ r_{1} }{ left( r_{s} – r_{1} right) } \
&= dfrac{ 0.75mathrm{~kg} times 0.25 mathrm{~cm} }{ left( 0.9 mathrm{~cm} – 0.25 mathrm{~cm} right) } \
&= 0.288 mathrm{~kg}
end{align*}
$$

So, the mass $m$ is less than $m_{1}$.

As the textbook mentions, the balance point corresponds to zero net torque assigns that

$$
begin{align*}
tau &= m_{1} ~ g ~ r_{1} – m_{2} ~ g ~ r_{2} \
&= 0 \
m_{1} ~ r_{1} &= m_{2} ~ r_{2} \
&= m ~ left( r_{s} – r_{1} right) \
end{align*}
$$

Rearrange and solve for the value of the mass $m_{1}$:

$$
begin{align*}
m &= dfrac{ m_{1} ~ r_{1} }{ r_{2} } \
&= dfrac{ m_{1} ~ r_{1} }{ left( r_{s} – r_{1} right) } \
&= dfrac{ 0.75mathrm{~kg} times 0.25 mathrm{~cm} }{ left( 0.9 mathrm{~cm} – 0.25 mathrm{~cm} right) } \
&= 0.288 mathrm{~kg}
end{align*}
$$

So, the mass $m =0.288 mathrm{~kg}$.