11.1.13.1: Illustrations

Illustration 1: Equilibrium on a Ramp

A uniform block of wood sits in equilibrium on a ramp as shown in the animation (position is given in meters). The slider allows you to control the height of the ramp and therefore the angle of the ramp. The red vector represents the weight of the block and the blue vector represents the normal force of the ramp on the block (Note: The normal force is shown with a fatter arrow so you can see it better). The force of static friction is an important feature of this problem, but its vector is not shown so we can focus on the weight and the normal force. Restart.

Consider the ramp when it is flat. Animation 1. All parts of the ramp are touching the bottom surface of the block. Therefore, the normal force does not act at just one point; actually, it is distributed across the bottom surface of the block. This is known as a distributed load. Despite this fact, we draw one vector, the normal force, to represent the resultant perpendicular component of the force of the ramp on the bottom surface of the block. But where should we draw the normal force of the ramp on the block? We draw the normal force at a location such that the torque due to this one force vector is equal to the net torque due to the distributed load of the ramp on the block. Therefore, when the ramp is flat, we draw the normal force of the ramp on the block as if it acts at the middle of the base of the block.

Before increasing the height of the ramp, predict what will happen to the location of the normal force of the ramp on the block. Will it stay in the same place or will it shift? If it shifts, will it shift toward the front edge or toward the back edge of the block? Now, increase the height of the block to \(0.35\text{ m}\). Notice the position where the normal force is drawn. Is this what you predicted?

There is a certain angle of the ramp that is too steep for the block to remain in equilibrium. If static friction is great enough to keep the block from sliding, then at this angle the block will tip; there will be an unbalanced torque to cause the block to "rotate." Where will the normal force on the block act when the ramp is at this angle? Increase the height of the ramp to its maximum value. Is this what you predicted?

At this angle the block would tip over. However, in Animation 2, you can increase the height of the ramp past the point where the block should tip. The animation will show you the normal force needed to keep the block from tipping. You can see how ridiculous this looks. Physicists would say that this animation is unphysical.

What do you notice about the point where the line of action of the normal force intersects the line of action of the weight? Try proving that the point where the line of action of the weight intersects the base of the block will be the same point where the normal force acts. Be sure to consider the conditions for static equilibrium.

Illustration authored by Aaron Titus.

Illustration 2: Center of Mass and Gravity

The first animation shows two blocks of equal mass (position is given in meters). The center of mass of the system is shown by a red dot and its position is calculated for you. Drag the right-hand green block to the right or to the left. What do you notice about the location of the center of mass as you change the block's position? Restart.

Now suppose the two blocks have unequal masses as shown in Animation 2. Is the center of mass at the center of the system? By viewing the location of the center of mass, you know which block is more massive. So which one is more massive?

How can we calculate the ratio of the mass of the blue block to the mass of the red block? If you only have two blocks, then the ratio of the distances of each block from the center of mass is related to the ratio of their masses. Therefore, by measuring the distance from each block to the center of mass, you can calculate the ratio of their masses.

In terms of the location of the two objects, the center of mass is located at:

\[X_{\text{cm}}=(x_{1}m_{1}+x_{2}m_{2})/(m_{1}+m_{2})\nonumber\]

for a one-dimensional, two-object system.

A concept similar to that of the center of mass is that of the center of gravity. In fact the two are often used interchangeably. The center of mass is defined above; the center of gravity is defined as the point on a system where gravity can be considered to act. The center of gravity takes into account the fact that the force of gravity—and therefore the acceleration due to gravity—is different for different heights above the surface of Earth. For this Illustration the center of mass is equivalent to the center of gravity. Only if the system is really large might the acceleration due to gravity be different at different parts of the system, thereby causing the center of gravity to differ from the center of mass.

Illustration authored by Aaron Titus and Mario Belloni.
Script authored by Aaron Titus.

Illustration 3: The Force and Torque for Equilibrium

A rigid rod of uniform mass is shown on top of a frictionless table as shown in Animation 1. The black circle is the location of the center of mass (position is given in meters and torque is given in newton meters). Restart.

Click Animation 2 to see the forces acting on the rod (the weight and normal force cancel each other out and are not shown; they are into and out of the page, respectively). Assume that the lengths of the force vectors are indicative of the magnitudes of the forces in newtons. If these are the only forces acting on the rod, is the rod in equilibrium?

If these are the only forces acting on the rod, then it is not in equilibrium because adding up the force vectors shows that the net force on the rod is not zero. Since the net force is not zero, the center of mass of the rod will have an acceleration, and therefore its velocity will change. In addition, adding up the torques on the rod shows that the net torque on the rod about the center of mass is not zero. The rod will have a changing angular velocity out of the page.

Suppose that we want the rod to be in equilibrium. What additional force must we apply to the rod?

Consider the conditions of static equilibrium. The net force on the rod must equal zero. If you add up all of the forces presently acting on the rod (as shown in Animation 2), the sum is not zero. Therefore, we must apply another force to the rod that is the negative of the sum of the other forces on the rod.

At what location should this additional force be applied?

To be in static equilibrium, the net torque on the rod must equal zero. Therefore, the torque due to this additional force on the rod must equal the negative of the sum of the torques of the other forces presently acting on the rod. Knowing the torque and the force needed for the rod to be in equilibrium, you can calculate the location where the force should be applied to the rod.

Now, in Animation 3 you get to add a force to get the rod into equilibrium. Adjust the magnitude and direction of the blue force vector and place it at the correct location on the beam. Then check your answer. You will see a red vector for the net force and a green calculation of the net torque (this is in the \(z\) direction, which is positive out of the page). If the net force is zero, then its vector will be zero and will not be seen. If the rod is in equilibrium, the net force vector will vanish and the torque calculation will be zero, and you calculated your answers correctly. If not, recheck your calculations, adjust the magnitude, direction, and location of the blue force vector and check your answer again.

Illustration authored by Aaron Titus.
Script authored by Aaron Titus and Mario Belloni.

Illustration 4: The Diving Board Problem

A \(2\text{-kg}\) box sits \(0.3\text{ m}\) from the right end of a board of negligible weight in Animation 1 and a board of weight \(10\text{ kg}\) in Animation 2. Two supports (support 1 and support 2) exert forces on the left and right ends of the board, as indicated by the two force vectors (position is given in meters). The arrows represent the relative sizes of the force vectors, but their length does not represent their actual magnitudes (the actual value of the forces, as well as the separation between the supports, is shown in the table). The board is \(6\text{ m}\) long and support 1 is \(0.3\text{ m}\) in from the left edge. Restart.

Consider the situation in Animation 1 in which the board has a negligible weight. How does the force of each support on the board (\(F_{1}\) corresponds to the right support and \(F_{2}\) corresponds to the left support) depend on where the box is located? You can drag the second support from left to right to view the forces of the supports on the board. When the board is of a negligible mass, there are three forces that act on the board: the weight of the box, and the forces of the supports. As you move support 2 around, what do you notice? When the movable support is as far right as it can go, the force of the first support is zero and the force of the second support cancels the force of the box. This seems logical, but why this force? An \(F_{1} = 10\text{ N}\) and an \(F_{2} = 9.6\text{ N}\), would work too, right? Well, yes and no. It would certainly make the sum of the forces on the board equal to zero, but what about the sum of the torques? This sum would not be equal to zero no matter where you measured the torques from. Only an equal and opposite force acting at the same position as the weight of the box will keep the board in equilibrium. As you move the second support to the left note that both forces exerted by the supports get bigger, and that the force that the first support exerts is negative. We can understand this effect by measuring the torque on the board from the first support. Since the force is in the \(y\) direction and the radius arm is in the \(x\) direction, the magnitude of the torque is \(rF\). The radius arm for the weight of the box is always \(5.4\text{ m}\). The radius arm for the second support changes as you drag it to the left. Therefore, as you drag it to the left the force the support exerts must increase in order to keep \(rF\) the same magnitude as (and in the opposite direction of) that of the box. Once the right force is determined from the torque, the force due to the first support can be determined by requiring the sum of the forces on the board to be zero.

Now, consider the situation in Animation 2 in which the board has a weight of \(10\text{ kg}\). How does the force of each support on the board depend on where the box is located? You can drag the second support from left to right to view the forces of the supports on the board. When the board has a mass, there are four forces that act on the board: the weight of the box, the weight of the board, and the forces of the supports. Much of the same analysis described above still is valid here; you just need to factor in one more force and therefore one more torque. Again drag the support around and pay close attention to the forces. What happens to the forces when the separation between the supports is \(3.15\text{ m}\)? How about \(2.7\text{ m}\)?

Illustratron authored by Aaron Titus.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.