Illustration 13.1: Equilibrium on a Ramp
Please wait for the animation to completely load.
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 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.