The minimum length of the wrench will assume that the maximum force is applied at an angle of . Therefore, we can use the simplified expression for torque:

Here,  is the length of the wrench.

Rearranging for length and plugging in our values, we get:

The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. Taking the fulcrum as the pivot point, the counterclockwise torque is due to the rod’s weight, gravitational force acting downwards at the center of the rod. If we use the pivot as our reference, then the center of the rod is 15cm from the reference.

Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.

.

Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.

The net torque on the pulley is zero. Remember that , 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 . Since both students will exert a downward force perpendicular to the length of the seesaw, . In our case, force is the force of gravity, given below, and is the distance from the center of the seesaw.

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.

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.

This a an example of rotational equilibrium involving torque. The formula for torque is , where  is the angle that the force vector makes with the object in equilibrium and  is the distance from the fulcrum to the point of the force vector. To achieve equilibrium, our torques must be equal.

Since the forces are applied perpendicular to the beam, becomes 1. The distance of the fulcrum from the left end is 1m and its distance from the right end is 2m.

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.

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

Substitute the formula for torque into this equation.

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

The acceleration form gravity cancels from each term.

This problem deals with torque and equilibrium. Noting that the string is between the two masses we can use the torque equation of . We can use the equation to find the torque. Since force is perpendicular to the distance we can use the equation  (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.

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

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:

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

Torque is defined as . In this case, is zero because Bob and the weight are sitting directly on top of the seesaw; all of their weight is projected directly downward. Therefore, the torque that the weight applies is:

In order for the seesaw to balance, the torque applied by Bob must be equal to .