Determining Torque and Static Equilibrium

1. Torque is what causes objects to change their rate of rotation. Like Ω, α, the moment of inertia, and angular momentum, torque is defined relative to a point. Choose a different point and you get a different value for your torque. For convenience, let's call the point the pivot point or the axis of rotation.

2. The magnitude of the torque depends both on the size of the force involved and on the distance from the pivot point.

3. The portion of the force that is directed through the pivot point, the radial component of the force F_rad, does not cause a change in the rate of rotation. Only the tangential component of the force, F_tan, causes a change in the rate of rotation. The two components are related to the total F by and tanφ = F_rad/F_tan. Note that F_rad is in the same direction as l, and that F_tan is perpendicular to l
   

4. For the above diagram, the torque is given by τ = lF_tan = lFsinφ.

5. The direction of the torque is up out of the plane formed by F and l if the torque makes the object rotate in a counterclockwise fashion. This is usually referred to as a positive or counterclockwise (CCW) torque. The direction of the torque is down out of the plane if the torque makes the object rotate in a clockwise fashion. This is usually referred to as a negative or clockwise (CW) torque.

6. An alternate method to determine the torque is to make use of the lever arm l_tan. To find the lever arm, draw a line along the direction of the force. Draw a line from the axis of rotation to that line so that the two lines are perpendicular to one another.
   

7. In the diagram above, the torque is given by τ = Fl_tan = Flsinφ.

8. In the typical static equilibrium problem, one must use geometry to find l and φ. This is illustrated in the following examples.

9. Quite often it is easier to break F and l into their i and j components. From τ = r × F, we know τ_z = xF_y - yF_x. This result can be seen from the diagram below where we see that x is the lever arm of F_y and y is the lever arm for F_x. The minus sign occurs because of the sign convention for CW and CCW torques.
   
   

Example 1

The diagram below shows a uniform rectangular sign attached to two walls by a hinge and two wires. The sign has a mass m = 75.0 kg. The tension in the upper cable is twice that of the side cable. Find the tension in the two cables. Find the vertical and horizontal components of the hinge force.

The sign is uniform so the centre of mass is at the geometric centre of the sign. The hinge force is unknown. Unlike the tension in the wires, we do not know it direction. We therefore break the hinge force into its i and j components, h_x and h_y. Since there are two unknowns at the hinge, it is a good idea to place our pivot point there. Next we draw the extended free-body diagram.

For static equilibrium, we apply Newton's Second Law to the above free body diagram

which gives us two force equations. As well we are told T₁ = 2T₂.

We will determine the torques from h_x, h_y, T₁, T₂, and W. First we choose the pivot point (labeled A in the following diagrams) from which to calculate all our torques. The hinge is a good place to choose the pivot point since we have two unknowns there. Hence the torques due to h_x, and h_y are zero since the distance l is zero. Let's look at our diagram with all the distances noted.

Putting our results into a table:

Force	distance	angle	Torque
hx	0	-	0
hy	0	-	0
T1
W
T2

Our last equation for static equilibrium is therefore

Στ = 0       ->      

We now have four equations with four unknowns T₁, T₂, h_x, and h_y. In theory, the problem is finished but there is some messy algebra left to complete.

An alternate way to solve this problem is to calculate the torque using the i and j components of each force and the x and y coordinates of the location each force with the pivot point as the origin (0,0).

Putting the results into a table, we have:

The torque equation for static equilibrium call also be written as

Note that this result is identical to previous result. You may wish to confirm that r₁sin(θ+α) = 3sinθ - 4cosθ and that r₂sin(γ-φ) = 6sinφ - 2cosφ.

Example 2

The diagram below shows a 'P'-shaped sign attached to a wall by a hinge and a wire. The sign is constructed of uniform material and has a mass m = 100.0 kg. The centre of the 'P' is a square 1 m × 1 m hole. Find the tension in the wire and the components of the hinge force.

The sign in this problem does not have a simple shape. Just by looking at the sign we cannot determine where the Centre of Mass is. We are told, however, that the sign is uniform which means that the any two pieces of the sign with the same area have the same mass. A quick calculation shows that the area of the sign is 10 m². Since the sign has a total mass of 100 kg, each 1 m² has a mass of 10 kg. Let's break the sign into simpler pieces as in the diagram below.

The masses of the four pieces 1, 2, 3, and 4, are 50 kg, 20 kg, 10 kg, and 20 kg respectively. And for these simple pieces we know where each Centre of Mass is. So next we construct an extended free body diagram.

We apply Newton's Second Law,

The second method for determining torque is the easiest here since the locations of the weights are easy to find and since these forces are purely vertical. Taking the hinge as the pivot point, we have:

Our last equation for static equilibrium is therefore

We have three unknowns, H_x, H_y, and T, and three equations so the problem is solved except for the algebra.

To see how much faster this method is, you may wish to do the above problem by the first method.

Questions? mike.coombes@kpu.ca