The net torque on the pulley is zero. Remember that $\displaystyle \tau=Fr$, assuming the force acts perpendicular to the radius. Because the pulley is symmetrical in this problem (meaning the r is the same) and the tension throughout the entire rope is the same (meaning F is the same), we know that the counterclockwise torque cancels out the clockwise torque, thus, the net torque is zero.

In the image below, T₁ (due to the platform with the 4 0.5kg weights) = T₂ (the 2kg mass).

Torque is defined by the equation $\displaystyle \tau = Frsin\Theta$. Since both students will exert a downward force perpendicular to the length of the seesaw, $\displaystyle sin\Theta=1$. In our case, force is the force of gravity, given below, and $\displaystyle r$ is the distance from the center of the seesaw.

$\displaystyle F=mg=m(10\frac{m}{s^2})$

Since the torque must be zero in order for the seesaw to stay parallel (not move), the lighter student on the right must make his torque on the right equal to the torque of the student on the left. We can determine the required distance by setting their torques equal to each other.

$\displaystyle F_{1}r_{1} = F_{2}r_{2}$

$\displaystyle (60kg)(10\frac{m}{s^{2}})(3m) = (45kg)(10\frac{m}{s^{2}})r_{2}$

$\displaystyle r_{2} = 4 m$

Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation. As a result, both students moving forward by one meter will cause a nonzero torque on the seesaw. This is because the heavier student's ratio of force and distance will result in less torque on his side than the lighter student.

$\displaystyle \tau=Fd$

$\displaystyle F_1d_1=F_2d_2$

$\displaystyle m_1gd_1=m_2gd_2$

$\displaystyle (60)gd_1=(45)gd_2$

$\displaystyle (60)g(d_1-1)\neq (45)g(d_2-1)$

This a an example of rotational equilibrium involving torque. The formula for torque is $\displaystyle \tau = Fdsin\left ( \Theta \right )$, where $\displaystyle \Theta$ is the angle that the force vector makes with the object in equilibrium and $\displaystyle d$ is the distance from the fulcrum to the point of the force vector. To achieve equilibrium, our torques must be equal.

$\displaystyle F_1d_1sin(\Theta_1)=F_2d_2sin(\Theta_2)$

Since the forces are applied perpendicular to the beam, $\displaystyle sin(\Theta)$ becomes 1. The distance of the fulcrum from the left end is 1m and its distance from the right end is 2m.

$\displaystyle F_1d_1=F_2d_2$

$\displaystyle (50N)(2m)=x(1m)$

$\displaystyle \frac{(50N)(2m)}{1m}=100N=x$

Since the 50N force is twice as far from the fulcrum as the force that must be applied on the left side, it must be half as strong as the force on the left. The force on the left can be found to be 100N.

For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides.

$\displaystyle \tau=Fd=mgd$

The total torque must be equal on both sides in order for the net torque to be zero.

$\displaystyle \tau_1+\tau_2=\tau_3$

Substitute the formula for torque into this equation.

$\displaystyle m_1gd_1+m_2gd_2=m_3gd_3$

Now we can use the given values to solve for the missing mass.

$\displaystyle (21kg)g(4m)+(25.5kg)g(2m)=m_3g(9m)$

The acceleration form gravity cancels from each term.

$\displaystyle (21kg)(4m)+(25.5kg)(2m)=m_3(9m)$

$\displaystyle 84kg\cdot m+51kg\cdot m=m_3(9m)$

$\displaystyle 135kg\cdot m=m_3(9m)$

$\displaystyle m_3=15kg$

This problem deals with torque and equilibrium. Noting that the string is between the two masses we can use the torque equation of $\displaystyle \tau_{ccw} = \tau _{cw}$. We can use the equation $\displaystyle \tau = rFsin\theta$ to find the torque. Since force is perpendicular to the distance we can use the equation $\displaystyle \tau = rF$ (sine of 90^o is 1). Force presented in this situation is gravity, therefore F=mg, and using the variable x as a placement for the string we can find r.

$\displaystyle \tau_{ccw} = \tau _{cw}$

$\displaystyle r_{1} (M_1) = r_{2}(M_2)$

$\displaystyle (x-10)15 = (60-x)30$

x=43, thus the string is placed at the 43cm mark.

In the instant the rope is cut right above the 2kg mass, the pulley now has net torque in the counterclockwise direction because the platform with the masses is still connected to the rope that winds around the pulley. Knowing that $\displaystyle \tau=Fr$, we can determine the torque by knowing that the force (F) on the pulley is due to the tension (T) by the platform. The tension is 19.6N  (equal to the weight) and knowing that the radius of the pulley is 0.25m, we can solve for torque.

$\displaystyle \tau=Fr=(19.6N)(0.25m)=4.9N*m$

Choice 2 is correct.  Think about trying to change a tire…you apply the force on the tire iron as close to perpendicular as possible in order to generate the most force. 

Since the sine of 90 degrees is 1, then you are applying the maximum torque you can generate according to the equation τ = F sin θ.

We are trying to find what force needs to be applied to the rope to result in a net of zero torque on the beam.

Torque applied by the car:

$\displaystyle T = mgd = (500kg)(10\frac{m}{s^2})(5m) = 25,000 J$

We can use this to find the mass of a student that will create the same amount of torque while hanging from the rope:

$\displaystyle 25,000 J = mgd = m(10\frac{m}{s^2})(35m)$

$\displaystyle m = 71.4 kg$

The only two factors that affect a pendulum's frequency are the acceleration due to gravity (g) and the length of the pendulum's string (L). This can be seen in the following formula: $\displaystyle 2\pi f = \sqrt{\frac{g}{L}}$.