For a rigid body to be in a complete state of equilibrium it must first be in a state of translational equilibrium where the sum of all of the forces equals zero.

 

1. 
2. 

 

Then, we must also place it in a state of rotational equilibrium where the sum of all of the torques equals zero. For horizontal beams in the plane of the page, the torques will either result in a clockwise (cw) rotation or a counterclockwise (ccw) rotation.

 

3. 

 

An important item to note is that the location of the pivot point is absolutely arbitrary - that is, you can calculate your torques about any convenient point. To simplify your work, it is recommended that you choose a point where the torques of as many of the forces as possible will be zeroed out - that is, a common point through which their "lines of action" will pass. As you become more familiar with these problems, you will develop a sense of the best locations to use.

 

We will now illustrate the steps you should follow to solve problems dealing with rotational equilibrium by examining an assortment of beam and force configurations.

 

 

 

Our first type of problem involves horizontal beams and vertical forces.

 

This CP workbook page will provide you with further practice calculating torques produced on horizontal beams by vertical forces.

 

Now let's examine a second type of problem in which the beam remains horizontal but the forces have a mixture of diagonal, horizontal, and vertical orientations.

 

 

Since the beam is a rigid body, we must satisfy all three of the following conditions if we wish to place it into a state of complete equilibrium:

1. 
2. 
3. 

 

Before constructing any of these equations, we must first take components of the tension in the cable and remember that there is (1) a horizontal force pushing outward on the beam and (2) a vertical force supporting the beam where it contacts the wall.

 

 

If the pivot is placed at the hinge where the boom meets the wall, H, V, and T cos 30º have no moment arms since they pass through the pivot. Listed below are our three equations. We will not simplify them any further since we do not know a specific value for the beam's mass.

 

	H = T cos 30º
	V + T sin 30º = Mg (where Mg represents the weight of the beam)
	T sin30º(L) = Mg(½L)

 

A third type of problem involves diagonal beams as well as diagonal, horizontal, and vertical forces. In this type of problem you may be required to take components not only of the applied forces but also of the beam in order to determine the moments arms of each respective force.

 

 

The first step in this solution is to resolve any diagonal forces into their components. This is illustrated in the following diagram.

 

Step two is to determine the moment arm for each force. We will be using the hinge as our pivot point.

 

 

As you can see in the above diagram, when a force is oriented horizontally, its moment arm is oriented vertically.

• the horizontal component, T₁ cos 37º, has a vertical moment arm equal to L sin 53º 
• the vertical component, T₁ sin 37º, has a horizontal moment arm equal to L cos 53º

• the vertical weight, mg, has a horizontal moment arm equal to ½L cos 53º
• the vertical tension, T₂, has a horizontal moment arm equal to L cos 53º

• the horizontal and vertical components of the hinge, H and V, do not have any moment arms since their lines of action pass through our pivot point

The equations that satisfy our three conditions become:

 

	H = T1 cos 37º
	V = mg + T1sin 37º + T2
	mg(½ L cos 53º) + T1 sin 37º(L cos 53º) + T2(L cos 53º) = T1 cos 37º(L sin 53º)

 

Substituting in mg = 500 N and T₂ = 4000 N give us that T₁ = 9100 N. Knowing the value of T₁ will now allow you to calculate the value of the vertical component of the hinge, V,  in the equation for and the value of the horizontal component of the hinge, H,  in the equation for .