The magnitude of the first force is $F_{1} = 72 mathrm{~N}$. The wheel’s radius is $r = 0.78 mathrm{~m}$. The angle that the first force makes with the radius is $theta_{1} = 50^{circ}$. The angle that the second force makes with the radius is $theta_{2} = 90^{circ}$. The net torque is equal to zero $tau_{net} = 0 mathrm{~N cdot m}$.

$textbf{Required: }$

Finding the magnitude of the second force $vec{F_{2}}$ that makes the net torque on the wheel equals zero.

Since the only force that acts on the meterstick is the weight. Then the direction of the net force is downward. As the textbook mentions that the torque is given by the product of force and distance

$$
begin{align*}
tau &= r ~ F \
&=r ~ F ~ sin left( theta right) \
end{align*}
$$

According to superposition principle, the net torque is given by

$$
begin{align*}
tau_{net} &= tau_{1} +tau_{2} \
&= r_{1} ~ F_{1} ~ sin left( theta_{1} right) – r_{2} ~ F_{2} ~ sin left( theta_{2} right) \
&= 0
end{align*}
$$

Rearrange and solve for the magnitude of the second force:

$$
begin{align*}
F_{2} &= dfrac{ r_{1} ~ F_{1} ~ sin left( theta_{1} right) }{ r_{2} ~ sin left( theta_{2} right) } \
&= dfrac{ F_{1} ~ sin left( theta_{1} right) }{ sin left( theta_{2} right) } \
&= dfrac{ 72 mathrm{~N} times sin left( 50^{circ}right) }{ sin left( 90^{circ} right) } \
&= 55.155 mathrm{~N}
end{align*}
$$

So, the magnitude of the second force is $55.155 mathrm{~N}$.

The applied torque is $tau = 15 mathrm{~N cdot m}$. The length of the wrench is $r = 25 mathrm{~cm}$.

$textbf{Required: }$

Finding the necessary force to desire the torque.

Since the only force that acts on the meterstick is the weight. Then the direction of the net force is downward. As the textbook mentions that the torque is given by the product of force and distance

$$
begin{align*}
tau &= r ~ F \
&=r ~ F ~ sin left( theta right) \
&= 0.41 mathrm{~m} times 8.8 mathrm{~N} times sin left( 22^{circ} right) \
&= 1.352 mathrm{~N cdot m}
end{align*}
$$

Rearrange and solve for the acting force:

$$
begin{align*}
F &= dfrac{ tau }{ r } \
&= dfrac{ 15 mathrm{~N cdot m} }{ 0.25 mathrm{~m} } \
&= 60 mathrm{~N}
end{align*}
$$

So, the necessary force to desire the torque is $60 mathrm{~N}$.