The equation for torque is . Looking at this equation we can infer that the maximum distance from the center would give maximum torque. Also, any angle besides  or  will give an absolute value of a number below one, giving us a smaller torque.

To calculate the torque on the pendulum, we need to know the position of the pendulum. We can find this using the following expression:

Note that we are using the cosine function because the pendulum begins at it's maximum angle. Plugging in our values:

The pendulum is still horizontal, but now on the other side. Now we can directly calculate the torque placed on the pendulum

Where the radius is the length of the pendulum and the force is the weight of the block (since the pendulum is horizontal).

For this question, we're given a number of scenarios in which an equal magnitude of force is applied to a rotating bar at a variety of different orientations and locations on the bar. We're then asked to identify which one would generate the most amount of torque.

First, let's recall that torque is a twisting force. That is, it is a force that causes an object to rotate about a pivot point, such as a sea-saw. We can write an expression for torque as follows.

Where  is the magnitude of the applied force,  is the distance of the applied force from the pivot point, and  is the angle between the applied force vector and the surface upon which the force is being applied.

Sometimes, the equation for torque is also expressed as follows.

Where  stands for the lever arm, which takes into account both  and . Thus, for any given magnitude of force, the torque will be the highest when  is greater and when  approaches .

With this expression in mind, we can look at each image and make a qualitative assessment of which one will have the greatest torque. We're looking for a diagram in which the force vector is furthest from the pivot point, and is also oriented as close to  with respect to the surface of the rotating object. This situation is described by the following picture, thus making it the correct answer.

Converting  to  and plugging in values

First, we must recall the formula for torque, which is 

 is the distance from the pivot, called the moment arm.  is the force, and  is the angle relative to the normal of the object.) Since all the forces are being applied perpendicular to the surface of the rod, . Thus,

The sum of the torques must equal zero, so David's torque plus Marc's torque must be the same as Paul's torque because David and Marc are applying torques in the opposite direction as Paul. This gives us 

 Dividing both sides by , we get Paul's force  to be 

First let's go on and define each of these terms...

Force: influence exerted 

Torque: a way for us to measure of the effectiveness of a force which consists of a force and its perpendicular distance from the line of action amongst the axis of rotation

Each reference in which you find these definition may vary, but they should all have a common gist. Force is some type of implication on an object that can cause that object's mobility. Torque, however, is the capability of a specific force to create rotation. Therefore, torque is not a force, it's a characteristic of a force.

The forces shown in the diagram each have a reciprocal. In other words,  is countered by, as is countered by . They each are equal, pulling in an opposite direction. With that said, the net force is zero. How? Well if we had a magazine (rectangle) and you and three of your friends each pulled on a corner as shown in the picture, the magazine wouldn't move. If you and a friend pulled on the same corners as and ,the magazine still wouldn't move (same goes for and ).

However, torques are slightly different. Torque is not (+) or (-), it is measured as clockwise or counter clockwise. So, looking at our diagram, all of the forces are in the same direction around the central pivot point (black circle). If three out of four of the forces shown were taken away, any of the remaining would cause the rectangle to rotate in the same manner (counter clockwise) around the pivot point. Therefore the net torque, or the total of the torques are all in the same direction and will NOT have a net value of zero, but rather a grand total of each force's torque separately.

It is important to remember that torque is not a force, it's a characteristic of a force.

F is the force applied, r is the distance from the force's contact point and the object's center of mass,  is the angle between  and .

The trick to this problem is in two steps:

1) Find the  using the Pythagorean theorem. 

2) What's our ? Well we know that every square has 4  angles. If we drew a line from the corner of the square to the very middle (pivot point) of the pinwheel, we've just cut one of those angles in half!  

Solving:

Knowing our trig functions we can plug in for 

This question deals with torques: 

It's important to note that torques are not (+) or (-), they are a direction, usually denoted as clockwise (CW) or counter clockwise (CCW). If you drew a FBD for each Deanna and Rishi, and included the pivot point, we'd see that each is in an opposite direction. That means that these torques oppose each other. If we want the seesaw to remain still, we must have the same torque on either side, this will result in a ZERO net torque. With that said, let's set this up!

The  on either side can cancel eachother out. Now let's solve for the distance Rishi is from the fulcrum ()

Remember that the equation for torque is:

where  is the force applied,  is the distance from the force's contact point to the object's (wrench's) COM and  is the angle between  and 

We are provided with a whole bunch of numbers in the equation, however the explanation is much simpler than that. Looking at the direction of the force in reference to the COM,  is zero.

Knowing our trig functions, we know that  (regardless if we use radians or degrees).

The answer is 

Knowing the centripetal force on the mass and the radius of the circle, we can calculate its velocity:

Rearranging for velocity:

We can use this to find the period of the mass:

Rearranging for period, we get: