We begin with a description of force. A force (often denoted F) is a push or pull exerted on an object. The magnitude of the force is a measure of how hard the push or pull is. In addition to magnitude, force has a direction. We will use two different units for force, pounds (lb) and Newtons (N). (Would you like a lesson or review of units before continuing?) Forces can be put into two categories.
Contact forces are those created through direct contact, such as the force you exert on a door knob when you push or pull it; or the force the tires of a car exert on the road; or the force the chair exerts on you as you sit down.
There are a few non-contact or action-at-a-distance forces as well. These are forces that are exerted without actual contact. You are subjected to one right now! The Earth's gravitational force, commonly referred to as your weight, is such a force. (This does not apply to students who have logged in from deep space.) You do not have to be touching the Earth for the force of gravity to be there. If you jump off a wall, the force of gravity (denoted w for weight) accelerates you until you hit the ground. And, of course, it's still there after you hit the ground. (The electrical force is another important example of a non-contact force. This comes later in the semester.)
Static Equilibrium - No Net Force:
Static equilibrium refers to an object that is not moving. When an object is not moving from one place to another, we also say it is not translating. As you lay there crumpled, motionless on the ground (your mother told you not to jump off walls!) you are in static equilibium. And what creates this state of affairs? Gravity (w) is still acting, but now the ground (N) is also exerting a force on you. In fact, the force of the ground is equal to and opposite your weight. This is an important requirement for static equilibrium.
The sum of all the forces acting on an object in static equilibrium must add to zero.
where the Greek symbol S means sum. There can be many forces acting on the object. Those forces have directions. In our example above, the directions are simple; your weight points downward (towards the center of the earth) and the force of the ground points upward. The arrows shown in the diagram represent the force vectors. A vector has magnitude (size of the arrow) and direction given by the direction of the arrow. In this case, the two vectors have equal magnitude and point in opposite directions. It is easy to see that they add to zero.
Newton's 3rd Law
Newton's 3rd Law represents a useful concept in understanding how forces work. It is very simple
The opposite forces are called action-reaction pairs. In the example above, the ground pushes upward on the unfortunate wall jumper with a force equal to his weight. The jumper is pushing downward on the ground with this exact same force. Either force could be called the action and the other the reaction. Action-reaction forces always come in pairs, but they never act on the same object. So they never "cancel" each other. (The upward force from the ground cancels the other force acting on the jumper, namely his weight.)
Forces, Directions and Components
Forces can still add to zero even if none of them are in exactly opposite directions. Consider the forces from the ropes that are pulling on the box shown. These forces can add to zero if the force vectors add to zero. How does that work? Actually, it's easy. Simply add the vectors head to tail (don't change the magnitudes or directions!) and see if they add up to zero. (Remember that the size of the force vectors represent the size of the forces, not the length of the ropes.)
There is another useful concept for forces ... components. Each force can be thought of as being composed of a horizontal component and a vertical component. (Actually, any two mutually perpendicular directions will work.) The components of any force are the perpendicular projections onto the horizontal and vertical lines. Look at the two diagrams and convince yourself that components are merely an extension of the previous rule for adding vectors. Any vector is the sum of its components.
Components are particularly useful for those forces for which you do not know the direction. Simply represent the force as two forces, one vertical and one horizontal. And there is another benefit to this method. The sum of all the horizontal components must be zero, and the sum of all the vertical components must be zero, independently. In other words,
Consider the example below.
Problem: A 10 kg box is stationary on a horizontal table. A workman is leaning on the box with a horizontal force of 50 N. What force does the table exert on the box? Identify all the action-reaction pairs.
Solution: First, we find the force of gravity, or the weight of the box. It is w = mg = 10 kg x 9.8 m/s2 = 98 N. It points downward. (Need a review of weight?) It should be clear from the diagram that the force exerted by the table cannot be just horizontal or just vertical, since it must "cancel" both the weight and the workman's force. We could solve this problem graphically, and that would be a good exercise. But components make it easy to solve analytically. The force exerted by the table is broken into horizontal and vertical components. The vertical force is called the normal force, because it is normal or perpendicular to the table surface. It is often denoted N. The horizontal force also has a name. You undoubtedly know it already. Why doesn't the box slide to the right when the workman leans on it? It would, if it weren't for the frictional force, which we denote as f. Friction is the force between any two surfaces that tends to keep the surfaces from sliding across one another. Now look at the diagram. Since the horizontal and vertical forces must sum to zero independently, it's easy to see that f = 50 N and N = 98 N. Since f and N are perpendicular, we can use the Pythagorean theorem to find the magnitude of the total force the table exerts, Ftable = (502 + 982) = 110 N.
And the action-reaction pairs? The workman pushes on the box to the right with a 50 N force. The box pushes on the workman with a 50 N force to the left. The table pushes on the block upwards with a force of 98 N and to the left with a force of 50 N. So the block pushes on the table with these same forces but downward and to the right respectively. What is the reaction force to the force of gravity (weight)? Think about it! This is a good one to discuss in class.
Static Equilibrium - No Net Moments:
There is one other important condition for static equilibrium. Not only should the object not be translating, but it should not be rotating either. If the forces add to zero, the object will not be translating. However, it is possible for the forces to add to zero, yet the object is not in equilibium. Not if it is rotating! Look at the equal but opposite forces acting on the box. In the first case we have static equilibrium. But common sense tells us what is going to happen in the second case. The two forces are not aligned and the box will twist or rotate. In the second case, the sum of the moments (also called torques) is not zero. In symbols, we denote a moment as M.
The sum of all the moments acting on an object in static equilibrium must add to zero.
And the defining moment is ... The moment of a force is a measure of how the force tends to rotate an object. There is a common sense equation for the moment created by a force. Hold a meter stick horizontal with your hand at the center. You can feel the weight of the stick, but notice there's no tendency for the stick to rotate. You can hold it with your hand opened. The force of gravity acts at the center of mass of the stick, which is at the geometric center of the stick where you are holding it. This force is acting at zero distance from your hand and it creates zero moment about your hand.
Now slide the stick, keeping it horizontal, so that you are no longer holding it at the center. In addition to the weight, you can feel the stick trying to twist out of your hand. You cannot hold it with your hand opened. In this case, the weight of the stick creates a moment about your hand. As you slide the stick from the center out to the end, you can feel this moment increasing. It increases directly with the distance of the force from your hand. We define the moment created by a force as:
The distance d is measured from the "pivot" point, your hand in this case, to where the force acts. The most common units for moments are N.m or ft.lb. However, in the fields of structural and mechanical engineering, it is common to find other combinations such as N.cm or lb.in.
Problem: Assume you are holding a stick with weight .3 N at the 20 cm mark. What moment does the weight of the stick create about your hand?
Solution: The weight of the stick acts at the center, the 50 cm mark. The moment the weight creates about your hand would be M = d F = (50 cm - 20 cm) x .3 N = 90 Ncm or (.50 m - .20 m) x .3 N = .09 Nm. If you are holding the stick stationary, the sum of the moments acting on the stick should be zero! Your hand is supplying an equal but opposite moment! Just like forces, moments also have direction.
It is easiest to think of these directions in terms of which way the force tends to rotate the object. A common convention is
In the picture, notice that the weight tends rotate the stick clockwise, a positive moment. Your hand must be supplying an equal moment in the opposite direction (ccw) to keep the stick stationary.
Holding the stick horizontal hides a complication that we must consider. We have not accounted for the line of action along which the force acts. Notice that if you rotate the stick to the vertical, the twisting or moment disappears! That's because the line action of the weight now goes through your hand, the pivot. So we modify the meaning of d just a bit. It should be d = the perpendicular distance to the line of action of the force. Look at how the value of d changes as the angle of the stick changes in the pictures below.
Where is the pivot point in each case? For the example above, the hand seemed the natural place to choose. But one of the convenient aspects of having a zero net moment is that the total moment is zero about any and every point. Draw all the forces acting on the object in a diagram, pick a "convenient" pivot point and find d for each force. The choice of a "convenient" pivot point can make the calculations much easier. Another example will help you see how to choose the pivot point.
The common playground teeter totter is usually a simple plank balanced on a fulcrum (shown as a triangle below). Two identical twins (weighing 80 lb each) sit at equal distances (6 ft) from the fulcrum. We know that the plank just hangs there. (Provided the twins don't start pushing off the ground.) The sum of moments must be zero. But if one of the twins moves inward, we know what will happen. The twin farthest from the fulcrum will move downward and the other will move up. The sum of moments is not zero and the plank rotates ccw. Let's see what our equation for moments tells us.
The center of the plank seems an obvious choice for our pivot point. In the first case, the moment created by the weight of the right-side child about the fulcrum is +(80 lb)(6ft), while that of the left-side child is -(80 lb)(6ft). Note that the weight of child on the right rotates the totter cw, and that of the child on the left rotates the totter ccw. Hence the signs are opposite and sum of moments is zero. But if the child on the right moves inward, the distance d will be less and we will have a net negative moment. The totter rotates in the ccw direction.
There are two other forces acting on the plank that we have ignored. The weight of the plank and the force the fulcrum exerts on the plank. Both of these forces act at the center of the plank, where we chose the pivot. Hence, the distance from the pivot to the each of these forces is zero. They create no moment about the pivot. Well, isn't that convenient! That is why the fulcrum was a good choice for the pivot point.
Problem: The twin on the left weighs 80 lb and the twin on the right has been lifting weights and has bulked up to 100 lb. We want the system balanced! Our playground sense tells us that the 100 lb child must move closer to the pivot. But how far?
Solution: Let's use our equation for moments. We take our pivot point at the fulcrum again. The 80 lb force on the totter from the first child acts a distance of 6 ft from the pivot. The 100 lb force acts at a distance x, the value we're trying to find. The sum of the moments about the pivot (at the center of the totter) should be zero, so
|S M = +x (100lb) - 6ft (80 lb) = 0|
Solving for x we get x = 6(80)/100 = 4.8 ft.
Problem: Here's a trickier problem. What if 100 lb child stays at the end and we move the plank instead instead? How far do we have to slide the plank to establish equilibrium? Ignore the weight of the plank.
Solution: Draw the diagram and try it. Let x be the distance from the fulcrum to the center of the plank. Choose the fulcrum for the pivot point. The childrens' weights will act at distances 6+x and 6-x from the pivot. The answer is x = .67 ft (more precisely, 2/3 ft)
Problem: Second, try a more realistic version where the plank has a weight of 30 lb. Although the force from the fulcrum still does not appear in our moments equation, the weight of the plank does!
Solution:See if you can get an answer of x = .57 ft
Problem: Take another look at the problem with the workman leaning on the box. Does it look like the box would rotate? Although the sums of the forces are zero, what must be different in the diagram to assure the sum of moments is also zero?
Solution: This is good problem for a class discussion. But here's an experiment that might help you figure it out. Find a floor lamp (or a tall box) and place it on a hard floor. Push horizontally near the base until the lamp slides. Now move your hand up a bit and try again. Keep doing this until the lamp will no longer slide, but tips instead. Think about about the moments created by your hand, the lamp's weight and the forces from the floor.
When you understand how both forces and moments must sum to zero for static equilibrium, then you can construct a house that does not translate and does not rotate. (And no jokes about mobile homes.)
Here are a few PROBLEMS to try that you might encounter when building a house.
When your are ready, the next section is Shear and Stress.